Progressive waves — Paper 2 · A Level Physics 9702

5 Fig. 5.1 shows the variation with time t of the displacements xA and xB at a point P of two For sound waves A and B. Examiner’s Use wave A 3 xA / 10 – 4 cm 2 1 0 0 1 2 3 4 5 6 t / ms –1 –2 –3 wave B 2 xB / 10 – 4 cm 1 0 0 1 2 3 4 5 6 t / ms –1 –2 Fig. 5.1 (a) By reference to Fig. 5.1, state one similarity and one difference between these two waves. similarity: … difference: … [2] (b) State, with a reason, whether the two waves are coherent. … … [1] (c) The intensity of wave A alone at point P is I. For Examiner’s (i) Show that the intensity of wave B alone at point P is 4 I. Use 9 [2] (ii) Calculate the resultant intensity, in terms of I, of the two waves at point P. resultant intensity = ……………………………… I [2] (d) Determine the resultant displacement for the two waves at point P (i) at time t = 3.0 ms, resultant displacement = ……………………………… cm [1] (ii) at time t = 4.0 ms. resultant displacement = ……………………………… cm [2]

10 marks

Mark scheme: 5 (a) similarity: e.g. same wavelength/frequency/period, constant phase difference B1 difference: e.g. different amplitude/phase B1 [2] (do not allow a reference to phase for both similarity and difference) (b) constant phase difference so coherent B1 [1] (c) (i) intensity ∝ amplitude2 C1 I ∝ 32 and IB ∝ 22 leading to M1 4 IB = I A0 [2] 9 (ii) resultant amplitude = 1.0 × 10-4 cm C1 1 resultant intensity = I A1 [2] 9 (d) (i) displacement = 0 B1 [1] (ii) xA = -2.6 × 10-4 cm and xB = +1.7 × 10-4 cm C1 allow ± 0.5 × 10-4 cm) resultant displacement = (-) 0.9 × 10-4 cm A1 [2]

This question in 9702/21 Oct/Nov 2005

6 A long tube, fitted with a tap, is filled with water. A tuning fork is sounded above the top of the tube as the water is allowed to run out of the tube, as shown in Fig. 6.1. tuning fork 512 Hz 32.4 cm Fig. 6.1 Fig. 6.2 A loud sound is first heard when the water level is as shown in Fig. 6.1, and then again when the water level is as shown in Fig. 6.2. Fig. 6.1 illustrates the stationary wave produced in the tube. (a) On Fig. 6.2, (i) sketch the form of the stationary wave set up in the tube, [1] (ii) mark, with the letter N, the positions of any nodes of the stationary wave. [1] Use (b) The frequency of the fork is 512 Hz and the difference in the height of the water level for the two positions where a loud sound is heard is 32.4 cm. Calculate the speed of sound in the tube. speed = …………………… m s–1 [3] (c) The length of the column of air in the tube in Fig. 6.1 is 15.7 cm. Suggest where the antinode of the stationary wave produced in the tube in Fig. 6.1 is likely to be found. … … … [2]

7 marks

Mark scheme: 6 (a) (i) correct shape drawn B1 [1] (ii) two nodes marked correctly B1 [1] (b) ½λ = 0.324 m C1 v = fλ C1 = 512 × 2 × 0.324 = 332 m s–1 A1 [3] (c) ¼λ = 16.2 cm C1 either antinode is 0.5 cm above top of tube or antinode is 16.2 cm above water surface A1 [2]

This question in 9702/21 May/June 2006

5 (a) Fig. 5.1 shows the variation with time t of the displacement y of a wave W as it passes a point P. The wave has intensity I. y 0 t wave W Fig. 5.1 A second wave X of the same frequency as wave W also passes point P. This wave has intensity I. The phase difference between the two waves is 60°. On Fig. 5.1, sketch the variation with time t of the displacement y of wave X. [3] (b) In a double-slit interference experiment using light of wavelength 540 nm, the separation of the slits is 0.700 mm. The fringes are viewed on a screen at a distance of 2.75 m from the double slit, as illustrated in Fig. 5.2 (not to scale). coherent light 0.700 mm wavelength 540 nm screen 2.75 m Fig. 5.2 Examiner’s Use Calculate the separation of the fringes observed on the screen. separation = … mm [3] (c) State the effect, if any, on the appearance of the fringes observed on the screen when the following changes are made, separately, to the double-slit arrangement in (b). (i) The width of each slit is increased but the separation remains constant. … … … … [3] (ii) The separation of the slits is increased. … … … [2]

11 marks

Mark scheme: 5 (a) amplitude between 6.5 squares and 7.5 squares on 3 peaks B2 (allow 1 mark if outside this range but between 6.0 and 8.0 squares) correct phase (ignore lead/lag, look at x-axis only and allow ±½ square B1 [3] (b) λ = ax / D C1 540 × 10–9 = (0.700 × 10–3 x) / 2.75 C1 x = 2.12 mm A1 [3] (c) (i) same separation B1 bright areas brighter (1) dark areas, no change (1) (allow ‘contrast greater’ for 1 mark if dark/light areas not discussed) fewer fringes observed (1) any two, 1 each B2 [3] (ii) smaller separation of fringes B1 no change in brightness B1 [2]

This question in 9702/21 Oct/Nov 2007

1 Make reasonable estimates of the following quantities. (a) the frequency of an audible sound wave frequency = … Hz [1] (b) the wavelength, in nm, of ultraviolet radiation wavelength = … nm [1] (c) the mass of a plastic 30 cm ruler mass = … g [1] (d) the density of air at atmospheric pressure density = … kg m–3 [1]

4 marks

Mark scheme: 1 (a) allow anything in range 20 Hz → 20 kHz B1 [1] (b) allow anything in range 10 nm → 400 nm B1 [1] (c) allow anything in range 10 g → 100 g B1 [1] (d) allow anything in range 0.1 kg m–3 → 10 kg m–3 B1 [1]

This question in 9702/21 May/June 2008

5 (a) State what is meant by For Examiner’s (i) the frequency of a progressive wave, Use … … … [2] (ii) the speed of a progressive wave. … … [1] (b) One end of a long string is attached to an oscillator. The string passes over a frictionless pulley and is kept taut by means of a weight, as shown in Fig. 5.1. string pulley oscillator weight Fig. 5.1 The frequency of oscillation is varied and, at one value of frequency, the wave formed on the string is as shown in Fig. 5.1. (i) Explain why the wave is said to be a stationary wave. … … [1] (ii) State what is meant by an antinode. … … [1] (iii) On Fig. 5.1, label the antinodes with the letter A. [1] (c) A weight of 4.00 N is hung from the string in (b) and the frequency of oscillation is For adjusted until a stationary wave is formed on the string. The separation of the antinodes Examiner’s on the string is 17.8 cm for a frequency of 125 Hz. Use The speed v of waves on a string is given by the expression T v = , m where T is the tension in the string and m is its mass per unit length. Determine the mass per unit length of the string. mass per unit length = … kg m–1 [5]

11 marks

Mark scheme: 5 (a) (i) frequency: number of oscillations per unit time M1 of the source / of a point on the wave A1 [2] (ii) speed: speed at which energy is transferred / speed of wavefront B1 [1] (b) (i) does not transfer energy (along the wave) B1 [1] (ii) position (along wave) where amplitude of vibration is a maximum B1 [1] (iii) all three positions marked B1 [1] (c) wavelength = 2 × 17.8 = 35.6 cm C1 v = fλ C1 v = 125 × 0.356 = 44.5 m s–1 C1 44.52 = 4.00 / m C1 m = 2.0 × 10–3 kg m–1 A1 [5] GCE A/AS LEVEL – May/June 2008 9702 02 2

This question in 9702/21 May/June 2008

5 (a) State what is meant by a progressive wave. For Examiner’s … Use … … [2] (b) The variation with distance x along a progressive wave of a quantity y, at a particular time, is shown in Fig. 5.1. y 0 0 x Fig. 5.1 (i) State what the quantity y could represent. … … [1] (ii) Distinguish between the quantity y for 1. a transverse wave, … … [1] 2. a longitudinal wave. … … [1] (c) The wave nature of light may be demonstrated using the phenomena of diffraction and For interference. Examiner’s Use Outline how diffraction and how interference may be demonstrated using light. In each case, draw a fully labelled diagram of the apparatus that is used and describe what is observed. diffraction … … … interference … … … [6]

11 marks

Mark scheme: 5 (a) transfer / propagation of energy … M1 as a result of oscillations / vibrations … A1 [2] (b) (i) displacement / velocity / acceleration (of particles in the wave) … B1 [1] (ii) displacement etc. is normal to direction of energy transfer / travel of wave / propagation of wave ……(not ‘wave motion’) … B1 [1] (iii) displacement etc. along / same direction of energy transfer / travel of wave / propagation of wave ……(not ‘wave motion’) … B1 [1] (c) diffraction: suitable object, means of observation … M1 either laser or lamp and aperture or distant source … M1 light region where darkness expected … A1 interference: suitable object, means of observation and illumination … B1 light and dark fringes observed … B1 appropriate reference to a dimension for diffraction or for interference … B1 [6] [Total: 11]

This question in 9702/21 Oct/Nov 2009

4 (a) State two features of a stationary wave that distinguish it from a progressive wave. For Examiner’s 1. … Use … 2. … … [2] (b) A long tube is open at one end. It is closed at the other end by means of a piston that can be moved along the tube, as shown in Fig. 4.1. tube piston loudspeaker L Fig. 4.1 A loudspeaker producing sound of frequency 550 Hz is held near the open end of the tube. The piston is moved along the tube and a loud sound is heard when the distance L between the piston and the open end of the tube is 45 cm. The speed of sound in the tube is 330 m s–1. (i) Show that the wavelength of the sound in the tube is 60 cm. [1] (ii) On Fig. 4.1, mark all the positions along the tube of 1. the displacement nodes (label these with the letter N), 2. the displacement antinodes (label these with the letter A). [3] (c) The frequency of the sound produced by the loudspeaker in (b) is gradually reduced. For Examiner’s Determine the lowest frequency at which a loud sound will be produced in the tube of Use length L = 45 cm. frequency = … Hz [3]

9 marks

Mark scheme: 4 (a) e.g. no energy transfer amplitude varies along its length/nodes and antinodes neighbouring points (in inter-nodal loop) vibrate in phase, etc. (any two, 1 mark each to max 2 ………………………………………………………..B2 [2] GCE AS/A LEVEL – May/June 2010 9702 22 (b) (i) λ = (330 × 102)/550 ………………………………………………………….. M1 λ = 60 cm ……………………………………………………………………… A0 [1] (ii) node labelled at piston ………………………………………………………. B1 antinode labelled at open end of tube ……………………………………… B1 additional node and antinode in correct positions along tube …………… B1 [3] (c) at lowest frequency, length = λ/4 ………………………………………………... C1 λ = 1.8 m frequency = 330/1.8 ………………………………………………………………. C1 = 180 Hz ………………………………………………………………………… … A1 [3]

This question in 9702/22 May/June 2010

5 (a) A source of sound has frequency f. Sound of wavelength λ is produced by the source. For Examiner’s (i) State Use 1. what is meant by the frequency of the source, … … [1] 2. the distance moved, in terms of λ, by a wavefront during n oscillations of the source. distance = … [1] (ii) Use your answers in (i) to deduce an expression for the speed v of the wave in terms of f and λ. [2] (b) The waveform of a sound wave produced on the screen of a cathode-ray oscilloscope (c.r.o.) is shown in Fig. 5.1. 1 cm 1 cm Fig. 5.1 The time-base setting of the c.r.o. is 2.0 ms cm–1. For Examiner’s (i) Determine the frequency of the sound wave. Use frequency = … Hz [2] (ii) A second sound wave has the same frequency as that calculated in (i). The amplitude of the two waves is the same but the phase difference between them is 90°. On Fig. 5.1, draw the waveform of this second wave. [1]

7 marks

Mark scheme: 5 (a) (i) 1 number of oscillations per unit time (not per second) B1 [1] 2 nλ A1 [1] (ii) v = distance / time = nλ/ t M1 n / t = f hence v= fλ A1 or f oscillations per unit time so fλ is distance per unit time M1 distance per unit time is v so v = fλ A1 [2] (b) (i) 1.0 period is 3 × 2 = 6.0 ms C1 frequency = 1 / (6 × 10–3) = 170 Hz A1 [2] (ii) wave (with approx. same amplitude and) with correct phase difference B1 [1]

This question in 9702/23 May/June 2010

5 A student is studying a water wave in which all the wavefronts are parallel to one another. For The variation with time t of the displacement x of a particular particle in the wave is shown Examiner’s in Fig. 5.1. Use +3 x / mm +2 +1 0 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 t / s –1 –2 –3 Fig. 5.1 The distance d of the oscillating particles from the source of the waves is measured. At a particular time, the variation of the displacement x with this distance d is shown in Fig. 5.2. +3 x / mm +2 +1 0 0 1 2 3 4 5 6 7 d / cm –1 –2 –3 Fig. 5.2 (a) Define, for a wave, what is meant by (i) displacement, … … [1] (ii) wavelength. … … [1] (b) Use Figs. 5.1 and 5.2 to determine, for the water wave, For Examiner’s (i) the period T of vibration, Use T = … s [1] (ii) the wavelength k, k = … cm [1] (iii) the speed v. v = … cm s–1 [2] (c) (i) Use Figs. 5.1 and 5.2 to state and explain whether the wave is losing power as it moves away from the source. … … … [2] (ii) Determine the ratio . ratio = … [3]

11 marks

Mark scheme: 5 (a) (i) distance (of point on wave) from rest / equilibrium position B1 [1] (ii) distance moved by wave energy / wavefront during one cycle of the source or minimum distance between two points with the same phase or between adjacent crests or troughs B1 [1] (b) (i) T = 0.60 s B1 [1] (ii) λ = 4.0 cm B1 [1] (iii) either v = λ/T or v = fλ and f = 1/T C1 v = 6.7 cm s–1 A1 [2] (c) (i) amplitude is decreasing M1 so, it is losing power A1 [2] (ii) intensity ~ (amplitude)2 C1 ratio = 2.02 / 1.12 C1 = 3.3 A1 [3]

This question in 9702/22 Oct/Nov 2010

1 Make estimates of the following quantities. (a) the thickness of a sheet of paper thickness = … mm [1] (b) the time for sound to travel 100 m in air time = … s [1] (c) the weight of 1000 cm3 of water weight = … N [1]

3 marks

Mark scheme: 1 (a) allow 0.05 mm → 0.15 mm B1 [1] (b) allow 0.25 s → 0.5 s B1 [1] (c) allow 8 N → 12 N B1 [1] ignore number of significant figures

This question in 9702/23 Oct/Nov 2010

3 A loudspeaker produces a sound wave of constant frequency. For Examiner’s Outline how a cathode-ray oscilloscope (c.r.o.) may be used to determine this frequency. Use … … … … … [4]

4 marks

Mark scheme: 3 connect microphone / (terminals of) loudspeaker to Y-plates of c.r.o. B1 adjust c.r.o. to produce steady wave of 1 (or 2) cycles / wavelengths on screen B1 measure length of cycle / wavelength λ and note time-base b M1 frequency = 1 / λb A1 [4] (assume b is measured as s cm–1, unless otherwise stated) (if statement is ‘measure T, f = 1/T’ then last two marks are lost)

This question in 9702/23 Oct/Nov 2010

6 (a) A transverse progressive wave travels along a stretched string from left to right. The For shape of part of the string at a particular instant is shown in Fig. 6.1. Examiner’s Use 8.0 P 6.0 displacement / mm 4.0 2.0 0 0 20 40 60 80 100 120 –2.0 distance along string / cm –4.0 –6.0 Q –8.0 Fig. 6.1 The frequency of the wave is 15 Hz. For this wave, use Fig. 6.1 to determine (i) the amplitude, amplitude = … mm [1] (ii) the phase difference between the points P and Q on the string, phase difference = … [1] (iii) the speed of the wave. speed = … m s–1 [2] (b) The period of vibration of the wave is T. The wave moves forward from the position shown in Fig 6.1 for a time 0.25 T. On Fig. 6.1, sketch the new position of the wave. [2] (c) Another stretched string is used to form a stationary wave. Part of this wave, at a For particular instant, is shown in Fig. 6.2. Examiner’s Use X Y Fig. 6.2 The points on the string are at their maximum displacement. (i) State the phase difference between the particles labelled X and Y. phase difference = … [1] (ii) Explain the following terms used to describe stationary waves on a string: antinode: … node: … [1] (iii) State the number of antinodes shown on Fig. 6.2 for this wave. number of antinodes = … [1] (iv) The period of vibration of this wave is τ. On Fig. 6.2, sketch the stationary wave 0.25 τ after the instant shown in Fig. 6.2. [1]

10 marks

Mark scheme: 6 (a) (i) amplitude = 7.6 mm allow 7.5 mm A1 [1] (ii) 180° / π rad A1 [1] (iii) v = f × λ = 15 × 0.8 C1 = 12 m s–1 A1 [2] (b) correct sketch with peak moved to the right B1 curve moved by the correct phase angle / time period of 0.25 T B1 [2] (c) (i) zero (rad) A1 [1] (ii) antinode maximum amplitude, node zero amplitude / displacement A1 [1] GCE AS/A LEVEL – May/June 2011 9702 23 (iii) 3 A1 [1] (iv) horizontal line through central section of wave B1 [1]

This question in 9702/23 May/June 2011

6 (a) State the principle of superposition. For Examiner’s … Use … … [2] (b) An arrangement that can be used to determine the speed of sound in air is shown in Fig. 6.1. S L microphone loudspeaker c.r.o. Fig. 6.1 Sound waves of constant frequency are emitted from the loudspeaker L and are reflected from a point S on a hard surface. The loudspeaker is moved away from S until a stationary wave is produced. Explain how sound waves from L give rise to a stationary wave between L and S. … … … [2] (c) A microphone connected to a cathode ray oscilloscope (c.r.o.) is positioned between L and S as shown in Fig. 6.1. The trace obtained on the c.r.o. is shown in Fig. 6.2. 1 cm 1 cm Fig. 6.2 The time-base setting on the c.r.o. is 0.10 ms cm–1. (i) Calculate the frequency of the sound wave. For Examiner’s Use frequency = … Hz [2] (ii) The microphone is now moved towards S along the line LS. When the microphone is moved 6.7 cm, the trace seen on the c.r.o. varies from a maximum amplitude to a minimum and then back to a maximum. 1. Use the properties of stationary waves to explain these changes in amplitude. … … … [1] 2. Calculate the speed of sound. speed of sound = … m s–1 [3] Please turn over for Question 7.

10 marks

Mark scheme: 6 (a) waves overlap B1 (resultant) displacement is the sum of the displacements of each of the waves B1 [2] GCE AS/A LEVEL – October/November 2011 9702 22 (b) waves travelling in opposite directions overlap / incident and reflected waves overlap (allow superpose or interfere for overlap here) B1 waves have the same speed and frequency B1 [2] (c) (i) time period = 4 × 0.1 (ms) C1 f = 1 / T = 1 / 4 × 10–4 = 2500 Hz A1 [2] (ii) 1. the microphone is at an antinode and goes to a node and then an antinode / maximum amplitude at antinode and minimum amplitude at node B1 [1] 2. λ / 2 = 6.7 (cm) C1 v = fλ C1 v = 2500 × 13.4 × 10–2 = 335 m s–1 A1 [3] incorrect λ then can only score second mark

This question in 9702/22 Oct/Nov 2011

5 (a) By reference to vibrations of the points on a wave and to its direction of energy transfer, For distinguish between transverse waves and longitudinal waves. Examiner’s Use … … … … [2] (b) Describe what is meant by a polarised wave. … … … [2] (c) The variation with distance x of the displacement y of a transverse wave is shown in Fig. 5.1. 3.0 AA 2.0 y / cm 1.0 BB 0 0 0.2 0.4 0.6 0.8 1.0 1.2 x / m –1.0 –2.0 –3.0 Fig. 5.1 (i) Use Fig. 5.1 to determine 1. the amplitude of the wave, amplitude = … cm [1] 2. the phase difference between the points labelled A and B. phase difference = … [2] (ii) Determine the amplitude of a wave with twice the intensity of that shown in For Fig. 5.1. Examiner’s Use amplitude = … cm [1]

8 marks

Mark scheme: 5 (a) transverse waves have vibrations that are perpendicular / normal to the direction of energy travel B1 longitudinal waves have vibrations that are parallel to the direction of energy travel B1 [2] (b) vibrations are in a single direction M1 either applies to transverse waves or normal to direction of wave energy travel or normal to direction of wave propagation A1 [2] (c) (i) 1. amplitude = 2.8 cm B1 [1] 2. phase difference = 135° or 0.75π rad or ¾π rad or 2.36 radians (three sf needed) numerical value M1 unit A1 [2] (ii) amplitude = 3.96 cm (4.0 cm) A1 [1]

This question in 9702/23 Oct/Nov 2011

5 (a) State three conditions required for maxima to be formed in an interference pattern For produced by two sources of microwaves. Examiner’s Use 1. … … 2. … … 3. … … [3] (b) A microwave source M emits microwaves of frequency 12 GHz. Show that the wavelength of the microwaves is 0.025 m. [3] (c) Two slits S1 and S2 are placed in front of the microwave source M described in (b), as shown in Fig 5.1. P m 0.75 m S1 0.90 O M microwave detector S2 Fig. 5.1 (not to scale) The distances S1O and S2O are equal. A microwave detector is moved from O to P. The distance S1P is 0.75 m and the distance S2P is 0.90 m. The microwave detector gives a maximum reading at O. For Examiner’s State the variation in the readings on the microwave detector as it is moved slowly along Use the line from O to P. … … … … … [3] (d) The microwave source M is replaced by a source of coherent light. State two changes that must be made to the slits in Fig. 5.1 in order to observe an interference pattern. 1. … 2. … [2]

11 marks

Mark scheme: 5 (a) waves overlap / meet / superpose (B1) coherence / constant phase difference (not constant λ or frequency) (B1) path difference = 0, λ, 2λ or phase difference = 0, 2π, 4π (B1) same direction of polarisation/unpolarised (B1) max. 3 [3] (b) λ = v / f C1 f = 12 × 109 Hz C1 λ = 3 × 108 / 12 × 109 (any subject) M1 = 0.025 m A0 [3] (c) maximum at P B1 several minima or maxima between O and P B1 5 maxima / 6 minima between O and P or 7 maxima / 6 minima including O and P B1 [3] (d) slits made narrower B1 slits put closer together B1 [2] (not just ‘make slits smaller’) Allow tilting the slits M1 and explanation of axes of rotation A1 GCE AS/A LEVEL – May/June 2013 9702 21

This question in 9702/21 May/June 2013

5 Fig. 5.1 shows a string stretched between two fixed points P and Q. For Examiner’s string Use P Q wall vibrator Fig. 5.1 A vibrator is attached near end P of the string. End Q is fixed to a wall. The vibrator has a frequency of 50 Hz and causes a transverse wave to travel along the string at a speed of 40 m s–1. (a) (i) Calculate the wavelength of the transverse wave on the string. wavelength = … m [2] (ii) Explain how this arrangement may produce a stationary wave on the string. … … … [2] (b) The stationary wave produced on PQ at one instant of time t is shown on Fig. 5.2. Each point on the string is at its maximum displacement. P Q Fig. 5.2 (not to scale) (i) On Fig. 5.2, label all the nodes with the letter N and all the antinodes with the letter A. [2] (ii) Use your answer in (a)(i) to calculate the length of string PQ. For Examiner’s Use length = … m [1] (iii) On Fig. 5.2, draw the stationary wave at time (t + 5.0 ms). Explain your answer. … [3]

10 marks

Mark scheme: 5 (a) (i) v = fλ C1 λ = 40 / 50 = 0.8(0) m A1 [2] (ii) waves (travel along string and) reflect at Q / wall / fixed end B1 incident and reflected waves interfere / superpose B1 [2] (b) (i) nodes labelled at P, Q and the two points at zero displacement B1 antinodes labelled at the three points of maximum displacement B1 [2] (ii) (1.5λ for PQ hence PQ = 0.8 × 1.5) = 1.2 m A1 [1] (iii) T = 1 / f = 1/50 = 20 ms C1 5 ms is ¼ of cycle A1 horizontal line through PQ drawn on Fig. 5.2 B1 [3]

This question in 9702/22 May/June 2013

5 (a) Explain the principle of superposition. For Examiner’s … Use … … [2] (b) Sound waves travel from a source S to a point X along two paths SX and SPX, as shown in Fig. 5.1. P reflecting surface m 3.0 4.0 m S X Fig. 5.1 (i) State the phase difference between these waves at X for this to be the position of 1. a minimum, phase difference = … unit … [1] 2. a maximum. phase difference = … unit … [1] (ii) The frequency of the sound from S is 400 Hz and the speed of sound is 320 m s–1. Calculate the wavelength of the sound waves. wavelength = … m [2] (iii) The distance SP is 3.0 m and the distance PX is 4.0 m. The angle SPX is 90°. Suggest whether a maximum or a minimum is detected at point X. Explain your answer. … … [2]

8 marks

Mark scheme: 5 (a) when waves overlap / meet B1 the resultant displacement is the sum of the individual displacements of the waves B1 [2] (b) (i) 1. phase difference = 180 º / (n + ½) 360 º (allow in rad) B1 [1] 2. phase difference = 0 / 360 º / (n360 º) (allow in rad) B1 [1] (ii) v = f λ C1 λ = 320 / 400 = 0.80 m A1 [2] (iii) path difference = 7 – 5 = 2 (m) = 2.5 λ M1 hence minimum or maximum if phase change at P is suggested A1 [2]

This question in 9702/23 May/June 2013

2 A source of radio waves sends a pulse towards a reflector. The pulse returns from the For reflector and is detected at the same point as the source. The emitted and reflected pulses Examiner’s are recorded on a cathode-ray oscilloscope (c.r.o.) as shown in Fig. 2.1. Use 1 cm 1 cm Fig. 2.1 The time-base setting is 0.20 μs cm–1. (a) Using Fig. 2.1, determine the distance between the source and the reflector. distance = … m [4] (b) Determine the time-base setting required to produce the same separation of pulses on the c.r.o. when sound waves are used instead of radio waves. The speed of sound is 300 m s–1. … … … [3]

7 marks

Mark scheme: 2 (a) d = v × t C1 t = 0.2 × 4 (allow t = 0.2 × 2) C1 d = 3 × 108 × 0.8 × 10–6 OR 3 × 108 × 0.4 × 10–6 C1 d = 240 m hence distance from source to reflector = 120 m A1 [4] (b) speed of sound 300 cf speed of light 3 × 108 OR time = 240 / 300 (= 0.8) OR time = 120 / 300 (= 0.4) C1 sound slower by factor of 106 OR time for one division 0.8 / 4 OR time for one division 0.4 / 2 C1 time base setting 0.2 s cm–1 [unit required] A1 [3]

This question in 9702/21 Oct/Nov 2013

5 A long rope is held under tension between two points A and B. Point A is made to vibrate For vertically and a wave is sent down the rope towards B as shown in Fig. 5.1. Examiner’s Use direction of travel of wave B A Fig. 5.1 (not to scale) The time for one oscillation of point A on the rope is 0.20 s. The point A moves a distance of 80 mm during one oscillation. The wave on the rope has a wavelength of 1.5 m. (a) (i) Explain the term displacement for the wave on the rope. … … [1] (ii) Calculate, for the wave on the rope, 1. the amplitude, amplitude = … mm [1] 2. the speed. speed = … m s–1 [3] (b) On Fig. 5.1, draw the wave pattern on the rope at a time 0.050 s later than that shown. [2] (c) State and explain whether the waves on the rope are (i) progressive or stationary, … … [1] (ii) longitudinal or transverse. … … [1]

9 marks

Mark scheme: 5 (a) (i) displacement is the distance the rope / particles are (above or below) from the equilibrium / mean / rest / undisturbed position (not ‘distance moved’) B1 [1] (ii) 1. amplitude (= 80 / 4) = 20 mm B1 [1] 2. v = fλ or v = λ / T C1 f = 1 / T = 1 / 0.2 (5 Hz) C1 v = 5 × 1.5 = 7.5 m s–1 A1 [3] (b) point A of rope shown at equilibrium position B1 same wavelength, shape, peaks / wave moved ¼λ to right B1 [2] (c) (i) progressive as energy OR peaks OR troughs is/are transferred/moved /propagated (by the waves) B1 [1] (ii) transverse as particles/rope movement is perpendicular to direction of travel /propagation of the energy/wave velocity B1 [1]

This question in 9702/21 Oct/Nov 2013

2 A source of radio waves sends a pulse towards a reflector. The pulse returns from the For reflector and is detected at the same point as the source. The emitted and reflected pulses Examiner’s are recorded on a cathode-ray oscilloscope (c.r.o.) as shown in Fig. 2.1. Use 1 cm 1 cm Fig. 2.1 The time-base setting is 0.20 μs cm–1. (a) Using Fig. 2.1, determine the distance between the source and the reflector. distance = … m [4] (b) Determine the time-base setting required to produce the same separation of pulses on the c.r.o. when sound waves are used instead of radio waves. The speed of sound is 300 m s–1. … … … [3]

7 marks

Mark scheme: 2 (a) d = v × t C1 t = 0.2 × 4 (allow t = 0.2 × 2) C1 d = 3 × 108 × 0.8 × 10–6 OR 3 × 108 × 0.4 × 10–6 C1 d = 240 m hence distance from source to reflector = 120 m A1 [4] (b) speed of sound 300 cf speed of light 3 × 108 OR time = 240 / 300 (= 0.8) OR time = 120 / 300 (= 0.4) C1 sound slower by factor of 106 OR time for one division 0.8 / 4 OR time for one division 0.4 / 2 C1 time base setting 0.2 s cm–1 [unit required] A1 [3]

This question in 9702/22 Oct/Nov 2013

5 A long rope is held under tension between two points A and B. Point A is made to vibrate For vertically and a wave is sent down the rope towards B as shown in Fig. 5.1. Examiner’s Use direction of travel of wave B A Fig. 5.1 (not to scale) The time for one oscillation of point A on the rope is 0.20 s. The point A moves a distance of 80 mm during one oscillation. The wave on the rope has a wavelength of 1.5 m. (a) (i) Explain the term displacement for the wave on the rope. … … [1] (ii) Calculate, for the wave on the rope, 1. the amplitude, amplitude = … mm [1] 2. the speed. speed = … m s–1 [3] (b) On Fig. 5.1, draw the wave pattern on the rope at a time 0.050 s later than that shown. [2] (c) State and explain whether the waves on the rope are (i) progressive or stationary, … … [1] (ii) longitudinal or transverse. … … [1]

9 marks

Mark scheme: 5 (a) (i) displacement is the distance the rope / particles are (above or below) from the equilibrium / mean / rest / undisturbed position (not ‘distance moved’) B1 [1] (ii) 1. amplitude (= 80 / 4) = 20 mm B1 [1] 2. v = fλ or v = λ / T C1 f = 1 / T = 1 / 0.2 (5 Hz) C1 v = 5 × 1.5 = 7.5 m s–1 A1 [3] (b) point A of rope shown at equilibrium position B1 same wavelength, shape, peaks / wave moved ¼λ to right B1 [2] (c) (i) progressive as energy OR peaks OR troughs is/are transferred/moved /propagated (by the waves) B1 [1] (ii) transverse as particles/rope movement is perpendicular to direction of travel /propagation of the energy/wave velocity B1 [1]

This question in 9702/22 Oct/Nov 2013

5 (a) (i) Define, for a wave, For Examiner’s 1. wavelength λ, Use … … [1] 2. frequency f. … … [1] (ii) Use your definitions to deduce the relationship between λ, f and the speed v of the wave. [1] (b) Plane waves on the surface of water are represented by Fig. 5.1 at one particular instant For of time. Examiner’s Use direction of travel of waves A B 8.0 mm 18 cm Fig. 5.1 (not to scale) The waves have frequency 2.5 Hz. Determine, for the waves, (i) the amplitude, amplitude = … mm [1] (ii) the speed, speed = … m s–1 [2] (iii) the phase difference between points A and B. phase difference = … unit … [1] (c) The wave in (b) was produced in a ripple tank. Describe briefly, with the aid of a sketch For diagram, how the wave may be observed. Examiner’s Use … … … [2]

9 marks

Mark scheme: 5 (a) (i) 1. wavelength: minimum distance between two points moving in phase OR distance between neighbouring or consecutive peaks or troughs OR wavelength is the distance moved by a wavefront in time T or one oscillation/cycle or period (of source) B1 [1] 2. frequency: number of wavefronts / (unit) time OR number of oscillations per unit time or oscillations/time B1 [1] (ii) speed = distance / time = wavelength / time period M1 speed = λ / T = λf A0 [1] (b) (i) amplitude = 4.0 mm (allow 1 s.f.) A1 [1] (ii) wavelength = 18 / 3.75 (= 4.8) C1 speed = 2.5 × 4.8 × 10–2 = 12 × 10–2 m s–1 unit consistent with numerical answer, e.g. in cm s–1 if cm used for λ and unit changed on answer line A1 [2] [if 18 cm = 3.5λ used giving speed 13 (12.9) cm s–1 allow max. 1]. (iii) 180º or π rad A1 [1] (c) light and screen and correct positions above and below ripple tank B1 strobe or video camera B1 [2]

This question in 9702/23 Oct/Nov 2013

5 (a) Explain what is meant by the following quantities for a wave on the surface of water: (i) displacement and amplitude, displacement … amplitude … [2] (ii) frequency and time period. frequency … time period … [2] (b) Fig. 5.1 represents waves on the surface of water in a ripple tank at one particular instant of time. direction of travel of waves vibrator 25 cm 15 mm 12 mm water side view ripple tank Fig. 5.1 (not to scale) A vibrator moves the surface of the water to produce the waves of frequency f. The speed of the waves is 7.5 cm s−1. Where the waves travel on the water surface, the maximum depth of the water is 15 mm and the minimum depth is 12 mm. (i) Calculate, for the waves, 1. the amplitude, amplitude = … mm [1] 2. the wavelength. wavelength = … m [2] (ii) Calculate the time period of the oscillations of the vibrator. time period = … s [2] (c) State and explain whether the waves on the surface of the water shown in Fig. 5.1 are (i) progressive or stationary, … … [1] (ii) transverse or longitudinal. … … [1]

11 marks

Mark scheme: 5 (a) (i) displacement is the distance from the equilibrium position / undisturbed position / midpoint / rest position B1 amplitude is the maximum displacement B1 [2] (ii) frequency is the number of wavefronts / crests passing a point per unit time / number of oscillations per unit time B1 time period is the time between adjacent wavefronts or time for one oscillation B1 [2] (b) (i) 1. amplitude = 1.5 mm A1 [1] 2. wavelength = 25 / 6 C1 = 4.2 cm or 4.2 × 10–2 m A1 [2] (ii) v = λ / T or v = f λ and T= 1 / f C1 T = 4.2 / 7.5 = 0.56 s A1 [2] (c) (i) progressive M0 wavefront / crests moving / energy is transferred by the waves A1 [1] (ii) transverse M0 the vibration is perpendicular to the direction of energy transfer / wave velocity or travel of the wave / wavefronts A1 [1]

This question in 9702/21 May/June 2014

7 A laser is placed in front of a double slit, as shown in Fig. 7.1. P double slit 12 mm laser Q bright fringes 2.8 m screen Fig. 7.1 (not to scale) The laser emits light of frequency 670 THz. Interference fringes are observed on the screen. (a) Explain how the interference fringes are formed. … … … … … … [3] (b) Show that the wavelength of the light is 450 nm. [2] (c) The separation of the maxima P and Q observed on the screen is 12 mm. The distance between the double slit and the screen is 2.8 m. Calculate the separation of the two slits. separation = … m [3] (d) The laser is replaced by a laser emitting red light. State and explain the effect on the interference fringes seen on the screen. … … … … [2]

10 marks

Mark scheme: 7 (a) waves from the double slit are coherent / constant phase difference B1 waves (from each slit) overlap / superpose / meet (not interfere) B1 maximum / bright fringe where path difference is nλ or phase difference is n360U / 2πn rad 1 or minimum / dark fringe where path difference is (n + )λ 2 or phase difference is (2n + 1) 180U / (2n + 1)π rad B1 [3] (b) v = fλ C1 λ = (3 × 108) / 670 × 1012 = 448 (or 450) (nm) M1 [2] (c) w = 12 / 9 C1 a (= Dλ / w) = (2.8 × 450 × 10–9) / (12 / 9 × 10–3) [allow nm, mm] C1 = 9.5 × 10–4 m [9.4 × 10–4 m using λ = 448 nm] A1 [3] (d) (red light has) larger / higher / longer wavelength (must be comparison) M1 fringes further apart / larger separation A1 [2]

This question in 9702/22 May/June 2014

6 A hollow tube is used to investigate stationary waves. The tube is closed at one end and open at the other end. A loudspeaker connected to a signal generator is placed near the open end of the tube, as shown in Fig. 6.1. L loudspeaker Q P signal generator hollow tube Fig. 6.1 The tube has length L. The frequency of the signal generator is adjusted so that the loudspeaker produces a progressive wave of frequency 440 Hz. A stationary wave is formed in the tube. A representation of this stationary wave is shown in Fig. 6.1. Two points P and Q on the stationary wave are labelled. (a) (i) Describe, in terms of energy transfer, the difference between a progressive wave and a stationary wave. … … [1] (ii) Explain how the stationary wave is formed in the tube. … … … … [3] (iii) State the direction of the oscillations of an air particle at point P. … … [1] (b) On Fig. 6.1 label, with the letter N, the nodes of the stationary wave. [1] (c) State the phase difference between points P and Q on the stationary wave. phase difference = … [1] (d) The speed of sound in the tube is 330 m s–1. Calculate (i) the wavelength of the sound wave, wavelength = … m [2] (ii) the length L of the tube. length = … m [2]

11 marks

Mark scheme: 6 (a) (i) progressive wave transfers energy, stationary wave no transfer of energy / keeps energy within wave B1 [1] (ii) (progressive) wave / wave from loudspeaker reflects at end of tube B1 reflected wave overlaps (another) progressive wave B1 same frequency and speed hence stationary wave formed B1 [3] (iii) (side to side) along length of tube / along axis of tube B1 [1] (b) all three nodes clearly marked with N / clearly labelled at cross-over points B1 [1] (c) phase difference = 0 A1 [1] (d) (i) v = fλ C1 λ = 330 / 440 = 0.75 m A1 [2] (ii) L = 5/4 λ C1 = 5/4 × 0.75 = 0.94 m A1 [2]

This question in 9702/23 May/June 2014

8 (a) Explain how stationary waves are formed. … … … [2] (b) The arrangement of apparatus used to determine the wavelength of a sound wave is shown in Fig. 8.1. microphone loudspeaker metal plate signal generator c.r.o. Fig. 8.1 The loudspeaker emits sound of one frequency. The microphone is connected to a cathode-ray oscilloscope (c.r.o.). The waveform obtained on the c.r.o. for one position of the microphone is shown in Fig. 8.2. 1.0 cm 1.0 cm Fig. 8.2 The time-base setting of the c.r.o. is 0.20 ms cm−1. (i) Use Fig. 8.2 to show that the frequency of the sound is approximately 1300 Hz. [2] (ii) Explain how the apparatus is used to determine the wavelength of the sound. … … … … [2] (iii) The wavelength of the sound wave is 0.26 m. Calculate the speed of sound in this experiment. speed = … m s−1 [2]

8 marks

Mark scheme: 8 (a) two waves (of the same kind) travelling in opposite directions overlap B1 waves have same frequency / wavelength and speed B1 [2] (b) (i) T = 0.8 (ms) C1 f = 1 / (0.8 × 10–3) = 1250 (Hz) A1 [2] (ii) microphone is moved from plate to loudspeaker or vice versa B1 wavelength is the twice the distance between adjacent maxima or minima (seen on c.r.o.) B1 [2] (iii) v = fλ C1 = 1250 × 0.26 = 330 (325) m s–1 A1 [2]

This question in 9702/21 Oct/Nov 2014

7 (a) (i) Explain what is meant by a progressive transverse wave. progressive: … … transverse: … … [2] (ii) Define frequency. … … [1] (b) The variation with distance x of displacement y for a transverse wave is shown in Fig. 7.1. 2.0 R y / cm 1.0 Q S 0 0 0.4 0.8 1.2 1.6 2.0 x / cm –1.0 P T –2.0 Fig. 7.1 On Fig. 7.1, five points are labelled. Use Fig. 7.1 to state any two points having a phase difference of (i) zero, … [1] (ii) 270°. … [1] (c) The frequency of the wave in (b) is 15 Hz. Calculate the speed of the wave in (b). speed = … m s–1 [3] (d) Two waves of the same frequency have amplitudes 1.4 cm and 2.1 cm. Calculate the ratio intensity of wave of amplitude 1.4 cm . intensity of wave of amplitude 2.1 cm ratio = … [2]

10 marks

Mark scheme: 7 (a) (i) progressive: energy is moved / transferred / propagated from one place to another (without the bulk movement of the medium) B1 transverse: (particles) oscillate / vibrate at right angles to the direction of travel of the energy / wavefront B1 [2] (ii) number of oscillations per unit time / number of wavefronts passing a point per unit time B1 [1] (b) (i) P and T B1 [1] (ii) P and S or Q and T B1 [1] (c) λ = 1.2 × 10–2 (m) C1 v = fλ = 15 × 1.2 × 10–2 C1 = 0.18 m s–1 A1 [3] (d) ratio = (1.4)2 / (2.1)2 C1 = 0.44 A1 [2]

This question in 9702/23 Oct/Nov 2014

6 (a) State two differences between progressive waves and stationary waves. 1. … … 2. … … [2] (b) A source S of microwaves is placed in front of a metal reflector R, as shown in Fig. 6.1. metal reflector R microwave detector D microwave source S meter Fig. 6.1 A microwave detector D is placed between R and S. Describe (i) how stationary waves are formed between R and S, … … … … [3] (ii) how D is used to show that stationary waves are formed between R and S, … … … [2] (iii) how the wavelength of the microwaves may be determined using the apparatus in Fig. 6.1. … … … [2] (c) The wavelength of the microwaves in (b) is 2.8 cm. Calculate the frequency, in GHz, of the microwaves. frequency = … GHz [3] Please turn over for Question 7.

12 marks

Mark scheme: 6 (a) progressive waves transfer/propagate energy and stationary waves do not B1 amplitude constant for progressive wave and varies (from max/antinode to min/zero/node) for stationary wave B1 adjacent particles in phase for stationary wave and out of phase for progressive wave (B1) [2] (b) (i) wave / microwave from source/S reflects at reflector/R B1 reflected and (further) incident waves overlap/meet/superpose B1 waves have same frequency/wavelength/period and speed (so stationary waves formed) B1 [3] (ii) detector/D is moved between reflector/R and source/S (or v.v.) B1 maximum, minimum/zero, (maximum… etc.) observed on meter/deflections/readings/measurements/recordings B1 [2] (iii) determine/measure the distance between adjacent minima/nodes or maxima/antinodes or across specific number of nodes/antinodes B1 wavelength is twice distance between adjacent nodes/minima or maxima/ antinodes (or other correct method of calculation of wavelength from measurement) B1 [2] (c) v = fλ C1 f = 3.0 × 108 / (2.8 × 10–2) [= 1.07 × 1010 Hz] C1 11 (10.7) GHz A1 [3]

This question in 9702/22 May/June 2015

6 (a) Two overlapping waves of the same type travel in the same direction. The variation with distance x of the displacement y of each wave is shown in Fig. 6.1. 3.0 y / cm 2.0 1.0 0 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 x / m Fig. 6.1 The speed of the waves is 240 m s–1. The waves are coherent and produce an interference pattern. (i) Explain the meaning of coherence and interference. coherence: … … interference: … … [2] (ii) Use Fig. 6.1 to determine the frequency of the waves. frequency = … Hz [2] (iii) State the phase difference between the waves. phase difference = … ° [1] (iv) Use the principle of superposition to sketch, on Fig. 6.1, the resultant wave. [2] (b) An interference pattern is produced with the arrangement shown in Fig. 6.2. B S1 A laser 0.13 mm S2 85 cm screen Fig. 6.2 (not to scale) Laser light of wavelength λ of 546 nm is incident on the slits S1 and S2. The slits are a distance 0.13 mm apart. The distance between the slits and the screen is 85 cm. Two points on the screen are labelled A and B. The path difference between S1A and S2A is zero. The path difference between S1B and S2B is 2.5 λ. Maxima and minima of intensity of light are produced on the screen. (i) Calculate the distance AB. distance = … m [3] (ii) The laser is replaced by a laser emitting blue light. State and explain the change in the distance between the maxima observed on the screen. … … … [1]

11 marks

Mark scheme: 6 (a) (i) coherent: constant phase difference B1 interference is the (overlapping of waves and the) sum of/addition of displacement of two waves B1 [2] (ii) wavelength = 3.2 m (allow ± 0.05 m) M1 f (= v / λ = 240 / 3.2) = 75 Hz A1 [2] (iii) 90° (allow ± 2°) or π/2 rad A1 [1] (iv) sketch has amplitude 3.0 ± 0.1 cm M1 correct displacement values at previous peaks to produce correct shape A1 [2] (b) (i) λ = ax / D C1 x = (546 × 10–9 × 0.85) / 0.13 × 10–3 (= 3.57 × 10–3 m) C1 AB = 8.9 (8.93) × 10–3 m A1 [3] (ii) shorter wavelength for blue light so separation is less B1 [1]

This question in 9702/23 May/June 2015

2 The signal from a microwave detector is recorded on a cathode-ray oscilloscope (c.r.o.), as shown in Fig. 2.1. 1 cm 1 cm Fig. 2.1 The time-base setting on the c.r.o. is 50 ps cm–1. (a) Using Fig. 2.1, determine the wavelength of the microwaves. wavelength = … m [4] (b) The signal from a radio wave detector is recorded on the same c.r.o. The wavelength of the radio waves is 1.5 × 103 m. Determine the time-base setting required to display the same number of oscillations on the c.r.o. as shown in Fig. 2.1. time-base setting = … unit … [2]

6 marks

Mark scheme: 2 (a) ps = 10–12 (s) or T = 4 × 50 × 10–12 (s) B1 v = fλ or v = λ / T C1 λ = 3.0 × 108 × 4 × 50 × 10–12 C1 = 0.06(0) m A1 [4] (b) 1500 = 3.0 × 108 × 4 × time-base setting or T = 5 × 10–6 s C1 time-base setting = 1.3 (1.25) µs cm–1 A1 [2]

This question in 9702/21 Oct/Nov 2015

5 (a) A progressive wave transfers energy. A stationary wave does not transfer energy. State two other differences between progressive waves and stationary waves. 1. … … 2. … … [2] (b) A stationary wave is formed on a stretched string between two fixed points A and B. The variation of the displacement y of particles of the string with distance x along the string for the wave at time t = 0 is shown on Fig. 5.1. 10 position of particles at t = 0 5 y / mm A B 0 0 1.0 2.0 x / m –5 –10 Fig. 5.1 The wave has a period of 20 ms and a wavelength of 1.2 m. The maximum amplitude of the particles of the string is 5.0 mm. (i) On Fig. 5.1, draw a line to represent the position of the string at t = 5.0 ms. [2] (ii) State the phase difference between the particles of the string at x = 0.40 m and at x = 0.80 m. phase difference = … unit … [1] (iii) State and explain the change in the kinetic energy of a particle at an antinode between t = 0 and t = 5.0 ms. A numerical value is not required. … … … … [2]

7 marks

Mark scheme: 5 (a) progressive: all particles have same amplitude stationary: no nodes or antinodes or maximum to minimum/zero amplitude B1 progressive: adjacent particles are not in phase stationary: waves particles are in phase (between adjacent nodes) B1 [2] (b) (i) wavelength 1.2 m (zero displacement at 0.0, 0.60 m, 1.2 m, 1.8 m, 2.4 m) either peaks at 0.30 m and 1.5 m and troughs at 0.90 m and 2.1 m or vice versa (but not both) B1 maximum amplitude 5.0 mm B1 [2] (ii) 180° or π rad A1 [1] (iii) at t = 0 particle has kinetic energy as particle is moving B1 at t = 5.0 ms no kinetic energy as particle is stationary so decrease in kinetic energy (between t = 0 and t = 5.0 ms) B1 [2]

This question in 9702/21 Oct/Nov 2015

7 An arrangement that is used to demonstrate interference with waves on the surface of water is shown in Fig. 7.1. A wooden bar dipper D1 to d.c. power B supply motor dipper water D2 Fig. 7.1 (view from above) (a) Two dippers D1 and D2 are connected to a motor and a d.c. power supply. Initially only D1 vibrates on the water surface to produce waves. The variation with distance x from D1 of the displacement y of the water at one instant of time is shown in Fig. 7.2. 4.0 y / mm 2.0 0 0 10 20 30 40 50 x / mm –2.0 –4.0 Fig. 7.2 Using Fig. 7.2, determine (i) the amplitude of the wave, amplitude = … mm [1] (ii) the wavelength of the wave. wavelength = … mm [1] (b) The two dippers D1 and D2 are made to vibrate and waves are produced by both dippers on the water surface. (i) State and explain whether these waves are stationary or progressive. … … [1] (ii) Explain why D1 and D2 are connected to the same motor. … … [1] (c) The points A and B on Fig. 7.1 are at the distances from D1 and D2 shown in Fig. 7.3. D1A D2A D1B D2B 5.0 cm 7.0 cm 5.0 cm 6.0 cm Fig. 7.3 State and explain the variation with time of the displacement of the water on the surface at (i) A, … … … … [2] (ii) B. … … … [1]

7 marks

Mark scheme: 7 (a) (i) 3.2 mm A1 [1] (ii) 20 mm A1 [1] (b) (i) energy is transferred/propagated (through the water) or wave profile/wavefronts move (outwards from dipper) so progressive B1 [1] (ii) to produce waves with constant/zero phase difference/coherent waves B1 [1] (c) (i) path difference is λ B1 water vibrates/oscillates with amplitude about 2 × 3.2 mm B1 [2] (ii) path difference is λ / 2 so little/no motion/displacement/amplitude B1 [1]

This question in 9702/22 Oct/Nov 2015

2 A signal generator is connected to two loudspeakers L1 and L2, as shown in Fig. 2.1. C L1 B M signal c.r.o. A generator L2 Fig. 2.1 A microphone M, connected to the Y-plates of a cathode-ray oscilloscope (c.r.o.), detects the intensity of sound along the line ABC. The distances L1A and L2A are equal. The time-base of the c.r.o. is switched off. The traces on the c.r.o. when M is at A, then at B and then at C are shown on Fig. 2.2, Fig. 2.3 and Fig. 2.4 respectively. 1.0 cm M at A M at B M at C Fig. 2.2 Fig. 2.3 Fig. 2.4 For these traces, 1.0 cm represents 5.0 mV on the vertical scale. (a) (i) Explain why coherent waves are produced by the loudspeakers. … … … [1] (ii) Use the principle of superposition to explain the traces shown with M at 1. A, … … … [1] 2. B, … … … [1] 3. C. … … … [1] (b) The sound emitted from L1 and L2 has frequency 500 Hz. The time-base on the c.r.o. is switched on. The microphone M is placed at A. On Fig. 2.5, draw the trace seen on the c.r.o. On the vertical scale, 1.0 cm represents 5.0 mV. On the horizontal scale, 1.0 cm represents 0.10 ms. 1.0 cm 1.0 cm Fig. 2.5 [3]

7 marks

Mark scheme: 2 (a) (i) (the loudspeakers) are connected to the same signal generator B1 [1] (ii) 1. the waves (that overlap) have phase difference of zero or path difference of zero and so either constructive interference or displacement larger B1 [1] 2. the waves (that overlap) have phase difference of (n + ½) × 360° or (n + ½) × 2π rad or path difference of (n + ½)λ and so either destructive interference or displacements cancel/smaller B1 [1] 3. the waves (that overlap) are in phase or have phase difference of n360° or 2πn rad or path difference of nλ and so either constructive interference or displacement larger B1 [1] (b) time period = 0.002 s or 2 ms C1 wave drawn is half time period B1 amplitude 1.0 cm (same as Fig. 2.2) B1 [3] 2

This question in 9702/23 Oct/Nov 2015

4 (a) (i) By reference to the direction of propagation of energy, state what is meant by a transverse wave. … … [1] (ii) State the principle of superposition. … … … [2] (b) Circular water waves may be produced by vibrating dippers at points P and Q, as illustrated in Fig. 4.1. wavefront P 44 cm R 29 cm Q Fig. 4.1 (not to scale) The waves from P alone have the same amplitude at point R as the waves from Q alone. Distance PR is 44 cm and distance QR is 29 cm. The dippers vibrate in phase with a period of 1.5 s to produce waves of speed 4.0 cm s−1. (i) Determine the wavelength of the waves. wavelength = … cm [2] (ii) By reference to the distances PR and QR, explain why the water particles are at rest at point R. … … … … … [3] (c) A wave is produced on the surface of a different liquid. At one particular time, the variation of the vertical displacement y with distance x along the surface of the liquid is shown in Fig. 4.2. 1.0 y / cm 0.5 0 00 2 4 6 8 10 x / cm –0.5 –1.0 Fig. 4.2 (i) The wave has intensity I1 at distance x = 2.0 cm and intensity I2 at x = 10.0 cm. Determine the ratio intensity I2 . intensity I1 ratio = … [2] (ii) State the phase difference, with its unit, between the oscillations of the liquid particles at distances x = 3.0 cm and x = 4.0 cm. phase difference = … [1] [Total: 11]

11 marks

Mark scheme: 4 (a) (i) Displacement of particles perpendicular to direction of energy propagation B1 (ii) waves meet / overlap (at a point) B1 (resultant) displacement is sum of the individual displacements B1 (b) (i) λ = vT or λ = v / f and f = 1 / T C1 λ = 4.0 × 1.5 λ = 6.0 (cm) A1 (ii) path difference [= (44 cm – 29 cm) / 6 cm] = 2.5λ M1 either waves have path difference = (n + ½)λ or waves have phase difference = 180° M1 so destructive interference A1 (c) (i) intensity ∝ (amplitude)2 C1 ratio = (0.602 / 0.902) = 0.44 A1 (ii) phase difference = 90° A1

This question in 9702/22 Feb/March 2016

5 The variation with time t of the displacement y of a wave X, as it passes a point P, is shown in Fig. 5.1. 4.0 3.0 \ / cm ZDYH ; 2.0 1.0 0 0 1.0 2.0 3.0 4.0 5.0 W / ms –1.0 –2.0 –3.0 –4.0 Fig. 5.1 The intensity of wave X is I. (a) Use Fig. 5.1 to determine the frequency of wave X. frequency = … Hz [2] (b) A second wave Z with the same frequency as wave X also passes point P. Wave Z has intensity 2I. The phase difference between the two waves is 90°. On Fig. 5.1, sketch the variation with time t of the displacement y of wave Z. Show your working. [3] (c) A double-slit interference experiment is used to determine the wavelength of light emitted from a laser, as shown in Fig. 5.2. 0.45 mm laser light ' double slit screen Fig. 5.2 (not to scale) The separation of the slits is 0.45 mm. The fringes are viewed on a screen at a distance D from the double slit. The fringe width x is measured for different distances D. The variation with D of x is shown in Fig. 5.3. 5.0 4.0 [ / mm 3.0 2.0 1.0 0 1.5 2.0 2.5 3.0 3.5 ' / m Fig. 5.3 (i) Use the gradient of the line in Fig. 5.3 to determine the wavelength, in nm, of the laser light. wavelength = … nm [4] (ii) The separation of the slits is increased. State and explain the effects, if any, on the graph of Fig. 5.3. … … … [2] [Total: 11]

11 marks

Mark scheme: 5 (a) T = 4 (ms) or 4 × 10–3 (s) C1 f = 1 / T = 1 / 0.004 = 250 Hz A1 [2] (b) intensity ∝ (amplitude)2 and amplitude = 2.8 (2.83) (cm) B1 curve with same period and with amplitude 2.8 cm B1 curve shifted 1.0 ms to left or to right of wave X B1 [3] (c) (i) gradient = (4.5 – 2.4) × 10–3 / (3.25 – 1.75) [= 1.4 × 10–3] B1 wavelength = 0.45 × 10–3 × 1.4 × 10–3 C1 = 6.30 × 10–7 (m) C1 = 630 nm A1 [4] (ii) (gradient is equal to λ / a therefore) gradient of line is reduced B1 value of x will be reduced for all values of D or new line is completely below old line or intercept is less B1 [2]

This question in 9702/21 May/June 2016

4 (a) By reference to the direction of the propagation of energy, state what is meant by a longitudinal wave and by a transverse wave. longitudinal: … … … transverse: … … … [2] (b) The intensity of a sound wave passing through air is given by Ι = Kvρf 2A2 where Ι is the intensity (power per unit area), K is a constant without units, v is the speed of sound, ρ is the density of air, f is the frequency of the wave and A is the amplitude of the wave. Show that both sides of the equation have the same SΙ base units. [3] (c) (i) Describe the Doppler effect. … … [1] (ii) A distant star is moving away from a stationary observer. State the effect of the motion on the light observed from the star. … … … [1] (d) A car travels at a constant speed towards a stationary observer. The horn of the car sounds at a frequency of 510 Hz and the observer hears a frequency of 550 Hz. The speed of sound in air is 340 m s–1. Calculate the speed of the car. speed = … m s–1 [3] [Total: 10]

10 marks

Mark scheme: 4 (a) longitudinal: vibrations/oscillations (of the particles/wave) are parallel to the direction or in the same direction (of the propagation of energy) B1 transverse: vibrations/oscillations (of the particles/wave) are perpendicular to the direction (of the propagation of energy) B1 [2] (b) LHS: intensity = power / area units: kg m s–2 × m × s–1 × m–2 or kg m2 s–3 × m–2 B1 RHS: units: m s–1 × kg m–3 × s–2 × m2 M1 LHS and RHS both kg s–3 A1 [3] (c) (i) change/difference in the observed/apparent frequency when the source is moving (relative to the observer) B1 [1] (ii) wavelength increases/frequency decreases/red shift B1 [1] (d) observed frequency = vfS / (v – vS) C1 550 = (340 × 510) / (340 – vS) C1 vS = 25 (24.7) m s–1 A1 [3]

This question in 9702/22 May/June 2016

7 (a) Apparatus used to produce stationary waves on a stretched string is shown in Fig. 7.1. frequency light string generator pulley wheel vibrator masses Fig. 7.1 The frequency generator is switched on. (i) Describe two adjustments that can be made to the apparatus to produce stationary waves on the string. 1. … … 2. … … [2] (ii) Describe the features that are seen on the stretched string that indicate stationary waves have been produced. … [1] (b) The variation with time t of the displacement x of a particle caused by a progressive wave R is shown in Fig. 7.2. For the same particle, the variation with time t of the displacement x caused by a second wave S is also shown in Fig. 7.2. 4.0 R 3.0 x / cm 2.0 S 1.0 0 0 0.2 0.4 0.6 0.8 1.0 t / s Fig. 7.2 (i) Determine the phase difference between wave R and wave S. Include an appropriate unit. phase difference = … [1] (ii) Calculate the ratio intensity of wave R . intensity of wave S ratio = … [2] [Total: 6]

6 marks

Mark scheme: 7 (a) (i) alter distance from vibrator to pulley alter frequency of generator (change tension in string by) changing value of the masses any two B2 [2] (ii) points on string have amplitudes varying from maximum to zero/minimum B1 [1] (b) (i) 60° or π / 3 rad A1 [1] (ii) ratio = [3.4 / 2.2]2 C1 = 2.4 (2.39) A1 [2] +

This question in 9702/23 May/June 2016

4 (a) State what is meant by the frequency of a progressive wave. … … … [2] (b) A cathode-ray oscilloscope (c.r.o.) is used to determine the frequency of the sound emitted by a loudspeaker. The trace produced on the screen of the c.r.o. is shown in Fig. 4.1. 1 cm 1 cm Fig. 4.1 The time-base setting of the c.r.o. is 250 μs cm–1. Show that the frequency of the sound wave is 1600 Hz. [2] (c) The loudspeaker in (b) emits the sound in all directions. A person attaches the loudspeaker to a string and then swings the loudspeaker at a constant speed in a horizontal circle above his head. An observer, standing a large distance away from the loudspeaker, hears sound of maximum frequency 1640 Hz. The speed of sound in air is 330 m s–1. (i) Determine the speed of the loudspeaker. speed = … m s–1 [2] (ii) Describe and explain, qualitatively, the variation in the frequency of the sound heard by the observer. … … … … … [2] [Total: 8]

8 marks

Mark scheme: 4 (a) the number of oscillations per unit time M1 of the source/of a point on the wave/of a particle (in the medium) A1 [2] or the number of wavelengths/wavefronts per unit time (M1) passing a (fixed) point (A1) (b) T or period = 2.5 × 250 (µs) (= 625 µs) M1 frequency = 1 / (6.25 × 10–4) or 1 / (2.5 × 250 × 10–6) = 1600 Hz A1 [2] (c) (i) for maximum frequency: fo = fsv / (v – vs) 1640 = (1600 × 330) / (330 – vs) C1 vs = 8(.0) m s–1 (8.049 m s–1) A1 [2] (ii) loudspeaker moving towards observer causes rise in/higher frequency B1 loudspeaker moving away from observer causes fall in/lower frequency B1 [2] or repeated rise and fall/higher and then lower frequency (M1) caused by loudspeaker moving towards and away from observer (A1)

This question in 9702/21 Oct/Nov 2016

4 (a) State what is meant by the frequency of a progressive wave. … … … [2] (b) A cathode-ray oscilloscope (c.r.o.) is used to determine the frequency of the sound emitted by a loudspeaker. The trace produced on the screen of the c.r.o. is shown in Fig. 4.1. 1 cm 1 cm Fig. 4.1 The time-base setting of the c.r.o. is 250 µs cm–1. Show that the frequency of the sound wave is 1600 Hz. [2] (c) The loudspeaker in (b) emits the sound in all directions. A person attaches the loudspeaker to a string and then swings the loudspeaker at a constant speed in a horizontal circle above his head. An observer, standing a large distance away from the loudspeaker, hears sound of maximum frequency 1640 Hz. The speed of sound in air is 330 m s–1. (i) Determine the speed of the loudspeaker. speed = … m s–1 [2] (ii) Describe and explain, qualitatively, the variation in the frequency of the sound heard by the observer. … … … … … [2] [Total: 8]

8 marks

Mark scheme: 4 (a) the number of oscillations per unit time M1 of the source/of a point on the wave/of a particle (in the medium) A1 [2] or the number of wavelengths/wavefronts per unit time (M1) passing a (fixed) point (A1) (b) T or period = 2.5 × 250 (µs) (= 625 µs) M1 frequency = 1 / (6.25 × 10–4) or 1 / (2.5 × 250 × 10–6) = 1600 Hz A1 [2] (c) (i) for maximum frequency: fo = fsv / (v – vs) 1640 = (1600 × 330) / (330 – vs) C1 vs = 8(.0) m s–1 (8.049 m s–1) A1 [2] (ii) loudspeaker moving towards observer causes rise in/higher frequency B1 loudspeaker moving away from observer causes fall in/lower frequency B1 [2] or repeated rise and fall/higher and then lower frequency (M1) caused by loudspeaker moving towards and away from observer (A1)

This question in 9702/23 Oct/Nov 2016

5 (a) Define the frequency of a sound wave. … … [1] (b) A sound wave travels through air. Describe the motion of the air particles relative to the direction of travel of the sound wave. … … [1] (c) The sound wave emitted from the horn of a stationary car is detected with a microphone and displayed on a cathode-ray oscilloscope (c.r.o.), as shown in Fig. 5.1. 1.0 cm 1.0 cm Fig. 5.1 The y-axis setting is 5.0 mV cm–1. The time-base setting is 0.50 ms cm–1. (i) Use Fig. 5.1 to determine the frequency of the sound wave. frequency = … Hz [2] (ii) The horn of the car sounds continuously. Describe the changes to the trace seen on the c.r.o. as the car travels at constant speed 1. directly towards the stationary microphone, … … 2. directly away from the stationary microphone. … … [3] [Total: 7]

7 marks

Mark scheme: 5(a) frequency is the number of vibrations/oscillations per unit time or the number of wavefronts passing a point per unit time B1 5(b) vibrations/oscillation of the air particles are parallel to the direction of it (the direction of travel of the sound wave) B1 5(c)(i) T = 2(.0) (ms) C1 f = 500 Hz A1 5(c)(ii) 1. amplitude increases (time) period decreases 2. amplitude decreases (time) period increases any 3 points B3

This question in 9702/22 May/June 2017

4 (a) By reference to the direction of propagation of energy, explain what is meant by a longitudinal wave. … … [1] (b) A car horn emits a sound wave of frequency 800 Hz. A microphone and a cathode-ray oscilloscope (c.r.o.) are used to analyse the sound wave. The waveform displayed on the c.r.o. screen is shown in Fig. 4.1. 1 cm 1 cm Fig. 4.1 Determine the time-base setting, in s cm–1, of the c.r.o. time-base setting = … s cm–1 [3] (c) The intensity I of the sound at a distance r from the car horn in (b) is given by the expression k I = 2 r where k is a constant. Fig. 4.2 shows the car in (b) on a road. O Y X road 30 m 120 m Fig. 4.2 An observer stands at point O. Initially the car is parked at point X which is 120 m away from point O. The car then moves directly towards the observer and stops at point Y, a distance of 30 m away from O. The car horn continuously emits sound when the car is moving between points X and Y. (i) The sound wave at point O has amplitude AX when the car is at X and has amplitude AY when the car is at Y. AY Calculate the ratio . AX ratio = … [3] (ii) When the car is parked at X, the frequency of the sound from the horn that is detected by the observer is 800 Hz. As the car moves from X to Y, the maximum change in the detected frequency is 16 Hz. The speed of the sound in air is 330 m s–1. Determine, to two significant figures, 1. the minimum wavelength of the sound detected by the observer, wavelength = … m [2] 2. the maximum speed of the car. speed = … m s–1 [2] [Total: 11]

11 marks

Mark scheme: 4(a) displacement of particles/vibration(s)/oscillation(s) is parallel to/along the direction of energy/propagation B1 4(b) period = 1 / 800 (= 1.25 × 10–3 s) C1 time-base setting = 1.25 × 10–3 / 2.5 C1 = 5.0 × 10–4 s cm–1 A1 4(c)(i) I ∝ A2 C1 (IX / IY =) [rY / rX] 2 = [AX / AY]2 C1 ratio AY / AX = 120 / 30 = 4.0 A1 Question Answer Marks 4(c)(ii) 1. v = f λ C1 minimum λ = 330 / (800 + 16) = 0.40 m A1 2. fo / fs = v / (v – vs) 816 / 800 = 330 / (330 – vs) C1 vs = 6.5 m s–1 A1

This question in 9702/23 Oct/Nov 2017

4 (a) For a progressive wave, state what is meant by (i) the period, … … [1] (ii) the wavelength. … … [1] (b) Fig. 4.1 shows the variation with time t of the displacement x of two progressive waves P and Q passing the same point. 4.0 3.0 x / mm wave P 2.0 1.0 0 0 0.20 0.40 0.60 0.80 t / s –1.0 wave Q –2.0 –3.0 –4.0 Fig. 4.1 The speed of the waves is 20 cm s–1. (i) Calculate the wavelength of the waves. wavelength = … cm [2] (ii) Determine the phase difference between the two waves. phase difference = … ° [1] (iii) Calculate the ratio intensity of wave Q . intensity of wave P ratio = … [2] (iv) The two waves superpose as they pass the same point. Use Fig. 4.1 to determine the resultant displacement at time t = 0.45 s. displacement = … mm [1] [Total: 8]

8 marks

Mark scheme: 4(a)(i) time for one oscillation/one vibration/one cycle or time between adjacent wavefronts/points in phase or shortest time between two wavefronts/points in phase B1 4(a)(ii) distance moved by wavefront/energy during one cycle/oscillation/period (of source) or minimum distance between two wavefronts or distance between two adjacent wavefronts or minimum distance between two points having the same displacement and moving in the same direction B1 4(b)(i) v = λ / T or v = fλ and f = 1 / T C1 λ = 20 × 0.60 = 12 cm A1 4(b)(ii) phase difference = 360° × (0.20 / 0.60) or 360° × (0.40 / 0.60) = 120° or 240° A1 4(b)(iii) I ∝ A2 C1 IQ / IP = AQ 2 / AP 2 = 2.02 / 3.02 = 0.44 A1 4(b)(iv) displacement = 1.00 – 3.00 = –2.00 mm A1

This question in 9702/21 May/June 2018

4 (a) (i) Define the wavelength of a progressive wave. … … [1] (ii) State what is meant by an antinode of a stationary wave. … … [1] (b) A loudspeaker producing sound of constant frequency is placed near the open end of a pipe, as shown in Fig. 4.1. pipe piston loudspeaker speed 0.75 cm s–1 x Fig. 4.1 A movable piston is at distance x from the open end of the pipe. Distance x is increased from x = 0 by moving the piston to the left with a constant speed of 0.75 cm s–1. The speed of the sound in the pipe is 340 m s–1. (i) A much louder sound is first heard when x = 4.5 cm. Assume that there is an antinode of a stationary wave at the open end of the pipe. Determine the frequency of the sound in the pipe. frequency = … Hz [3] (ii) After a time interval, a second much louder sound is heard. Calculate the time interval between the first louder sound and the second louder sound being heard. time interval = … s [2] [Total: 7]

7 marks

Mark scheme: 4(a)(i) distance moved by wavefront/energy during one cycle/oscillation/period (of source) or minimum distance between two wavefronts or distance between two adjacent wavefronts B1 4(a)(ii) (position where) maximum amplitude B1 4(b)(i) λ = 4 × 0.045 ( = 0.18 (m) or 18 (cm)) C1 v = fλ C1 f = 340 / 0.18 = 1900 Hz A1 4(b)(ii) distance = λ / 2 ( = 0.09 (m) or 9 (cm)) C1 time = 0.09 / 0.0075 = 12 s A1 or t = 4.5 / 0.75 and t = 13.5 / 0.75 (C1) time = 18 – 6 = 12 s (A1)

This question in 9702/22 May/June 2018

5 (a) State the relationship between the intensity and the amplitude of a wave. … … [1] (b) Microwaves of the same amplitude and wavelength are emitted in phase from two sources P and Q. The sources are arranged as shown in Fig. 5.1. P 1.840 m X 2.020 m path of detector Q Fig. 5.1 A microwave detector is moved along a path that is parallel to the line joining P and Q. A series of intensity maxima and intensity minima are detected. When the detector is at a point X, the distance PX is 1.840 m and the distance QX is 2.020 m. The microwaves have a wavelength of 6.0 cm. (i) Calculate the frequency of the microwaves. frequency = … Hz [2] (ii) Describe and explain the intensity of the microwaves detected at X. … … … … … [3] (iii) Describe the effect on the interference pattern along the path of the detector due to each of the following separate changes. 1. The wavelength of the microwaves decreases. … … 2. The phase difference between the microwaves emitted from the sources changes to 180°. … … [2] [Total: 8]

8 marks

Mark scheme: 5(a) B1 5(b)(i) v = fλ or c = fλ C1 f = 3.00 × 108 / 0.060 = 5.0 × 109 Hz A1 5(b)(ii) (at X path) difference = 3λ M1 (at X phase) difference = 0 or 1080° M1 so intensity is at a maximum/it is an intensity maximum A1 5(b)(iii) 1. decrease in the distance between (adjacent intensity) maxima/minima B1 2. (intensity) maxima and minima exchange places B1

This question in 9702/23 May/June 2018

4 (a) On Fig. 4.1, complete the two graphs to illustrate what is meant by the amplitude A, the wavelength λ and the period T of a progressive wave. Ensure that you label the axes of each graph. 0 0 Fig. 4.1 [3] (b) A horizontal string is stretched between two fixed points X and Y. A vibrator is used to oscillate the string and produce a stationary wave. Fig. 4.2 shows the string at one instant in time. string X Y Fig. 4.2 The speed of a progressive wave along the string is 30 m s–1. The stationary wave has a period of 40 ms. (i) Explain how the stationary wave is formed on the string. … … … … [2] (ii) A particle on the string oscillates with an amplitude of 13 mm. At time t, the particle has zero displacement. Calculate 1. the displacement of the particle at time (t + 100 ms), displacement = … mm 2. the total distance moved by the particle from time t to time (t + 100 ms). distance = … mm [3] (iii) Determine 1. the frequency of the wave, frequency = … Hz [1] 2. the horizontal distance from X to Y. distance = … m [3] [Total: 12]

12 marks

Mark scheme: 4(a) B1 graph with x-axis labelled ‘time’ and period/T correctly shown B1 graph with y-axis labelled ‘displacement’ and amplitude/A correctly shown B1 4(b)(i) wave (moves along string and) reflects at fixed point/Y/X/end/wall/boundary B1 the incident and reflected waves interfere/superpose B1 4(b)(ii) 100 / 40 or 2.5 (cycles/periods/T) C1 1. displacement = 0 B1 2. distance = 130 mm A1 4(b)(iii) 1. f = 1 / 40 × 10–3 = 25 Hz A1 2. v = fλ or λ = vT C1 λ = 30 / 25 or 30 × 40 × 10–3 (= 1.2 m) C1 distance = 1.2 × 1.5 = 1.8 m A1

This question in 9702/23 Oct/Nov 2018

5 (a) A loudspeaker oscillates with frequency f to produce sound waves of wavelength λ. The loudspeaker makes N oscillations in time t. (i) State expressions, in terms of some or all of the symbols f, λ and N, for: 1. the distance moved by a wavefront in time t distance = … 2. time t. time t = … [2] (ii) Use your answers in (i) to deduce the equation relating the speed v of the sound wave to f and λ. [1] (b) The waveform of a sound wave is displayed on the screen of a cathode-ray oscilloscope (c.r.o.), as shown in Fig. 5.1. 1.0 cm 1.0 cm Fig. 5.1 The time-base setting is 0.20 ms cm−1. Determine the frequency of the sound wave. frequency = … Hz [2] (c) Two sources S1 and S2 of sound waves are positioned as shown in Fig. 5.2. S1 X L Q S2 L 7.40 m Q L Y Fig. 5.2 (not to scale) The sources emit coherent sound waves of wavelength 0.85 m. A sound detector is moved parallel to the line S1S2 from a point X to a point Y. Alternate positions of maximum loudness L and minimum loudness Q are detected, as illustrated in Fig. 5.2. Distance S1X is equal to distance S2X. Distance S2Y is 7.40 m. (i) Explain what is meant by coherent waves. … … [1] (ii) State the phase difference between the two waves arriving at the position of minimum loudness Q that is closest to point X. phase difference = … ° [1] (iii) Determine the distance S1Y. distance = … m [2] [Total: 9]

9 marks

Mark scheme: 5(a)(i) B1 2. N / f B1 5(a)(ii) v (= distance / time) = Nλ / (N / f) so v = fλ B1 5(b) T = 4.0 × 0.20 = 0.80 (ms) or 8.0 × 10–4 (s) C1 f = 1 / 8.0 × 10–4 = 1300 Hz A1 5(c)(i) constant phase difference (between the waves) B1 5(c)(ii) 180° A1 5(c)(iii) path difference = 2λ or S1Y – S2Y = 2λ C1 distance = 7.40 + (0.85 × 2) = 9.1 m A1

This question in 9702/21 May/June 2019

4 (a) For a progressive water wave, state what is meant by: (i) displacement … … [1] (ii) amplitude. … … [1] (b) Two coherent waves X and Y meet at a point and superpose. The phase difference between the waves at the point is 180°. Wave X has an amplitude of 1.2 cm and intensity I. Wave Y has an amplitude of 3.6 cm. Calculate, in terms of I, the resultant intensity at the meeting point. intensity = … [2] (c) (i) Monochromatic light is incident on a diffraction grating. Describe the diffraction of the light waves as they pass through the grating. … … … [2] (ii) A parallel beam of light consists of two wavelengths 540 nm and 630 nm. The light is incident normally on a diffraction grating. Third-order diffraction maxima are produced for each of the two wavelengths. No higher orders are produced for either wavelength. Determine the smallest possible line spacing d of the diffraction grating. d = … m [3] (iii) The beam of light in (c)(ii) is replaced by a beam of blue light incident on the same diffraction grating. State and explain whether a third-order diffraction maximum is produced for this blue light. … … … [2] [Total: 11]

11 marks

Mark scheme: 4(a)(i) distance (in a specified direction of particle/point on wave) from the equilibrium position B1 4(a)(ii) the maximum distance (of particle/point on wave) from the equilibrium position or the maximum displacement (of particle/point on wave) B1 4(b) I ∝ A2 C1 IR / I = (3.6 – 1.2)2 / (1.2)2 resultant intensity = 4.0I A1 4(c)(i) as wave(s) pass through the slit(s) B1 wave(s) spread (into geometric shadow) B1 4(c)(ii) nλ = d sin θ C1 3λ = d sin 90° or 3λ = d C1 d = 3 × 630 × 10–9 = 1.9 × 10–6 m A1 4(c)(iii) wavelength of blue light is shorter (than 540 nm/630 nm/wavelengths of original light) M1 (so) third order diffraction maximum is produced A1

This question in 9702/22 May/June 2019

5 (a) State what is meant by the wavelength of a progressive wave. … … [1] (b) A cathode-ray oscilloscope (CRO) is used to analyse a sound wave. The screen of the CRO is shown in Fig. 5.1. 1 cm 1 cm Fig. 5.1 The time-base setting of the CRO is 2.5 ms cm–1. Determine the frequency of the sound wave. frequency = … Hz [2] (c) The source emitting the sound in (b) is at point A. Waves travel from the source to point C along two different paths, AC and ABC, as shown in Fig. 5.2. 20.8 m C A 8.0 m reflecting B surface Fig. 5.2 (not to scale) Distance AB is 8.0 m and distance AC is 20.8 m. Angle ABC is 90°. Assume that there is no phase change of the sound wave due to the reflection at point B. The wavelength of the waves is 1.6 m. (i) Show that the waves meeting at C have a path difference of 6.4 m. [1] (ii) Explain why an intensity maximum is detected at point C. … … … [2] (iii) Determine the difference between the times taken for the sound to travel from the source to point C along the two different paths. time difference = … s [2] (iv) The wavelength of the sound is gradually increased. Calculate the wavelength of the sound when an intensity maximum is next detected at point C. wavelength = … m [1] [Total: 9]

9 marks

Mark scheme: 5(a) distance moved by wavefront/energy during one cycle/oscillation/period (of source) or minimum distance between two wavefronts or distance between two adjacent wavefronts B1 5(b) (T =) 2.0 × 2.5 (= 5.0 ms) or 2.0 × 2.5 × 10–3 (= 5.0 × 10–3 s) C1 f = 1 / (5.0 × 10–3) = 200 Hz A1 5(c)(i) (path difference =) 8.0 + (20.82 – 8.02)0.5 – 20.8 = 6.4 (m) A1 5(c)(ii) • path difference = 4λ • waves (meet at C) in phase • constructive interference (of waves) any two points, one mark each B2 5(c)(iii) v = 200 × 1.6 v = 320 (m s–1) C1 ∆t = 6.4 / 320 or 27.2 / 320 – 20.8 / 320 = 0.020 s A1 5(c)(iv) 3λ = 6.4 3λ = 2.1 m A1

This question in 9702/22 Oct/Nov 2019

4 (a) For a progressive wave, state what is meant by: (i) the wavelength … … [1] (ii) the amplitude. … … [1] (b) A beam of red laser light is incident normally on a diffraction grating. (i) Diffraction of the light waves occurs at each slit of the grating. The light waves emerging from the slits are coherent. Explain what is meant by: 1. diffraction … … [1] 2. coherent. … … [1] (ii) The wavelength of the laser light is 650 nm. The angle between the third order diffraction maxima is 68°, as illustrated in Fig. 4.1. third order diffraction maximum laser light 68° wavelength 650 nm third order diffraction diffraction maximum grating Fig. 4.1 (not to scale) Calculate the separation d between the centres of adjacent slits of the grating. d = … m [3] (iii) The red laser light is replaced with blue laser light. State and explain the change, if any, to the angle between the third order diffraction maxima. … … … [2] [Total: 9]

9 marks

Mark scheme: 4(a)(i) distance moved by wavefront / energy during one cycle / vibration / oscillation / period (of source) or minimum distance between two wavefronts or distance between two adjacent wavefronts B1 4(a)(ii) maximum displacement (of particle / point on wave) B1 4(b)(i) 1 light / waves spread (at each slit) B1 2 constant phase difference (between light / waves) B1 4(b)(ii) nλ = d sinθ C1 d = 3 × 650 × 10–9 / sin34° C1 d = 3.5 × 10–6 m A1 4(b)(iii) wavelength of blue light is shorter (than 650 nm / red light) M1 so angle (between third order diffraction maxima) decreases A1

This question in 9702/22 Feb/March 2020

4 (a) State the difference between progressive waves and stationary waves in terms of the transfer of energy along the wave. … … [1] (b) A progressive wave travels from left to right along a stretched string. Fig. 4.1 shows part of the string at one instant. R direction of wave travel Q P string 0.48 m Fig. 4.1 P, Q and R are three different points on the string. The distance between P and R is 0.48 m. The wave has a period of 0.020 s. (i) Use Fig. 4.1 to determine the wavelength of the wave. wavelength = … m [1] (ii) Calculate the speed of the wave. speed = … m s–1 [2] (iii) Determine the phase difference between points Q and R. phase difference = … ° [1] (iv) Fig. 4.1 shows the position of the string at time t = 0. Describe how the displacement of point Q on the string varies with time from t = 0 to t = 0.010 s. … … … … [2] (c) A stationary wave is formed on a different string that is stretched between two fixed points X and Y. Fig. 4.2 shows the position of the string when each point is at its maximum displacement. W X Y Z Fig. 4.2 (i) Explain what is meant by a node of a stationary wave. … [1] (ii) State the number of antinodes of the wave shown in Fig. 4.2. number = … [1] (iii) State the phase difference between points W and Z on the string. phase difference = … ° [1] (iv) A new stationary wave is now formed on the string. The new wave has a frequency that is half of the frequency of the wave shown in Fig. 4.2. The speed of the wave is unchanged. On Fig. 4.3, draw a position of the string, for this new wave, when each point is at its maximum displacement. X Y Fig. 4.3 [1] [Total: 11]

11 marks

Mark scheme: 4(a) progressive waves transfer energy or stationary waves do not transfer energy B1 4(b)(i) 0.32 m A1 4(b)(ii) v = λ / T or v = fλ and f = 1 / T C1 v = 0.32 / 0.020 or 50 × 0.32 = 16 m s–1 A1 4(b)(iii) 450° or 90° A1 4(b)(iv) (P has) maximum downward displacement at 0.005 s B1 returns to original position/point (at 0.010 s) B1 4(c)(i) (position where) zero amplitude B1 4(c)(ii) 2 A1 4(c)(iii) 180° A1 4(c)(iv) string drawn between X and Y with one antinode midway along the string B1

This question in 9702/22 May/June 2020

4 Two progressive sound waves Y and Z meet at a fixed point P. The variation with time t of the displacement x of each wave at point P is shown in Fig. 4.1. 6 4 wave Y x / μm 2 0 0 1.0 2.0 3.0 tt // msms 4.0 –2 wave Z –4 –6 Fig. 4.1 (a) Use Fig. 4.1 to state one quantity of waves Y and Z that is: (i) the same … [1] (ii) different. … [1] (b) State and explain whether waves Y and Z are coherent. … … [1] (c) Determine the phase difference between the waves. phase difference = … ° [1] (d) The two waves superpose at P. Use Fig. 4.1 to determine the resultant displacement at time t = 0.75 ms. resultant displacement = … μm [1] (e) The intensity of wave Y at point P is I. Determine, in terms of I, the intensity of wave Z. intensity = … [2] (f) The speed of wave Z is 330 m s–1. Determine the wavelength of wave Z. wavelength = … m [3] [Total: 10]

10 marks

Mark scheme: 4(a)(i) frequency or period B1 4(a)(ii) amplitude B1 4(b) constant phase difference so coherent B1 4(c) 120° B1 4(d) resultant displacement = 4.0 μm – 1.0 μm = 3.0 μm B1 4(e) I ∝ A2 C1 intensity of Z = (22 / 42) I = 0.25 I A1 4(f) v = λ / T or v = fλ and f = 1 /T C1 330 = λ / 3.0 × 10–3 C1 λ = 0.99 m A1

This question in 9702/23 May/June 2020

5 A progressive wave Y passes a point P. The variation with time t of the displacement x for the wave at P is shown in Fig. 5.1. 6.0 4.0 x / mm 2.0 0 0 0.1 0.2 0.3 0.4 0.5 t / s –2.0 –4.0 –6.0 Fig. 5.1 The wave has a wavelength of 8.0 cm. (a) Determine the speed of the wave. speed = … m s–1 [2] (b) A second wave Z has wavelength 8.0 cm and amplitude 2.0 mm at point P. Waves Y and Z have the same speed. For the waves at point P, calculate the ratio intensity of wave Z . intensity of wave Y ratio = … [3] [Total: 5]

5 marks

Mark scheme: 5(a) v = λ / T or v = fλ and f = 1 / T v = 8.0 × 10–2 / 0.40 = 0.20 m s–1 A1 5(b) I ∝ A2 C1 ratio = 22 / 42 C1 = 0.25 A1

This question in 9702/21 Oct/Nov 2020

5 Microwaves with the same wavelength and amplitude are emitted in phase from two sources X and Y, as shown in Fig. 5.1. path of detector X A position of central maximum B position of adjacent minimum Y Fig. 5.1 (not to scale) A microwave detector is moved along a path parallel to the line joining X and Y. An interference pattern is detected. A central intensity maximum is located at point A and there is an adjacent intensity minimum at point B. The microwaves have a wavelength of 0.040 m. (a) Calculate the frequency, in GHz, of the microwaves. frequency = … GHz [3] (b) For the waves arriving at point B, determine: (i) the path difference path difference = … m [1] (ii) the phase difference. phase difference = … ° [1] (c) The amplitudes of the waves from the sources are changed. This causes a change in the amplitude of the waves arriving at point A. At this point, the amplitude of the wave arriving from source X is doubled and the amplitude of the wave arriving from source Y is also doubled. Describe the effect, if any, on the intensity of the central maximum at point A. … … … [2] (d) Describe the effect, if any, on the positions of the central intensity maximum and the adjacent intensity minimum due to the following separate changes. (i) The separation of the sources X and Y is increased. … … [1] (ii) The phase difference between the microwaves emitted by the sources X and Y changes to 180°. … … [1] [Total: 9]

9 marks

Mark scheme: 5(a) v = fλ or c = fλ C1 f = 3.0 × 108 / 0.040 C1 = 7.5 × 109 (Hz) = 7.5 GHz A1 5(b)(i) path difference = 0.020 m A1 5(b)(ii) phase difference = 180° A1 5(c) (intensity) increases C1 (intensity) increases by a factor of 4 A1 5(d)(i) minimum moves closer to the maximum or decrease in separation of maximum and minimum B1 5(d)(ii) the maximum and minimum exchange places or the maximum becomes a minimum and the minimum becomes a maximum B1

This question in 9702/22 Oct/Nov 2020

5 (a) A sound wave is detected by a microphone that is connected to a cathode-ray oscilloscope (CRO). The trace on the screen of the CRO is shown in Fig. 5.1. 1.0 cm 1.0 cm Fig. 5.1 The time-base setting of the CRO is 2.0 × 10–5 s cm–1. (i) Determine the frequency of the sound wave. frequency = … Hz [2] (ii) The intensity of the sound wave is now doubled. The frequency is unchanged. Assume that the amplitude of the trace is proportional to the amplitude of the sound wave. On Fig. 5.1, sketch the new trace shown on the screen. [2] (iii) The time-base is now switched off. Describe the trace seen on the screen. … … [1] (b) A beam of light of a single wavelength is incident normally on a diffraction grating, as illustrated in Fig. 5.2. diffraction second order grating 16° zero order 16° light beam second order Fig. 5.2 (not to scale) Fig. 5.2 does not show all of the emerging beams from the grating. The angle between the second-order emerging beam and the central zero-order beam is 16°. The grating has a line spacing of 3.4 × 10–6 m. (i) Calculate the wavelength of the light. wavelength = … m [2] (ii) Determine the highest order of emerging beam from the grating. highest order = … [2] [Total: 9]

9 marks

Mark scheme: 5(a)(i) T = 2.0 × 10–5 × 6.0 (= 1.2 × 10–4 s) C1 f = 1 / (2.0 × 10–5 × 6.0) = 8300 Hz A1 5(a)(ii) new trace shows the same period B1 new trace shows amplitude of 10 small squares B1 5(a)(iii) (trace is a) vertical line B1 5(b)(i) nλ = d sin θ C1 λ = (3.4 × 10–6 × sin 16°) / 2 = 4.7 × 10–7 m A1 5(b)(ii) n = 3.4 × 10–6 (× sin 90°) / 4.7 × 10–7 or 2 (× sin 90°) / sin 16° (= 7.2 or 7.3) C1 highest order = 7 A1

This question in 9702/23 Oct/Nov 2020

4 (a) For a progressive wave, state what is meant by wavelength. … … [1] (b) A light wave from a laser has a wavelength of 460 nm in a vacuum. Calculate the period of the wave. period = … s [3] (c) The light from the laser is incident normally on a diffraction grating. Describe the diffraction of the light waves at the grating. … … … [2] (d) A diffraction grating is used with different wavelengths of visible light. The angle θ of the fourth-order maximum from the zero-order (central) maximum is measured for each wavelength. The variation with wavelength λ of sin θ is shown in Fig. 4.1. sin θ 0 0 400 700 λ/ nm Fig. 4.1 (i) The gradient of the graph is G. Determine an expression, in terms of G, for the distance d between the centres of two adjacent slits in the diffraction grating. d = … [2] (ii) On Fig. 4.1, sketch a graph to show the results that would be obtained for the second-order maxima. [2] [Total: 10]

10 marks

Mark scheme: 4(a) distance moved by wavefront/energy during one cycle/oscillation/period (of source) or minimum distance between two wavefronts or distance between two adjacent wavefronts B1 4(b) v = λ / T or v = fλ and f = 1 / T C1 T = 460 × 10–9 / 3.00 × 108 C1 = 1.5 × 10–15 s A1 4(c) waves pass through/enter the slit(s) B1 waves spread (into geometric shadow) B1 4(d)(i) nλ = d sinθ C1 G = sinθ / λ d = 4 / G A1 4(d)(ii) straight line from 400 nm to 700 nm that is always below printed line M1 straight line has smaller gradient than printed line and is 5 small squares high at wavelength of 700 nm A1

This question in 9702/21 May/June 2021

4 (a) For a progressive wave, state what is meant by its period. … … [1] (b) State the principle of superposition. … … … [2] (c) Electromagnetic waves of wavelength 0.040 m are emitted in phase from two sources X and Y and travel in a vacuum. The arrangement of the sources is shown in Fig. 4.1. X path of detector 1.380 m Z 1.240 m Y Fig. 4.1 (not to scale) A detector moves along a path that is parallel to the line XY. A pattern of intensity maxima and minima is detected. Distance XZ is 1.380 m and distance YZ is 1.240 m. (i) State the name of the region of the electromagnetic spectrum that contains the waves from X and Y. … [1] (ii) Calculate the period, in ps, of the waves. period = … ps [3] (iii) Show that the path difference at point Z between the waves from X and Y is 3.5 λ, where λ is the wavelength of the waves. [1] (iv) Calculate the phase difference between the waves at point Z. phase difference = … ° [1] (v) The waves from X alone have the same amplitude at point Z as the waves from Y alone. State the intensity of the waves at point Z. … [1] (vi) The frequencies of the waves from X and Y are both decreased to the same lower value. The waves stay within the same region of the electromagnetic spectrum. Describe the effect of this change on the pattern of intensity maxima and minima along the path of the detector. … … [1] [Total: 11]

11 marks

Mark scheme: 4(a) time for one oscillation/vibration/cycle or time between adjacent wavefronts (passing the same point) or shortest time between two wavefronts (passing the same point) B1 4(b) (when two or more) waves meet/overlap (at a point) B1 (resultant) displacement is sum of the individual displacements B1 4(c)(i) microwave(s) B1 4(c)(ii) v = λ / T or v = fλ and f = 1/T C1 T = 0.040 / 3.00 × 108 C1 = 1.33 × 10–10 (s) = 1.33 × 10–10 / 10–12 (ps) = 130 ps A1 4(c)(iii) (1.380 – 1.240) / 0.040 = 3.5 or 1.380 / 0.040 – 1.240 / 0.040 = 3.5 A1 4(c)(iv) phase difference = 1260° or 180° A1 4(c)(v) (always) zero A1 4(c)(vi) increase in distance between (adjacent intensity) maxima/minima A1

This question in 9702/22 May/June 2021

5 (a) For a progressive wave on a stretched string, state what is meant by amplitude. … … [1] (b) Light from a laser has a wavelength of 690 nm in a vacuum. Calculate the period of the light wave. period = … s [3] (c) A two-source interference experiment uses the arrangement shown in Fig. 5.1. a D light from laser, double slit screen wavelength λ Fig. 5.1 (not to scale) Light from a laser is incident normally on a double slit. A screen is parallel to the double slit. Interference fringes are seen on the screen at distance D from the double slit. The separation of the centres of the slits is a. The light has wavelength λ. The separation x of the centres of adjacent bright fringes is measured for different values of distance D. The variation with D of x is shown in Fig. 5.2. x 0 0 D Fig. 5.2 The gradient of the graph is G. (i) Determine an expression, in terms of G and λ, for the separation a of the slits. a = … [2] (ii) The experiment is repeated with slits of separation 2a. The wavelength of the light is unchanged. On Fig. 5.2, sketch a graph to show the results of this experiment. [2] [Total: 8]

8 marks

Mark scheme: 5(a) maximum displacement (of a point/particle on string/wave) B1 5(b) v = λ / T or v = fλ and f = 1 / T C1 T = 690 × 10–9 / 3.00 × 108 C1 = 2.3 × 10–15 s A1 5(c)(i) λ = ax / D C1 G = x / D (so) a = λ / G A1 5(c)(ii) straight line from origin always below printed line M1 line is half the height of printed line at maximum D A1

This question in 9702/23 Oct/Nov 2021

5 A horizontal string is stretched between two fixed points A and B. A vibrator is used to oscillate the string and produce an observable stationary wave. At one instant, the moving string is straight, as shown in Fig. 5.1. point P A B Fig. 5.1 The dots in the diagram represent the positions of the nodes on the string. Point P on the string is moving downwards. The wave on the string has a speed of 35 m s–1 and a period of 0.040 s. (a) Explain how the stationary wave is formed on the string. … … … … [2] (b) On Fig. 5.1, sketch a line to show a possible position of the string a quarter of a cycle later than the position shown in the diagram. [1] (c) Determine the horizontal distance from A to B. distance = … m [3] (d) A particle on the string has zero displacement at time t = 0. From time t = 0 to time t = 0.060 s, the particle moves through a total distance of 72 mm. Calculate the amplitude of oscillation of the particle. amplitude = … mm [2] [Total: 8]

8 marks

Mark scheme: 5(a) wave(s) (travel along string and) reflect at fixed point / A / B / end B1 incident and reflected waves superpose / interfere or two waves travelling / with speed in opposite directions superpose / interfere B1 5(b) line has an approximate sinusoidal shape with maximum downward displacement at P and zero displacement at each node B1 5(c) v =  / T or v = f and f = 1 / T C1  = 35  0.040 or 35 / 25 ( = 1.4 m) C1 distance = 1.4  2.5 = 3.5 m A1 5(d) (number of periods / cycles) (= t / T) = 0.060 / 0.040 ( = 1.5) C1 amplitude = 72 / 6 = 12 mm A1

This question in 9702/21 May/June 2022

5 Light from a laser is used to produce an interference pattern on a screen, as shown in Fig. 5.1. 0.44 mm O central bright fringe P dark fringe Q bright fringe light of R dark fringe wavelength 660 nm 1.8 m double screen slit Fig. 5.1 (not to scale) The light of wavelength 660 nm is incident normally on two slits that have a separation of 0.44 mm. The double slit is parallel to the screen. The perpendicular distance between the double slit and the screen is 1.8 m. The central bright fringe on the screen is formed at point O. The next dark fringe below point O is formed at point P. The next bright fringe and the next dark fringe below point P are formed at points Q and R respectively. (a) The light waves from the two slits are coherent. State what is meant by coherent. … … [1] (b) For the two light waves superposing at R, calculate: (i) the difference in their path lengths, in nm, from the slits path difference = … nm [1] (ii) their phase difference. phase difference = … ° [1] (c) Calculate the distance OQ. distance OQ = … m [3] (d) The intensity of the light incident on the double slit is increased without changing the frequency. Describe how the appearance of the fringes after this change is different from, and similar to, their appearance before the change. … … … … … [3] (e) The light of wavelength 660 nm is now replaced by blue light from a laser. State and explain the change, if any, that must be made to the separation of the two slits so that the fringe separation on the screen is the same as it was for light of wavelength 660 nm. … … … … [2] [Total: 11]

11 marks

Mark scheme: 5(a) constant phase difference (between the waves) B1 5(b)(i) path difference = 1.5  660 = 990 nm A1 5(b)(ii) phase difference = 360°  1.5 = 540° A1 5(c)  = ax / D C1 x = (660  10–9  1.8) / 0.44  10–3 C1 = 2.7  10–3 m A1 5(d) bright fringes are brighter B1 no change to dark fringes B1 no change to (fringe) separation / (fringe) spacing B1 5(e) (blue light has) shorter wavelength M1 (so) decrease (slit) separation A1

This question in 9702/22 May/June 2022

4 (a) A progressive longitudinal wave travels through a medium from left to right. Fig. 4.1 shows the positions of some of the particles of the medium at time t0 and a graph showing the particle displacements at the same time t0. direction of wave travel X Y Z displacement 0 distance Fig. 4.1 Particle displacements to the right of their equilibrium positions are shown as positive on the graph and particle displacements to the left are shown as negative on the graph. The period of the wave is T. (i) On Fig. 4.1, draw circles around two particles which are exactly one wavelength apart. [1] (ii) On Fig. 4.1, sketch a line on the graph to represent the displacements of the particles for T the longitudinal wave at time t0 + . [3] 4 T (iii) State the direction of motion of particle Z at time t0 + . 4 … [1] (b) The frequency of the wave in (a) is 16 kHz. The distance between particles X and Y is 0.19 m. Calculate the speed of the wave as it travels through the medium. speed = … m s–1 [3] (c) A longitudinal sound wave is travelling through a solid. The initial intensity of the wave is I0. The frequency of the wave remains constant and the amplitude falls to half of its original value. Determine, in terms of I0, the final intensity of the wave. intensity = … I0 [2] (d) The sound wave in (c) now meets another sound wave travelling in the opposite direction. (i) State a condition necessary for these two waves to form a stationary wave. … [1] (ii) State two ways in which a stationary wave differs from a progressive wave. 1 … … 2 … … [2] [Total: 13]

13 marks

Mark scheme: 4(a)(i) circles drawn around any two particles with seven (uncircled) particles in between A1 4(a)(ii) curve has an initial negative displacement and initial amplitude same as original curve B1 curve has same amplitude as original curve throughout B1 curve has same wavelength as original curve throughout, with constant (non-zero) phase difference B1 4(a)(iii) (to the) right / rightwards A1 4(b)  = 2  0.19 C1 = 0.38 m v = f C1 v = (16  103)  0.38 A1 v = 6100 m s–1 4(c) I  A2 C1 I = (1 / 2)2 I0 A1 intensity = 0.25 I0 4(d)(i) same frequency / wavelength / period B1 4(d)(ii) • a stationary wave has nodes/antinodes (and a progressive wave does not) B2 • a stationary wave does not transfer/propagate energy (and a progressive wave does transfer/propagate energy) • different points on a stationary wave have different amplitudes (and all points on a progressive wave have the same/constant amplitude) • stationary wave has adjacent particles that are in phase (and adjacent particles on progressive wave are out of phase) Any two points, 1 mark each. Allow the reverse statement for each marking point.

This question in 9702/23 Oct/Nov 2022

5 (a) A microphone and cathode-ray oscilloscope (CRO) are used to analyse a sound wave of frequency 5000 Hz. The trace that is displayed on the screen of the CRO is shown in Fig. 5.1. 1.0 cm 1.0 cm Fig. 5.1 (i) Determine the time-base setting, in s cm–1, of the CRO. time-base setting = … s cm–1 [2] (ii) The intensity of the sound detected by the microphone is now increased from its initial value of I to a new value of 3I. The frequency of the sound is unchanged. Assume that the amplitude of the trace on the CRO screen is proportional to the amplitude of the sound wave. On Fig. 5.1, sketch the new trace shown on the screen of the CRO. [3] (b) An arrangement for demonstrating interference using light is shown in Fig. 5.2. 3.6 × 10–4 m P light from laser, wavelength 630 nm D double slit screen Fig. 5.2 (not to scale) The wavelength of the light from the laser is 630 nm. The light is incident normally on the double slit. The separation of the two slits is 3.6 × 10–4 m. The perpendicular distance between the double slit and the screen is D. Coherent light waves from the slits form an interference pattern of bright and dark fringes on the screen. The distance between the centres of two adjacent bright fringes is 4.0 × 10–3 m. The central bright fringe is formed at point P. (i) Explain why a bright fringe is produced by the waves meeting at point P. … … [1] (ii) Calculate distance D. D = … m [3] (c) The wavelength λ of the light in (b) is now varied. This causes a variation in the distance x between the centres of two adjacent bright fringes on the screen. The distance D and the separation of the two slits are unchanged. On Fig. 5.3, sketch a graph to show the variation of x with λ from λ = 400 nm to λ = 700 nm. Numerical values of x are not required. x 0 400 700 λ/ nm Fig. 5.3 [1] [Total: 10]

10 marks

Mark scheme: 5(a)(i) period or T = 1 / 5000 (= 2  10–4 s) C1 time-base setting = 1.5  2  10–4 / 6.0 or 2  10–4 / 4.0 A1 = 5  10–5 s cm–1 5(a)(ii) new trace drawn with same period as original trace B1 new trace drawn with amplitude greater than 1.0 cm M1 new trace drawn with amplitude of 1.7 cm A1 5(b)(i) path difference (from slits to P) is zero or phase difference (between waves at P) is zero (so constructive interference) B1 5(b)(ii) = ax / D C1 D = (3.6  10–4  4.0  10–3) / 630  10–9 C1 = 2.3 m A1 5(c) upward sloping straight line starting from a non-zero value of x at = 400 nm B1

This question in 9702/22 Feb/March 2023

4 (a) For a progressive wave, state what is meant by the frequency. … … [1] (b) A loudspeaker, microphone and cathode-ray oscilloscope (CRO) are arranged as shown in Fig. 4.1. microphone loudspeaker CRO Fig. 4.1 The loudspeaker is emitting a sound wave which is detected by the microphone and displayed on the screen of the CRO as shown in Fig. 4.2. 1.0 cm 1.0 cm Fig. 4.2 The time-base on the CRO is set to 0.50 ms cm−1 and the y-gain is set to 0.20 V cm−1. Calculate: (i) the frequency of the sound wave frequency = … Hz [2] (ii) the amplitude of the signal received by the CRO. amplitude = … V [1] (c) The intensity of the sound wave in (b) is reduced to a quarter of its original intensity without a change in frequency. Assume that the amplitude of the signal received by the CRO is proportional to the amplitude of the sound wave. On Fig. 4.2, sketch the trace that is now seen on the screen of the CRO. [3] (d) A metal sheet is now placed in front of the loudspeaker in (b), as shown in Fig. 4.3. microphone metal sheet loudspeaker CRO Fig. 4.3 A stationary wave is formed between the loudspeaker and the metal sheet. (i) State the principle of superposition. … … … [2] (ii) The initial position of the microphone is such that the trace on the CRO has an amplitude minimum. It is now moved a distance of 1.05 m away from the loudspeaker along the line joining the loudspeaker and metal sheet. As the microphone moves, it passes through three positions where the trace has an amplitude maximum before ending at a position where the trace has an amplitude minimum. Determine the wavelength of the sound wave. wavelength = … m [2] (iii) Use your answers in (b)(i) and (d)(ii) to determine the speed of the sound in the air. speed = … m s−1 [2] [Total: 13]

13 marks

Mark scheme: 4(a) the number of wavefronts/crests/troughs passing a fixed point per unit time or the number of oscillations per unit time (of source / point on wave / particle of medium) B1 4(b)(i) T = 4  0.50  10–3 ( = 2.0  10–3 s) C1 f = 1 / 2.0  10–3 = 500 Hz A1 4(b)(ii) amplitude = 2.8  0.20 = 0.56 V A1 4(c) period same as original trace B1 sinusoidal wave of constant amplitude less than 2.8 cm throughout M1 amplitude 1.4 cm A1 Question Answer Marks 4(d)(i) when (two or more) waves meet (at a point) B1 (resultant) displacement is the sum of the individual displacements B1 4(d)(ii) node-to-node separation is  / 2 or microphone moves through 3 node-to-node separations or d = 1.5 C1  = 1.05 / 1.5 = 0.70 m A1 4(d)(iii) v = f   C1 = 500  0.70 = 350 m s–1 A1

This question in 9702/23 May/June 2023

6 A train travels at constant speed along a straight horizontal track towards an observer standing adjacent to the track, as shown in Fig. 6.1. train observer track Fig. 6.1 The train sounds its horn continuously as it approaches the observer. The horn emits a sound of constant frequency 251 Hz. The frequency of sound heard by the observer is 291 Hz. The speed of sound in air is 340 m s–1. (a) Calculate the speed of the train. speed = … m s–1 [2] (b) The train approaches and then passes the observer. The intensity I of the sound heard by the observer varies with the distance d of the horn from the observer. When the horn is at a distance x0 from the observer, the intensity I of the sound heard is I0 and the amplitude A of the sound wave at the observer is A0. Fig. 6.2 shows the variation with d / x0 of I / I0 as the train moves away from the observer. 1.0 1.0 I / I0 A / A0 0.8 0.8 0.6 0.6 0.4 0.4 0.2 0.2 0 0 1 2 3 4 1 2 3 4 d / x0 d / x0 Fig. 6.2 Fig. 6.3 (i) State the relationship between amplitude A and intensity I for a progressive wave. … [1] (ii) On Fig. 6.3, sketch the variation with d / x0 of A / A0. [2] [Total: 5]

5 marks

Mark scheme: 6(a) 291 = 251  340 / (340–v(s)) C1 vs = 47 m s–1 A1 6(b)(i) I  A2 B1 6(b)(ii) sketch: line starts at (1.0, 1.0) B1 approximately straight line drawn with negative gradient and line ends at (4.0, 0.25) B1

This question in 9702/23 Oct/Nov 2023

5 (a) By reference to the direction of propagation of energy, state what is meant by a transverse wave. … … [1] (b) A space telescope is designed to detect electromagnetic radiation with wavelengths in the range 12 μm to 28 μm. State the region of the electromagnetic spectrum for this radiation. … [1] (c) A detector on another space telescope detects an electromagnetic wave. The signal from the detector is transmitted to Earth and displayed on an oscilloscope as shown in Fig. 5.1. The frequency of the signal displayed on the oscilloscope is equal to the frequency of the detected electromagnetic wave. 1.0 cm 1.0 cm Fig. 5.1 The time-base setting on the oscilloscope is 5.0 × 10–15 s cm–1. Calculate the wavelength of the detected electromagnetic wave. wavelength = … m [3] [Total: 5]

5 marks

Mark scheme: 5(a) vibrations / oscillations (of the particles / wave) are perpendicular to the direction (of the propagation of energy) B1 5(b) infrared B1 5(c) T = 6  5.0  10–15 C1 T = 3.0  10–14  = c T or  = c / f and f = 1 / T C1  = 3.0  108  3.0  10-14 or  = 3.0  108 / 3.33  1013 A1 = 9.0  10–6 m

This question in 9702/22 Feb/March 2024

4 A progressive transverse wave travelling from left to right is shown at an instant in time in Fig. 4.1. R wave direction of travel T Fig. 4.1 R and T are points on the wave. (a) State the phase difference between the points R and T. phase difference = … ° [1] (b) On Fig. 4.1, draw an arrow at point T to show the direction of movement of point T at the instant shown. [1] (c) The horizontal distance between R and T is 0.62 cm, as shown in Fig. 4.2. 0.62 cm R T Fig. 4.2 (not to scale) The speed of the wave is 0.27 m s–1. Calculate the frequency of the wave. frequency = … Hz [3] (d) The wave is a water wave produced by a dipper S1 attached to a vibrator in a ripple tank. An identical dipper S2 is attached to the same vibrator. The two dippers produce an interference pattern on the water in the tank, as shown in Fig. 4.3. water P wave crests wave troughs S1 S2 Fig. 4.3 (not to scale) The wave crests from each source are represented by solid lines on Fig. 4.3 and the wave troughs are represented by dashed lines. At point P in Fig. 4.3, the wave from S1 has the same amplitude A as the wave from S2. Describe and explain the amplitude of the resultant wave at point P. … … … … [3] [Total: 8]

8 marks

Mark scheme: 4(a) 270° A1 4(b) arrow pointing vertically downwards at T A1 4(c) v = f or v =  / T and f = 1 / T C1 wavelength = 0.62  10–2  (4 / 3) ( = 0.83  10–2 m) C1 f = 0.27 / (0.83  10–2) = 33 Hz A1 4(d) resultant displacement is the sum of the displacements of the waves (from S1 and S2) or waves (from S1 and S2) superpose (at P) B1 Any one point from:  (at P) the waves (from the two sources) (always) destructively interfere  (at P) the waves have a path difference that is (always) an odd number of half-wavelengths / their path difference is one and a half wavelengths  (at P) the waves have a phase difference that is (always) 180° / they are in antiphase / crest of one wave meets trough of other wave B1 amplitude (of the resultant wave) is zero (at all times) B1

This question in 9702/23 May/June 2024

4 (a) With reference to the direction of transfer of energy, compare the oscillations of transverse and longitudinal progressive waves. … … … … [2] (b) A pipe is open at one end and closed at the other with a piston. The piston can slide freely and is at a distance of 4.5 × 10–2 m from the open end of the pipe. A loudspeaker is positioned near the open end of the pipe and emits a sound wave of a single constant frequency. A stationary wave is formed in the pipe, as illustrated in Fig. 4.1. stationary wave pipe loudspeaker piston 4.5 × 10–2 m Fig. 4.1 (i) On Fig. 4.1, draw a letter A at the position of an antinode. [1] (ii) The speed of sound in air is 340 m s–1. Determine the frequency of the sound wave. frequency = … Hz [3] (iii) The piston is moved to the left. The frequency of the sound wave emitted by the loudspeaker is then changed so that a stationary wave is formed with same number of antinodes as in Fig. 4.1. State and explain the change that is made to the frequency of the sound wave. … … … … [2] [Total: 8]

8 marks

Mark scheme: 4(a) longitudinal waves have oscillations parallel to the (direction of) transfer of energy B1 transverse waves have oscillations perpendicular to the (direction of) transfer of energy B1 4(b)(i) A marked at the open end of the pipe B1 4(b)(ii) f = v /  C1 = 4  4.5  10–2 C1 f = 340 / (4  4.5  10–2) A1 = 1900 Hz 4(b)(iii) the node–antinode distance is longer M1 or the wavelength (of the wave) is longer (the speed of sound is constant so) the frequency (of the wave) is lower A1

This question in 9702/22 Oct/Nov 2024

5 A stationary loudspeaker emits sound of constant frequency. A microphone is placed near to the loudspeaker and connected to a cathode-ray oscilloscope (CRO). The trace on the screen of the CRO is shown in Fig. 5.1. 1 cm 1 cm Fig. 5.1 The time-base of the CRO is set to 5.0 × 10– 4 s cm–1. (a) The speed of the sound emitted by the loudspeaker is 330 m s–1. Determine the wavelength of the sound. wavelength = … m [3] (b) The loudspeaker now moves in a straight line while emitting the same sound of constant frequency. The period of the trace on the CRO increases continuously. Describe the motion of the loudspeaker. … … … [2] [Total: 5]

5 marks

Mark scheme: 5(a) T = 5.8  5.0  10–4 C1 = 2.9  10–3  = vT or v = f and f = 1 / T C1  = 330  2.9  10–3 or = 330 / 345 A1 = 0.96 m 5(b) (loudspeaker) moves away (from the microphone) B1 at an increasing speed / whilst accelerating B1

This question in 9702/22 Feb/March 2025

3 Two progressive water waves X and Y travel along a straight line from point A to point B. The variation of displacement of the waves with distance from A at an instant in time is shown in Fig. 3.1. 20 displacement / cm 10 wave X 0 0 0.2 0.4 0.6 0.8 1.0 distance from A / m –10 wave Y –20 Fig. 3.1 (a) State the amplitude of wave X. amplitude = … cm [1] (b) Both waves have frequency 16 Hz. (i) Determine the speed of wave X. speed = … m s–1 [2] (ii) State and explain whether X and Y are coherent. … … … [1] (c) Wave X and wave Y superpose to form a resultant wave. On Fig. 3.2, sketch the variation of displacement of the resultant wave with distance from A at the instant of time shown in Fig. 3.1. 20 displacement / cm 10 0 0 0.2 0.4 0.6 0.8 1.0 distance from A / m –10 –20 Fig. 3.2 [2] (d) The intensity of wave X is IX. The intensity of wave Y is IY. IX Use Fig. 3.1 to determine the ratio . IY ratio = … [2] [Total: 8]

8 marks

Mark scheme: 3(a) 10.0 cm A1 3(b)(i) v = f C1 = 16  0.40 = 6.4 m s–1 A1 3(b)(ii) (X and Y have a) constant phase difference (of 180°) so (they are) coherent B1 3(c) A single wave of amplitude 10.0 cm B1 A single negative sine wave of wavelength 0.40 m B1 3(d) I  A2 C1 I X = 102 / 202 IY ratio = 0.25 A1

This question in 9702/22 May/June 2025