Stationary waves — Paper 2 · A Level Physics 9702

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 Light reflected from the surface of smooth water may be described as a polarised transverse wave. (a) By reference to the direction of propagation of energy, explain what is meant by (i) a transverse wave, … … [1] (ii) polarisation. … … [1] (b) A glass tube, closed at one end, has fine dust sprinkled along its length. A sound source is placed near the open end of the tube, as shown in Fig. 5.1. dust heap tube sound source 39.0 cm Fig. 5.1 The frequency of the sound emitted by the source is varied and, at one frequency, the dust forms small heaps in the tube. (i) Explain, by reference to the properties of stationary waves, why the heaps of dust are formed. … … … … [3] Examiner’s Use (ii) One frequency at which heaps are formed is 2.14 kHz. The distance between six heaps, as shown in Fig. 5.1, is 39.0 cm. Calculate the speed of sound in the tube. speed = … m s–1 [3] (c) The wave in the tube is a stationary wave. Explain, by reference to the formation of a stationary wave, what is meant by the speed calculated in (b)(ii). … … … … [3]

11 marks

Mark scheme: 5 (a) (i) vibrations (in plane) normal to direction of energy propagation B1 [1] (ii) vibrations in one direction (normal to direction of propagation) B1 [1] (b) (i) at (displacement) antinodes / where there are no heaps, wave has maximum amplitude (of vibration) B1 at (displacement) nodes/where there are heaps, amplitude of vibration is zero/minimum B1 dust is pushed to / settles at (displacement) nodes B1 [3] (ii) 2.5λ = 39 cm C1 v = fλ C1 v = 2.14 × 103 × 15.6 × 10-2 = 334 m s-1 (allow 330, not 340) A1 [3] (c) Stationary wave formed by interference / superposition / overlap of B1 either wave travelling down tube and its reflection or two waves of same (type and) frequency travelling in opposite directions B1 speed is the speed of the incident / reflected waves B1 [3] GCE A/AS LEVEL – May/June 2007 9702 2

This question in 9702/21 May/June 2007

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 uniform string is held between a fixed point P and a variable-frequency oscillator, as shown For in Fig. 5.1. Examiner’s Use L 1 8 L X P Y oscillator 1 8 L Fig. 5.1 The distance between point P and the oscillator is L. The frequency of the oscillator is adjusted so that the stationary wave shown in Fig. 5.1 is formed. Points X and Y are two points on the string. 1 Point X is a distance 8L from the end of the string attached to the oscillator. It vibrates with frequency f and amplitude A. 1 Point Y is a distance 8L from the end P of the string. (a) For the vibrations of point Y, state (i) the frequency (in terms of f ), frequency = … [1] (ii) the amplitude (in terms of A). amplitude = … [1] (b) State the phase difference between the vibrations of point X and point Y. phase difference = … [1] (c) (i) State, in terms of f and L, the speed of the wave on the string. For Examiner’s speed = … [1] Use (ii) The wave on the string is a stationary wave. Explain, by reference to the formation of a stationary wave, what is meant by the speed stated in (i). … … … … … [3]

7 marks

Mark scheme: 5 (a) (i) frequency f … B1 [1] (ii) amplitude A … B1 [1] (b) π rad or 180° ………(unit necessary) … B1 [1] (c) (i) speed = f × L … B1 [1] (ii) wave is reflected at end / at P … B1 either incident and reflected waves interfere or two waves travelling in opposite directions interfere … M1 speed is the speed of incident or reflected wave / one of these waves … A1 [3] [Total: 7]

This question in 9702/22 Oct/Nov 2009

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

6 (a) Use the principle of superposition to explain the formation of a stationary wave. For Examiner’s … Use … … … … [3] (b) Describe an experiment to determine the wavelength of sound in air using stationary waves. Include a diagram of the apparatus in your answer. … … … … … … … … … … [3] (c) The variation with distance x of the intensity I of a stationary sound wave is shown in For Fig. 6.1. Examiner’s Use 1.0 I / arbitrary units 0.5 00 20 40 60 x / cm Fig. 6.1 (i) On the x-axis of Fig. 6.1, indicate the positions of all the nodes and antinodes of the stationary wave. Label the nodes N and the antinodes A. [1] (ii) The speed of sound in air is 340 m s–1. Use Fig. 6.1 to determine the frequency of the sound wave. frequency = … Hz [3] Please turn over for Question 7.

10 marks

Mark scheme: 6 (a) two waves travelling (along the same line) in opposite directions overlap/meet M1 same frequency / wavelength A1 resultant displacement is the sum of displacements of each wave / produces nodes and antinodes B1 [3] (b) apparatus: source of sound + detector + reflection system B1 adjustment to apparatus to set up standing waves – how recognised B1 measurements made to obtain wavelength B1 [3] (c) (i) at least two nodes and two antinodes A1 [1] (ii) node to node = λ / 2 = 34 cm (allow 33 to 35 cm) C1 c = fλ C1 f = 340 / 0.68 = 500 (490 to 520) Hz A1 [3] GCE AS/A LEVEL – May/June 2012 9702 22

This question in 9702/22 May/June 2012

4 Fig. 4.1 shows an arrangement for producing stationary waves in a tube that is closed at one For end. Examiner’s Use signal generator loudspeaker tube Fig. 4.1 (a) Explain how waves from the loudspeaker produce stationary waves in the tube. … … … … … [3] (b) One of the stationary waves that may be formed in the tube is represented in Fig. 4.2. P S Fig. 4.2 (i) Describe the motion of the air particles in the tube at 1. point P, … [1] 2. point S. … [1] (ii) The speed of sound in the tube is 330 m s–1 and the frequency of the waves from the loudspeaker is 880 Hz. Calculate the length of the tube. length = … m [3]

8 marks

Mark scheme: 4 (a) waves (travels along tube) reflect at closed end / end of tube B1 incident and reflected waves or these two waves are in opposite directions M1 interfere or stationary wave formed if tube length equivalent to λ / 4, 3λ / 4, etc. A1 [3] (b) (i) 1. no motion (as node) / zero amplitude B1 [1] 2. vibration backwards and forwards / maximum amplitude along length B1 [1] (ii) λ = 330 / 880 (= 0.375 m) C1 L = 3λ / 4 C1 L = 3 / 4 × (0.375) = 0.28 (0.281) m A1 [3]

This question in 9702/22 Oct/Nov 2012

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

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

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 An arrangement for producing stationary waves in air in a tube that is closed at one end is shown in Fig. 6.1. loudspeaker signal generator tube of adjustable L length air Fig. 6.1 A loudspeaker produces sound waves of wavelength 0.680 m in the tube. For some values of the length L of the tube, stationary waves are formed. (a) Explain how stationary waves are formed in the tube. … … … [2] (b) The length L is adjusted between 0.200 m and 1.00 m. (i) Calculate two values of L for which stationary waves are formed. L = … m and L = … m [2] (ii) On Fig. 6.2, label the positions of the antinodes with an A and the nodes with an N for the least value of L for which a stationary wave is formed. Fig. 6.2 [1]

5 marks

Mark scheme: 6 (a) waves from loudspeaker (travel down tube and) are reflected at closed end B1 two waves (travelling) in opposite directions with same frequency/wavelength overlap B1 [2] (b) (i) 0.51 m A1 0.85 m A1 [2] (ii) A at open end, N at closed end, with an N and A in between, equally spaced (by eye) B1 [1]

This question in 9702/23 Oct/Nov 2015

4 (a) State the conditions required for the formation of stationary waves. … … … [2] (b) One end of a string is attached to a vibrator. The string is stretched by passing the other end over a pulley and attaching a load, as illustrated in Fig. 4.1. string pulley A B vibrator support for pulley load Fig. 4.1 The frequency of vibration of the vibrator is adjusted to 250 Hz and a transverse wave travels along the string with a speed of 12 m s–1. The wave is reflected at the pulley and a stationary wave forms on the string. Fig. 4.2 shows the string between points A and B at time t = t1. string A B Fig. 4.2 At time t = t1 the string has maximum displacement. (i) Calculate the distance AB. distance = … m [2] (ii) On Fig. 4.2, sketch the position of the string between A and B at times 1. t = t1 + 2.0 ms (label this line P), 2. t = t1 + 5.0 ms (label this line Q). [3] [Total: 7]

7 marks

Mark scheme: 4(a) (two) waves travelling (at same speed) in opposite directions overlap B1 waves (are same type and) have same frequency/wavelength B1 4(b)(i) λ = 12 / 250 (= 0.048 m) C1 distance = 1.5 × 0.048 = 0.072 m A1 4(b)(ii) T = 1 / 250 = 0.004 (s) or 4 (ms) C1 1. curve drawn is mirror image of that in Fig. 4.2 and labelled P A1 2. horizontal line drawn between A and B and labelled Q A1

This question in 9702/21 May/June 2017

3 (a) State the difference between a stationary wave and a progressive wave in terms of (i) the energy transfer along the wave, … … [1] (ii) the phase of two adjacent vibrating particles. … … [1] (b) A tube is open at both ends. A loudspeaker, emitting sound of a single frequency, is placed near one end of the tube, as shown in Fig. 3.1. tube A A A A loudspeaker 0.60 m Fig. 3.1 The speed of the sound in the tube is 340 m s–1. The length of the tube is 0.60 m. A stationary wave is formed with an antinode A at each end of the tube and two antinodes inside the tube. (i) State what is meant by an antinode of the stationary wave. … … [1] (ii) State the distance between a node and an adjacent antinode. distance = … m [1] (iii) Determine, for the sound in the tube, 1. the wavelength, wavelength = … m [1] 2. the frequency. frequency = … Hz [2] (iv) Determine the minimum frequency of the sound from the loudspeaker that produces a stationary wave in the tube. minimum frequency = … Hz [2] [Total: 9]

9 marks

Mark scheme: 3(a)(i) in a stationary wave energy is not transferred or in a progressive wave energy is transferred B1 3(a)(ii) in a stationary wave (adjacent) particles are in phase or in a progressive wave (adjacent) particles are out of phase/have a phase difference/not in phase B1 3(b)(i) (position where) maximum amplitude B1 3(b)(ii) distance = 0.10 m B1 3(b)(iii) 1. λ = 0.60 / 1.5 = 0.40 m A1 2. v = fλ C1 f = 340 / 0.40 = 850 Hz A1 3(b)(iv) λ = 2 × 0.60 or λ = 3 × 0.40 or f = 850 / 3 C1 f = 280 (283) Hz A1

This question in 9702/21 Oct/Nov 2017

4 (a) State the conditions required for the formation of a stationary wave. … … … … [2] (b) A horizontal string is stretched between two fixed points X and Y. The string is made to vibrate vertically so that a stationary wave is formed. At one instant, each particle of the string is at its maximum displacement, as shown in Fig. 4.1. string Q X Y P 2.0 m Fig. 4.1 P and Q are two particles of the string. The string vibrates with a frequency of 40 Hz. Distance XY is 2.0 m. (i) State the number of antinodes in the stationary wave. number = … [1] (ii) Determine the minimum time taken for the particle P to travel from its lowest point to its highest point. time taken = … s [2] (iii) State the phase difference, with its unit, between the vibrations of particle P and of particle Q. phase difference = … [1] (iv) Determine the speed of a progressive wave along the string. speed = … m s–1 [2] [Total: 8]

8 marks

Mark scheme: 4(a) B1 waves (are same type and) have same frequency/wavelength B1 4(b)(i) 5 A1 4(b)(ii) T = 1 / 40 (= 2.5 × 10–2) C1 time taken = 2.5 × 10–2 / 2 = 1.3 × 10–2 s (1.25 × 10–2 s) A1 4(b)(iii) 180° A1 4(b)(iv) v = fλ C1 λ = 2.0 / 2.5 (= 0.80 m) v = 0.80 × 40 = 32 m s–1 A1

This question in 9702/22 Oct/Nov 2017

4 (a) State the conditions required for the formation of a stationary wave. … … … … [2] (b) The sound from a loudspeaker is detected by a microphone that is connected to a cathode-ray oscilloscope (c.r.o.). Fig. 4.1 shows the trace on the screen of the c.r.o. 1 cm 1 cm Fig. 4.1 In air, the sound wave has a speed of 330 m s–1 and a wavelength of 0.18 m. (i) Calculate the frequency of the sound wave. frequency = … Hz [2] (ii) Determine the time-base setting, in s cm–1, of the c.r.o. time-base setting = … s cm–1 [2] (iii) The intensity of the sound from the loudspeaker is now halved. The wavelength of the sound is unchanged. Assume that the amplitude of the trace is proportional to the amplitude of the sound wave. On Fig. 4.1, sketch the new trace shown on the screen of the c.r.o. [2] (c) The loudspeaker in (b) is held above a vertical tube of liquid, as shown in Fig. 4.2. loudspeaker liquid level A level A tube level B level B liquid tap Fig. 4.2 Fig. 4.3 A tap at the bottom of the tube is opened so that liquid drains out at a constant rate. The wavelength of the sound from the loudspeaker is 0.18 m. The sound that is heard first becomes much louder when the liquid surface reaches level A. The next time that the sound becomes much louder is when the liquid surface reaches level B, as shown in Fig. 4.3. (i) Calculate the vertical distance between level A and level B. distance = … m [1] (ii) On Fig. 4.3, label with the letter N the positions of the nodes of the stationary wave that is formed in the air column when the liquid surface is at level B. [1] (iii) The mass of liquid leaving the tube per unit time is 6.7 g s–1. The tube has an internal cross-sectional area of 13 cm2. The density of the liquid is 0.79 g cm–3. Calculate the time taken for the liquid to move from level A to level B. time = … s [2] [Total: 12]

12 marks

Mark scheme: 4(a) (two) waves (travelling at same speed) in opposite directions overlap B1 (waves are same type and) have same frequency / wavelength B1 4(b)(i) v = fλ f = 330 / 0.18 C1 = 1800 Hz (1830 Hz) A1 4(b)(ii) T = 1 / 1800 (= 5.5 × 10–4) time-base setting = (1.5 × 5.5 ×10–4) / 8.0 or 1 / (1800 × 5.3) C1 = 1.0 × 10–4 s cm–1 A1 4(b)(iii) waveform drawn with same period as original waveform B1 waveform drawn with amplitude of 1.7 cm B1 4(c)(i) distance = λ / 2 = 0.18 / 2 = 0.090 m A1 Question Answer Marks 4(c)(ii) letter N shown at level B and at level A and not anywhere else. B1 4(c)(iii) m = ρAx = 0.79 × 13 × 9.0 (=92.4) or 790 × 13×10–4 × 0.090 (=0.0924) t = 92.4 / 6.7 or 0.0924 / 0.0067 C1 = 14 s A1

This question in 9702/22 Feb/March 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

4 (a) Sound waves are longitudinal waves. By reference to the direction of propagation of energy, state what is meant by a longitudinal wave. … … [1] (b) A stationary sound wave in air has amplitude A. In an experiment, a detector is used to determine A2. The variation of A2 with distance x along the wave is shown in Fig. 4.1. 4.0 3.0 A2 / arbitrary units 2.0 1.0 0 0 10 20 30 40 50 60 x / cm Fig. 4.1 (i) State the phase difference between the vibrations of an air particle at x = 25 cm and the vibrations of an air particle at x = 50 cm. phase difference = … ° [1] (ii) The speed of the sound in the air is 330 m s–1. Determine the frequency of the sound wave. frequency = … Hz [3] (iii) Determine the ratio amplitude A of wave at x = 20 cm . amplitude A of wave at x = 25 cm ratio = … [2]

7 marks

Mark scheme: 4(a) vibration(s)/oscillation(s) (of particles) parallel to direction of propagation of energy B1 4(b)(i) phase difference = 180° A1 4(b)(ii) v = fλ C1 λ / 2 = 25 (cm) or 0.25 (m) C1 f = 330 / 0.50 = 660 Hz A1 4(b)(iii) (readings from graph =) 2.6 and 4.0 C1 ratio = (2.6 / 4.0)1/2 = 0.81 A1

This question in 9702/22 Oct/Nov 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 vertical tube of length 0.60 m is open at both ends, as shown in Fig. 5.1. A tube N 0.60 m A direction of incident sound wave Fig. 5.1 An incident sinusoidal sound wave of a single frequency travels up the tube. A stationary wave is then formed in the air column in the tube with antinodes A at both ends and a node N at the midpoint. (a) Explain how the stationary wave is formed from the incident sound wave. … … … … [2] (b) On Fig. 5.2, sketch a graph to show the variation of the amplitude of the stationary wave with height h above the bottom of the tube. amplitude 0 0 0.20 0.40 0.60 h / m Fig. 5.2 [2] (c) For the stationary wave, state: (i) the direction of the oscillations of an air particle at a height of 0.15 m above the bottom of the tube … [1] (ii) the phase difference between the oscillations of a particle at a height of 0.10 m and a particle at a height of 0.20 m above the bottom of the tube. phase difference = … ° [1] (d) The speed of the sound wave is 340 m s−1. Calculate the frequency of the sound wave. frequency = … Hz [2] (e) The frequency of the sound wave is gradually increased. Determine the frequency of the wave when a stationary wave is next formed. frequency = … Hz [1] [Total: 9]

9 marks

Mark scheme: 5(a) (incident) wave reflects at end/top of tube B1 (incident) wave and reflected wave interfere/superpose B1 5(b) line has maximum value of amplitude at h = 0 and h = 0.60 m only B1 line has minimum/zero value of amplitude at h = 0.30 m only B1 5(c)(i) vertical/along length of tube/along axis of tube B1 5(c)(ii) phase difference = 0 A1 5(d) v = fλ C1 v = 340 / (2 × 0.60) = 280 Hz A1 5(e) f = 340 / 0.60 = 570 Hz A1

This question in 9702/23 May/June 2019

6 (a) Describe the conditions required for two waves to be able to form a stationary wave. … … … … [2] (b) A stationary wave on a string has nodes and antinodes. The distance between a node and an adjacent antinode is 6.0 cm. (i) State what is meant by a node. … [1] (ii) Calculate the wavelength of the two waves forming the stationary wave. wavelength = … cm [1] (iii) State the phase difference between the particles at two adjacent antinodes of the stationary wave. phase difference = … ° [1] [Total: 5]

5 marks

Mark scheme: 6(a) the waves (of the same type) move in opposite directions and overlap B1 the waves have the same (speed and) frequency/wavelength B1 6(b)(i) zero amplitude B1 6(b)(ii) distance = 6.0 × 4 = 24 cm A1 6(b)(iii) 180° A1

This question in 9702/21 Oct/Nov 2020

4 (a) State the principle of superposition. … … … [2] (b) A transmitter produces microwaves that travel in air towards a metal plate, as shown in Fig. 4.1. microwave metal transmitter microwave plate receiver X Fig. 4.1 The microwaves have a wavelength of 0.040 m. A stationary wave is formed between the transmitter and the plate. (i) Explain the function of the metal plate. … … [1] (ii) Calculate the frequency, in GHz, of the microwaves. frequency = … GHz [3] (iii) A microwave receiver is initially placed at position X where it detects an intensity minimum. The receiver is then slowly moved away from X directly towards the plate. 1. Determine the shortest distance from X of the receiver when it detects another intensity minimum. distance = … m 2. Determine the number of intensity maxima that are detected by the receiver as it moves from X to a position that is 9.1 cm away from X. number = … [2] [Total: 8]

8 marks

Mark scheme: 4(a) (two or more) waves meet (at a point) B1 (resultant) displacement is the sum of the individual displacements B1 4(b)(i) it is a (wave) reflector / it reflects (the wave) B1 4(b)(ii) v = fλ or c = fλ C1 f = 3.0 × 108 / 0.040 = 7.5 × 109 (Hz) = 7.5 × 109 / 109 (GHz) C1 = 7.5 GHz A1 4(b)(iii) 1 distance = 0.020 m A1 2 number = 5 A1

This question in 9702/22 Feb/March 2021

5 A tube is initially fully submerged in water. The axis of the tube is kept vertical as the tube is slowly raised out of the water, as shown in Fig. 5.1. loudspeaker surface of water air column water wall of tube Fig. 5.1 A loudspeaker producing sound of frequency 530 Hz is positioned at the open top end of the tube as it is raised. The water surface inside the tube is always level with the water surface outside the tube. The speed of the sound in the air column in the tube is 340 m s–1. (a) Describe a simple way that a student, without requiring any additional equipment, can detect when a stationary wave is formed in the air column as the tube is being raised. … … [1] (b) Determine the height of the top end of the tube above the surface of the water when a stationary wave is first produced in the tube. Assume that an antinode is formed level with the top of the tube. height = … m [3] (c) Determine the distance moved by the tube between the positions at which the first and second stationary waves are formed. distance = … m [1] [Total: 5]

5 marks

Mark scheme: 5(a) a (much) louder sound can be heard B1 5(b) v = fλ C1 λ = 340 / 530 ( = 0.64 m) C1 height = 0.64 / 4 = 0.16 m A1 5(c) distance = 0.64 / 2 or 0.16 × 2 or (¾ × 0.64 – ¼ × 0.64) = 0.32 m A1

This question in 9702/22 Oct/Nov 2021

5 (a) (i) State the conditions required for the formation of a stationary wave. … … … … [2] (ii) State the phase difference between any two vibrating particles in a stationary wave between two adjacent nodes. phase difference = … ° [1] (b) A motorcycle is travelling at 13.0 m s–1 along a straight road. The rider of the motorcycle sees a pedestrian standing in the road directly ahead and operates a horn to emit a warning sound. The pedestrian hears the warning sound from the horn at a frequency of 543 Hz. The speed of the sound in the air is 334 m s–1. (i) Calculate the frequency, to three significant figures, of the sound emitted by the horn. frequency = … Hz [2] (ii) The motorcycle rider passes the stationary pedestrian and then moves directly away from her. As the rider moves away, he operates the horn for a second time. The pedestrian now hears sound that is increasing in frequency. State the variation, if any, in the speed of the motorcycle when the rider operates the horn for the second time. … [1] (c) A beam of vertically polarised monochromatic light is incident normally on a polarising filter, as shown in Fig. 5.1. polarising filter vertically polarised 20° transmitted incident light beam, light beam, intensity I0 intensity IT transmission axis of filter Fig. 5.1 The filter is positioned with its transmission axis at an angle of 20° to the vertical. The incident light has intensity I0 and the transmitted light has intensity IT. IT (i) By considering the ratio , calculate the ratio I0 amplitude of transmitted light . amplitude of incident light Show your working. ratio = … [3] (ii) The filter is now rotated, about the direction of the light beam, from its starting position shown in Fig. 5.1. The direction of rotation is such that the angle of the transmission axis to the vertical initially increases. Calculate the minimum angle through which the filter must be rotated so that the intensity of the transmitted light returns to the value that it had when the filter was at its starting position. angle = … ° [1] [Total: 10]

10 marks

Mark scheme: 5(a)(i) (two) waves travelling (at same speed) in opposite directions overlap B1 waves (of the same type) have same frequency/wavelength B1 5(a)(ii) phase difference = 0 A1 Question Answer Marks 5(b)(i) fO = fS v / v–vS 543 = f × 334 / (334 – 13) C1 f = 522 Hz A1 5(b)(ii) (the speed is) decreasing B1 5(c)(i) I ∝ A2 B1 IT / I0 = cos2 20° or AT / A0 = cos 20° C1 ratio = 0.94 A1 5(c)(ii) angle = 140° A1

This question in 9702/22 Feb/March 2022

4 (a) Polarisation is a phenomenon associated with light waves but not with sound waves. (i) State the meaning of polarisation. … … … [1] (ii) State why light waves can be plane polarised but sound waves cannot. … … … [1] (b) Two polarising filters A and B are positioned so that their planes are parallel to each other and perpendicular to a central axis line XY, as shown in Fig. 4.1. filter filter A B direction of rotation I0 X Y unpolarised light vertical horizontal transmission axis transmission axis Fig. 4.1 The transmission axis of filter A is vertical and the transmission axis of filter B is horizontal. Unpolarised light of a single frequency is directed along the line XY from a source positioned at X. The light emerging from filter A is vertically plane polarised and has intensity I0. Filter B is rotated from its starting position about the line XY, as shown in Fig. 4.1. 1 After rotation, the intensity of the light emerging from filter B is I0. 4 Calculate the angle of rotation of filter B from its starting position. angle of rotation = … ° [3] (c) A microwave of intensity I0 and amplitude A0 meets another microwave of the same frequency 1 and of intensity I0 travelling in the opposite direction. Both microwaves are vertically plane 4 polarised and superpose where they meet. (i) Explain, without calculation, why these two waves cannot form a stationary wave with zero amplitude at its nodes. … … … [2] (ii) Determine, in terms of A0, the maximum amplitude of the wave formed. maximum amplitude = … A0 [3] [Total: 10]

10 marks

Mark scheme: 4(a)(i) oscillations are in a single direction, which is perpendicular to the direction of propagation (of the wave) B1 or oscillations are in a single plane, which contains the direction of propagation (of the wave) 4(a)(ii) light waves are transverse and sound waves are longitudinal B1 4(b) I = I0 cos2 C1 cos2 = 1 / 4 so cos = 1 / 2 C1 = 60° or 120° or 240° or 300° angle of rotation = (120° – 90°) or (240° – 90°) or (300° – 90°) A1 = 30° or 150° or 210° or 330° 4(c)(i) the waves have different amplitudes B1 cannot have resultant displacement that is always zero B1 or cannot have (complete) destructive interference (at nodes) or (at nodes resultant) amplitude is the difference of the amplitudes 4(c)(ii) I  A2 C1 A2 / A02 = (I0 / 4) / I0 C1 A = 0.5 A0 maximum amplitude = A0 + 0.5 A0 A1 = 1.5 A0

This question in 9702/21 Oct/Nov 2022

5 (a) Two progressive sound waves meet to form a stationary wave. The two waves have the same amplitude, wavelength, frequency and speed. State the other condition that must be fulfilled by the two waves in order for them to produce the stationary wave. … [1] (b) A stationary wave is formed on a string that is stretched between two fixed points A and B. Fig. 5.1 shows the string at time t = 0 when each point is at its maximum displacement. A B Fig. 5.1 Distance AB is 0.80 m. The period of the stationary wave is 0.016 s. (i) On Fig. 5.1, sketch a solid line to show the position of the string: ● at time t = 0.004 s (label this line P) ● at time t = 0.024 s (label this line Q). [2] (ii) Determine the speed of a progressive wave along the string. speed = … m s–1 [3] (c) A beam of vertically polarised light of intensity I0 is incident normally on a polarising filter that has its transmission axis at 30° to the vertical, as shown in Fig. 5.2. vertically polarised 30° incident light transmitted transmitted beam, beam, beam, intensity I0 intensity I1 intensity I2 polarising filter with second polarising filter transmission axis at with transmission axis at 30° to the vertical 90° to the vertical Fig. 5.2 The transmitted light from the first polarising filter has intensity I1. This light is then incident normally on a second polarising filter that has its transmission axis at 90° to the vertical. The transmitted light from the second filter has intensity I2. Calculate: I1 (i) the ratio I0 I1 = … [2] I0 I2 (ii) the ratio . I0 I2 = … [2] I0 [Total: 10]

10 marks

Mark scheme: 5(a) (they travel in) opposite directions B1 5(b)(i) straight line from A to B, labelled P B1 line that is ‘mirror image’ of given line, labelled Q B1 5(b)(ii) = 0.80 / 2 C1 ( = 0.40 m) v =  / T C1 or v = f and f = 1 / T v = 0.40 / 0.016 or 62.5  0.40 A1 = 25 m s–1 5(c)(i) I1 / I0 = cos2 30° C1 = 0.75 A1 5(c)(ii) I2 / I1 = cos2 60° C1 I2 / I0 = cos2 30°  cos2 60° or 0.75  cos2 60° = 0.19 A1

This question in 9702/22 Oct/Nov 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 progressive wave travels through a medium. The wave causes a particle of the medium to vibrate along a line P. The energy of the wave propagates along a line Q. Compare the directions of lines P and Q if the wave is: (i) a transverse wave … [1] (ii) a longitudinal wave. … [1] (b) A tube is closed at one end. A loudspeaker is placed near the other end of the tube, as shown in Fig. 5.1. tube A A loudspeaker L Fig. 5.1 (not to scale) The loudspeaker emits sound of frequency 1.7 kHz. The speed of sound in the air in the tube is 340 m s–1. A stationary wave is formed with an antinode A at the open end of the tube. There is only one other antinode A inside the tube, as shown in Fig. 5.1. Determine: (i) the wavelength of the sound wavelength = … m [2] (ii) the length L of the tube L = … m [1] (iii) the maximum wavelength of the sound from the loudspeaker that can produce a stationary wave in the tube. maximum wavelength = … m [1] (c) Two polarising filters are arranged so that their planes are vertical and parallel. The first filter has its transmission axis at an angle of 35° to the vertical and the second filter has its transmission axis at angle α to the vertical, as shown in Fig. 5.2. 35° α incident light beam, intensity 8.5 W m–2 intensity 5.2 W m–2 transmission first filter second filter axis of filter Fig. 5.2 Angle α is greater than 35° and less than 90°. A beam of vertically polarised light of intensity 8.5 W m–2 is incident normally on the first filter. (i) Show that the intensity of the light transmitted by the first filter is 5.7 W m–2. [1] (ii) The intensity of the light transmitted by the second filter is 5.2 W m–2. Calculate angle α. α = … ° [2] [Total: 9]

9 marks

Mark scheme: 5(a)(i) (they are) perpendicular B1 5(a)(ii) (they are) parallel B1 5(b)(i)  = v / f C1 = 340 / 1700 = 0.20 m A1 5(b)(ii) L = 3 4   = 3 4  0.20 = 0.15 m A1 5(b)(iii)  = 4  0.15 or 0.20  3 = 0.60 m A1 5(c)(i) (I =) 8.5  cos2 35° = 5.7 (W m–2) A1 5(c)(ii) 5.2 = 5.7 cos2  ( = 17°) C1  = 35° + 17° = 52° A1

This question in 9702/22 May/June 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) Coherent visible light of a single frequency is incident normally on a double slit. This produces a pattern of bright and dark interference fringes on a screen, as illustrated in Fig. 6.1. fringe pattern on screen screen double slit bright fringe X light 1.2 mm 10.2 mm dark fringe 3.1 m bright fringe Y Fig. 6.1 (not to scale) There are seven bright fringes. (i) Explain how the pattern of bright and dark interference fringes is formed. … … … … … … [3] (ii) The distance between the centres of bright fringe X and bright fringe Y in the pattern is 10.2 mm. The slit spacing is 1.2 mm. The distance from the slits to the screen is 3.1 m. Calculate the wavelength of the light incident on the slits. wavelength = … m [3] (iii) The light is replaced by different visible light with a shorter wavelength. State how the new fringe separation will compare to the original fringe separation. … [1] (b) A stationary wave is formed on a stretched string AB, as shown in Fig. 6.2. string P Q mean position of string A B R Fig. 6.2 P, Q and R are points on the string. (i) On Fig. 6. 2, draw a cross (×) to show the position of a node. [1] (ii) State the phase difference between P and Q. phase difference = … ° [1] (iii) State the phase difference between P and R. phase difference = … ° [1] [Total: 10]

10 marks

Mark scheme: 6(a)(i) Any three from: B3 • Light diffracts at the (two) slits. • Light (from each slit) meets / superposes (at the screen). • When the phase difference is 0 (degrees) a bright fringe / (intensity) maximum is formed. • When the phase difference is 180 (degrees) a dark fringe / (intensity) minimum is formed. 6(a)(ii)  = ax / D C1  = 1.2  10–3  (10.2  10–3 / 6) / 3.1 C1  = 6.6  10–7 m A1 6(a)(iii) (new fringe separation will be) smaller B1 6(b)(i) A cross at the intersection of the string and the mean position line. B1 6(b)(ii) 0 A1 6(b)(iii) 180˚ A1

This question in 9702/22 Feb/March 2024

5 A stretched string PQ has length 1.2 m. One end of the string is attached to a vibration generator and the other end is attached to a wall, as shown in Fig. 5.1. wall 1.2 m vibration generator Q P string Fig. 5.1 The vibration generator is switched on and a stationary wave is formed on the string. The string is shown at one instant of time in Fig. 5.2. P Q Fig. 5.2 (a) Explain how a stationary wave is formed between the vibration generator and the wall. … … … … [2] (b) Calculate the wavelength of the stationary wave shown in Fig. 5.2. wavelength = … m [1] (c) Fig. 5.3 shows the stationary wave at time t = 0 when all points on the wave are at their maximum displacements. P Q Fig. 5.3 The period of the wave is 0.16 s. On Fig. 5.3, sketch the shape of the stationary wave at time t = 0.24 s. [2] (d) Points R and T on the string are a horizontal distance of 0.30 m apart and in the positions shown in Fig. 5.4. 0.30 m R T Fig. 5.4 State the phase difference between the oscillations of points R and T. phase difference = … ° [1] (e) Calculate the speed of the progressive waves on the stretched string. speed = … m s–1 [2] [Total: 8]

8 marks

Mark scheme: 5(a) wave(s) (travel along string and) reflect at fixed point / wall / Q / end / vibration generator / P B1 incident and reflected waves superpose B1 5(b) 0.80 m A1 5(c) same wavelength as original throughout and passing through intersection of solid and dashed lines B1 reflected in dashed line and of same amplitude B1 5(d) 180° A1 5(e) v = fand f =1 / T or v = / T C1 v = 6.25  0.80 or 0.80 / 0.16 = 5.0 m s–1 A1

This question in 9702/21 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) A stationary wave is formed on a string XY that has a length of 0.48 m. Fig. 5.1 shows the string at one instant in time. 0.48 m X Y Fig. 5.1 The speed of the wave on the string is 1400 m s–1. (i) On Fig. 5.1, draw a cross (×) at one position that is a node and another cross at one position that is an antinode. Label the node N and the antinode A. [1] (ii) Show that the wavelength of the wave produced is 0.32 m. Explain your reasoning. [1] (iii) Calculate the frequency of the wave. frequency = … Hz [2] (b) A source of sound waves of frequency 780 Hz is on a rotating platform. The speed of the source is 39 m s–1. The sound is detected by an observer that is a large distance from the rotating platform, as shown in Fig. 5.2. source of sound, observer speed 39 m s–1 platform Fig. 5.2 (not to scale) (i) The speed of sound in air is 320 m s–1. Calculate the maximum frequency of the sound detected by the observer. maximum frequency = … Hz [2] (ii) At time t = 0, the observer detects the sound emitted by the source when it was in the position shown in Fig. 5.2. On Fig. 5.3, sketch the variation with t of the frequency f of the sound detected by the observer for one complete rotation of the platform. Calculations are not required. f 0 t Fig. 5.3 [2] [Total: 8]

8 marks

Mark scheme: 5(a)(i) cross labelled N marked at the intersection of the solid and dashed lines or at X or Y B1 and cross labelled A marked at a peak or a trough 5(a)(ii) (XY is 1.5 wavelengths, so) wavelength = 0.48  (2 / 3) = 0.32 m B1 or (wavelength is twice node–node distance so)  / 2 = 0.48 / 3 = 0.16 m and wavelength = 0.16  2 = 0.32 m 5(a)(iii) v = f C1 frequency = 1400 / 0.32 A1 = 4400 Hz 5(b)(i) fo = fsv / (v – vs) C1 fo = (780  320) / (320 – 39) maximum frequency = 890 Hz A1 5(b)(ii) line showing f varying both above and below a mean frequency and returning to the original start value of f B1 a single cycle of a smoothly oscillating curve of correct phase (starting at mean position, falling to a trough, then rising to a B1 peak, ending at mean position)

This question in 9702/23 Oct/Nov 2024

4 A device containing a microwave emitter and receiver is placed in front of a large metal sheet in a vacuum as shown in Fig. 4.1. X P Q Y microwave emitter and receiver metal sheet Fig. 4.1 (not to scale) The line XY is perpendicular to the metal sheet. The device emits microwaves of frequency 6.3 GHz. (a) When the device is at position P, a stationary wave is formed between the device and the sheet. Explain how the stationary wave, including the nodes and the antinodes, is formed. … … … … … … … … … [4] (b) (i) Calculate the wavelength of the microwaves. wavelength = … m [2] (ii) At point P the receiver detects a maximum amplitude of the stationary wave. The device is moved slowly from point P along the line XY and the receiver detects a series of minimum and maximum amplitudes. The first time a minimum amplitude is detected by the receiver is when the device is at point Q. Determine the distance between P and Q. distance = … m [1] (iii) The intensity of the microwaves emitted by the device is increased. The frequency of the microwaves is unchanged. The device is moved slowly along the line XY from point Q until the next maximum amplitude is detected at point R. State and explain whether the distance QR is greater than, less than or the same as distance PQ. … … … [1] [Total: 8]

8 marks

Mark scheme: 4(a) (micro)wave (from the transmitter) reflects (at metal/sheet) B1 The incident and reflected waves superpose B1 (resultant) amplitude is maximum at an antinode B1 (resultant) amplitude is minimum/zero at a node B1 4(b)(i)  = c / f C1 = 3  108 / 6.3  109 = 0.048 m A1 4(b)(ii) distance PQ = ¼ A1 = 0.048 / 4 = 0.012 m 4(b)(iii) (Distance QR is the) same (as PQ) and one of: B1 • Distance (between maxima / minima) does not depend on intensity • distance depends only on wavelength • wavelength is unchanged / constant

This question in 9702/22 Feb/March 2025

4 (a) State the principle of superposition. … … … [2] (b) An electromagnetic wave of wavelength 0.026 m in free space is incident normally on an aluminium sheet, as shown in Fig. 4.1. transmitter aluminium sheet electromagnetic wave Fig. 4.1 The wave reflects at the aluminium sheet and a stationary wave is formed in the region between the transmitter and the sheet. (i) Explain how the stationary wave, including its nodes and antinodes, is formed. … … … … … [3] (ii) Calculate the frequency of the electromagnetic wave. frequency = … Hz [2] (iii) State the principal region of the electromagnetic spectrum to which the wave belongs. … [1] (iv) Determine the distance between a node and an adjacent antinode. distance = … m [1] [Total: 9]

9 marks

Mark scheme: 4(a) when two (or more) waves meet M1 (resultant) displacement equals the sum of the displacements of the (two separate) waves A1 4(b)(i) the incident and reflected waves superpose B1 (the waves superpose so that the resultant) amplitude is maximum at an antinode B1 (the waves superpose so that the resultant) amplitude is minimum / zero at a node B1 4(b)(ii) c = f C1 f = (3.00 × 108) / 0.026 A1 = 1.2 × 1010 Hz 4(b)(iii) microwave B1 4(b)(iv) distance = 0.026 / 4 A1 = 6.5 × 10–3 m

This question in 9702/24 Oct/Nov 2025