Energy in simple harmonic motion — Paper 4 · A Level Physics 9702

4 A piston moves vertically up and down in a cylinder, as illustrated in Fig. 4.1. cylinder pivot piston pivot P wheel Fig. 4.1 The piston is connected to a wheel by means of a rod that is pivoted at the piston and at the wheel. As the piston moves up and down, the wheel is made to rotate. (a) (i) State the number of oscillations made by the piston during one complete rotation of the wheel. number = ………………………. [1] (ii) The wheel makes 2400 revolutions per minute. Determine the frequency of oscillation of the piston. frequency = ………………………. Hz [1] Use (b) The amplitude of the oscillations of the piston is 42 mm. Assuming that these oscillations are simple harmonic, calculate the maximum values for the piston of (i) the linear speed, speed = …………………………. m s–1 [2] (ii) the acceleration. acceleration = …………………………. m s–2 [2] (c) On Fig. 4.1, mark a position of the pivot P for the piston to have (i) maximum speed (mark this position S), [1] (ii) maximum acceleration (mark this position A). [1]

8 marks

Mark scheme: 4 (a) (i) 1.0 B1 [1] (ii) 40 Hz B1 [1] (b) (i) speed = 2πfa C1 = 2π × 40 × 42 × 10-3 = 10.6 m s-1 A1 [2] (ii) acceleration = 4π2f2 a C1 = (80π)2 × 42 × 10-3 = 2650 m s-2 A1 [2] (c) (i) S marked correctly (on ‘horizontal line through centre of wheel) B1 (ii) A marked correctly (on ‘vertical line’ through centre of wheel) B1 [2] GCE A/AS Level – May/June 2006 9702 04

This question in 9702/41 May/June 2006

3 A spring is hung from a fixed point. A mass of 130 g is hung from the free end of the spring, as shown in Fig. 3.1. spring mass 130 g Fig. 3.1 The mass is pulled downwards from its equilibrium position through a small distance d and is released. The mass undergoes simple harmonic motion. Fig. 3.2 shows the variation with displacement x from the equilibrium position of the kinetic energy of the mass. 3.0 kinetic energy / mJ 2.0 1.0 0 –1.0 –0.5 0 +0.5 +1.0 x / cm Fig. 3.2 Examiner’s Use (a) Use Fig. 3.2 to (i) determine the distance d through which the mass was displaced initially, d = … cm [1] (ii) show that the frequency of oscillation of the mass is approximately 4.0 Hz. [6] (b) (i) On Fig. 3.2, draw a line to represent the total energy of the oscillating mass. [1] (ii) After many oscillations, damping reduces the total energy of the mass to 1.0 mJ. For the oscillations with reduced energy, 1. state the frequency, frequency = … Hz 2. using the graph, or otherwise, state the amplitude. amplitude = … cm [2]

10 marks

Mark scheme: 3 (a) (i) 0.8 cm … B1 [1] (ii) (max.) kinetic energy = 2.56 mJ … C1 v(MAX) = ωa … C1 (max.) kinetic energy = ½mω2a2 or ½mω2 (a2 – x2) … C1 2.56 × 10-3 = ½ × 0.130 × ω2 × (0.8 × 10-2)2 … M1 ω = 24.8 rad s-1 … C1 f = ω/2π … M1 = 4.0 Hz (3.95 Hz) … A0 [6] (b) (i) line parallel to x-axis at 2.56 mJ … B1 [1] (ii) 1 4.0 Hz … B1 2 0.50 cm (allow ±0.03 cm) … B1 [2]

This question in 9702/41 Oct/Nov 2007

4 The variation with time t of the displacement x of the cone of a loudspeaker is shown in For Fig. 4.1. Examiner’s Use 0.3 x / mm 0.2 0.1 0 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 t / ms – 0.1 – 0.2 – 0.3 Fig. 4.1 (a) Use Fig. 4.1 to determine, for these oscillations, (i) the amplitude, amplitude = … mm [1] (ii) the frequency. frequency = … Hz [2] (b) State two times at which (i) the speed of the cone is maximum, time … ms and time … ms [1] (ii) the acceleration of the cone is maximum. time … ms and time … ms [1] (c) The effective mass of the cone is 2.5 g. For Examiner’s Use your answers in (a) to determine the maximum kinetic energy of the cone. Use kinetic energy = … J [3] (d) The loudspeaker must be designed so that resonance of the cone is avoided. (i) State what is meant by resonance. … … … [2] (ii) State and briefly explain one other situation in which resonance should be avoided. … … … … [2]

12 marks

Mark scheme: 4 (a) (i) amplitude = 0.2 mm … A1 [1] (ii) period = 1.2 ms … C1 frequency = 830 Hz … A1 [2] (b) (i) any two of zero, 0.6 ms and 1.2 ms … A1 [1] (ii) any two of 0.3 ms, 0.9 ms, 1.5 ms … A1 [1] (c) either v = ωx0 = 2πfx0 = 2π × 830 × 0.2 × 10-3 = 1.05 m s-1 or slope of graph = 1.0 m s-1 ……(allow ± 0.1 m s-1) … C1 EK = ½mv2 = ½ × 2.5 × 10-3 × 1.052 … C1 = 1.4 × 10-3 J … A1 [3] (d) (i) large / maximum amplitude of vibration … B1 when impressed frequency equals natural frequency of vibration … B1 [2] (ii) e.g. metal panels on machinery vibrate / oscillate … (M1) motor in machine impresses frequency on panel … (A1) e.g. car suspension system vibrates / oscillates … (M1) going over bumps would give large amplitude vibrations … (A1) any feasible example, M1 + A1 … [2] [Total: 12] GCE A/AS LEVEL – October/November 2009 9702 41

This question in 9702/41 Oct/Nov 2009

3 (a) Define simple harmonic motion. For Examiner’s … Use … … [2] (b) A horizontal plate is vibrating vertically, as shown in Fig. 3.1. cube, mass 5.8 g plate vertical oscillations frequency 4.5 Hz Fig. 3.1 The plate undergoes simple harmonic motion with a frequency of 4.5 Hz and amplitude 3.0 mm. A metal cube of mass 5.8 g rests on the plate. Calculate, for the cube, the energy of oscillation. energy = … J [3] (c) The amplitude of oscillation of the plate in (b) is gradually increased. The frequency remains constant. At one particular amplitude, the cube just loses contact momentarily with the plate. (i) State the position of the plate in its oscillation at the point when the cube loses contact. … … … [2] (ii) Calculate this amplitude of oscillation. For Examiner’s Use amplitude = … m [2]

9 marks

Mark scheme: 3 (a) acceleration proportional to displacement/distance from fixed point M1 and in opposite directions/directed towards fixed point A1 [2] (b) energy = ½mω 2x02 and ω = 2πf C1 = ½ × 5.8 × 10–3 × (2π × 4.5)2 × (3.0 × 10–3)2 C1 = 2.1 × 10–5 J A1 [3] (c) (i) at maximum displacement M1 above rest position A1 [2] (ii) acceleration = (–)ω 2x0 and acceleration = 9.81 or g C1 9.81 = (2π × 4.5)2 × x0 x0 = 1.2 × 10–2 m A1 [2]

This question in 9702/43 Oct/Nov 2011

4 A small metal ball is suspended from a fixed point by means of a string, as shown in Fig. 4.1. For Examiner’s Use string ball x Fig. 4.1 The ball is pulled a small distance to one side and then released. The variation with time t of the horizontal displacement x of the ball is shown in Fig. 4.2. 6 x / cm 4 2 0 0 0.2 0.4 0.6 0.8 tt // ss 1.0 – 2 – 4 – 6 Fig. 4.2 The motion of the ball is simple harmonic. (a) Use data from Fig. 4.2 to determine the horizontal acceleration of the ball for a displacement x of 2.0 cm. acceleration = … m s–2 [3] (b) The maximum kinetic energy of the ball is EK. For On the axes of Fig. 4.3, sketch a graph to show the variation with time t of the kinetic Examiner’s energy of the ball for the first 1.0 s of its motion. Use kinetic energy EK 0 0 0.2 0.4 0.6 0.8 t / s 1.0 Fig. 4.3 [3]

6 marks

Mark scheme: 4 (a) a = (–)ω2x and ω = 2π/T C1 T = 0.60 s C1 a = (4π2 × 2.0 × 10–2) / (0.6)2 = 2.2 m s–2 A1 [3] (b) sinusoidal wave with all values positive B1 all values positive, all peaks at EK and energy = 0 at t = 0 B1 period = 0.30 s B1 [3]

This question in 9702/41 May/June 2012

2 A ball of mass 37 g is held between two fixed points A and B by two stretched helical springs, For as shown in Fig. 2.1. Examiner’s Use ball mass 37 g A B Fig. 2.1 The ball oscillates along the line AB with simple harmonic motion of frequency 3.5 Hz and amplitude 2.8 cm. (a) Show that the total energy of the oscillations is 7.0 mJ. [2] (b) At two points in the oscillation of the ball, its kinetic energy is equal to the potential energy stored in the springs. Calculate the magnitude of the displacement at which this occurs. displacement = … cm [3] (c) On the axes of Fig. 2.2 and using your answers in (a) and (b), sketch a graph to show For the variation with displacement x of Examiner’s Use (i) the total energy of the system (label this line T), [1] (ii) the kinetic energy of the ball (label this line K), [2] (iii) the potential energy stored in the springs (label this line P). [2] 8 6 energy / mJ 4 2 0 –3 –2 –1 0 1 2 3 x / cm Fig. 2.2 (d) The arrangement in Fig. 2.1 is now rotated through 90° so that the line AB is vertical and the ball oscillates in a vertical plane. Suggest one form of energy, other than those in (c), that must be taken into consideration when plotting new graphs to show energy changes with displacement. … [1]

11 marks

Mark scheme: 2 (a) energy = ½mω2a2 and ω = 2πf C1 = ½ × 37 × 10–3 × (2π × 3.5)2 × (2.8 × 10–2)2 M1 = 7.0 × 10–3 J A0 [2] (allow 2π × 3.5 shown as 7π) Energy = ½ mv2 and v = rω (C1) Correct substitution (M1) Energy = 7.0 × 10–3 J (A0) (b) EK = EP ½mω2 (a2 – x2) = ½mω2x2 or EK or EP = 3.5 mJ C1 x = a/√2 = 2.8 /√2 or EK = ½mω2 (a2 – x2) or EP = ½mω2x2 C1 = 2.0 cm A1 [3] (EK or EP = 7.0 mJ scores 0/3) Allow: k = 17.9 (C1) E = ½ kx2 (C1) x = 2.0 cm (A1) GCE AS/A LEVEL – May/June 2012 9702 42 (c) (i) graph: horizontal line, y-intercept = 7.0 mJ with end-points of line at +2.8 cm and –2.8 cm B1 [1] (ii) graph: reasonable curve B1 with maximum at (0,7.0) end-points of line at (–2.8, 0) and (+2.8, 0) B1 [2] (iii) graph: inverted version of (ii) M1 with intersections at (–2.0, 3.5) and (+2.0, 3.5) A1 [2] (Allow marks in (iii), but not in (ii), if graphs K & P are not labelled) (d) gravitational potential energy B1 [1]

This question in 9702/42 May/June 2012

4 A small metal ball is suspended from a fixed point by means of a string, as shown in Fig. 4.1. For Examiner’s Use string ball x Fig. 4.1 The ball is pulled a small distance to one side and then released. The variation with time t of the horizontal displacement x of the ball is shown in Fig. 4.2. 6 x / cm 4 2 0 0 0.2 0.4 0.6 0.8 tt // ss 1.0 – 2 – 4 – 6 Fig. 4.2 The motion of the ball is simple harmonic. (a) Use data from Fig. 4.2 to determine the horizontal acceleration of the ball for a displacement x of 2.0 cm. acceleration = … m s–2 [3] (b) The maximum kinetic energy of the ball is EK. For On the axes of Fig. 4.3, sketch a graph to show the variation with time t of the kinetic Examiner’s energy of the ball for the first 1.0 s of its motion. Use kinetic energy EK 0 0 0.2 0.4 0.6 0.8 t / s 1.0 Fig. 4.3 [3]

6 marks

Mark scheme: 4 (a) a = (–)ω2x and ω = 2π/T C1 T = 0.60 s C1 a = (4π2 × 2.0 × 10–2) / (0.6)2 = 2.2 m s–2 A1 [3] (b) sinusoidal wave with all values positive B1 all values positive, all peaks at EK and energy = 0 at t = 0 B1 period = 0.30 s B1 [3]

This question in 9702/43 May/June 2012

3 A ball is held between two fixed points A and B by means of two stretched springs, as shown For in Fig. 3.1. Examiner’s Use A B ball Fig. 3.1 The ball is free to oscillate along the straight line AB. The springs remain stretched and the motion of the ball is simple harmonic. The variation with time t of the displacement x of the ball from its equilibrium position is shown in Fig. 3.2. 2.0 x / cm 1.0 0 0 0.2 0.4 0.6 0.8 1.0 t / s 1.2 –1.0 –2.0 Fig. 3.2 (a) (i) Use Fig. 3.2 to determine, for the oscillations of the ball, 1. the amplitude, amplitude = … cm [1] 2. the frequency. frequency = … Hz [2] (ii) Show that the maximum acceleration of the ball is 5.2 m s–2. For Examiner’s Use [2] (b) Use your answers in (a) to plot, on the axes of Fig. 3.3, the variation with displacement x of the acceleration a of the ball. a / m s–2 0 0 x / 10–2 m Fig. 3.3 [2] (c) Calculate the displacement of the ball at which its kinetic energy is equal to one half of For the maximum kinetic energy. Examiner’s Use displacement = … cm [3]

10 marks

Mark scheme: 3 (a) (i) 1. amplitude = 1.7 cm A1 [1] 2. period = 0.36 cm C1 frequency = 1/0.36 frequency = 2.8 Hz A1 [2] (ii) a = (–)ω2x and ω = 2π/T C1 acceleration = (2π/0.36)2 × 1.7 × 10–2 M1 = 5.2 m s–2 A0 [2] (b) graph: straight line, through origin, with negative gradient M1 from (–1.7 × 10–2, 5.2) to (1.7 × 10–2, –5.2) A1 [2] (if scale not reasonable, do not allow second mark) (c) either kinetic energy = ½mω2(x02 – x2) or potential energy = ½mω2x2 and potential energy = kinetic energy B1 ½mω2(x0 – x2) = ½ × ½mω2x02 or ½mω2x2 = ½ × ½mω2x02 C1 x02 = 2x2 x = x0 / √2 = 1.7 / √2 = 1.2 cm A1 [3]

This question in 9702/41 May/June 2013

3 A metal ball is suspended from a fixed point by means of a string, as illustrated in Fig. 3.1. For Examiner’s Use string ball x Fig. 3.1 The ball is given a small displacement and then released. The variation with time t of the displacement x of the ball is shown in Fig. 3.2. 2.0 x / cm 1.0 0 0 0.4 0.8 1.2 1.6 2.0 t / s –1.0 –2.0 Fig. 3.2 (a) (i) State two times at which the speed of the ball is a maximum. time = … s and time = … s [1] (ii) Show that the maximum speed of the ball is approximately 0.08 m s–1. [2] (b) The variation with displacement x of the potential energy EP of the oscillations of the For ball is shown in Fig. 3.3. Examiner’s Use 25 EP 20 energy / mJ 15 10 5 0 –1.5 –1.0 –0.5 0 0.5 1.0 1.5 x / cm Fig. 3.3 (i) On the axes of Fig. 3.3, sketch a graph to show the variation with displacement x of the kinetic energy of the ball. [2] (ii) The amplitude of the oscillations reduces over a long period of time. After many oscillations, the amplitude of the oscillations is 0.60 cm. Use Fig. 3.3 to determine the total energy of the oscillations of the ball for oscillations of amplitude 0.60 cm. Explain your working. energy = … J [2]

7 marks

Mark scheme: 3 (a) (i) any two from 0.3(0) s, 0.9(0) s, 1.50 s (allow 2.1 s etc.) B1 [1] (ii) either v = ωx and ω = 2π/T C1 either v = (2π/1.2) × 1.5 × 10–2 M1 either v = 0.079 m s–1 A0 [2] or gradient drawn clearly at a correct position (C1) working clear (M1) to give (0.08 ± 0.01) m s–1 (A0) GCE A LEVEL – October/November 2013 9702 41 (b) (i) sketch: curve from (±1.5, 0) passing through (0, 25) M1 sketch: reasonable shape (curved with both intersections between y = 12.0→13.0) A1 [2] (ii) at max. amplitude potential energy is total energy B1 total energy = 4.0 mJ B1 [2]

This question in 9702/41 Oct/Nov 2013

3 A metal ball is suspended from a fixed point by means of a string, as illustrated in Fig. 3.1. For Examiner’s Use string ball x Fig. 3.1 The ball is given a small displacement and then released. The variation with time t of the displacement x of the ball is shown in Fig. 3.2. 2.0 x / cm 1.0 0 0 0.4 0.8 1.2 1.6 2.0 t / s –1.0 –2.0 Fig. 3.2 (a) (i) State two times at which the speed of the ball is a maximum. time = … s and time = … s [1] (ii) Show that the maximum speed of the ball is approximately 0.08 m s–1. [2] (b) The variation with displacement x of the potential energy EP of the oscillations of the For ball is shown in Fig. 3.3. Examiner’s Use 25 EP 20 energy / mJ 15 10 5 0 –1.5 –1.0 –0.5 0 0.5 1.0 1.5 x / cm Fig. 3.3 (i) On the axes of Fig. 3.3, sketch a graph to show the variation with displacement x of the kinetic energy of the ball. [2] (ii) The amplitude of the oscillations reduces over a long period of time. After many oscillations, the amplitude of the oscillations is 0.60 cm. Use Fig. 3.3 to determine the total energy of the oscillations of the ball for oscillations of amplitude 0.60 cm. Explain your working. energy = … J [2]

7 marks

Mark scheme: 3 (a) (i) any two from 0.3(0) s, 0.9(0) s, 1.50 s (allow 2.1 s etc.) B1 [1] (ii) either v = ωx and ω = 2π/T C1 either v = (2π/1.2) × 1.5 × 10–2 M1 either v = 0.079 m s–1 A0 [2] or gradient drawn clearly at a correct position (C1) working clear (M1) to give (0.08 ± 0.01) m s–1 (A0) GCE A LEVEL – October/November 2013 9702 42 (b) (i) sketch: curve from (±1.5, 0) passing through (0, 25) M1 sketch: reasonable shape (curved with both intersections between y = 12.0→13.0) A1 [2] (ii) at max. amplitude potential energy is total energy B1 total energy = 4.0 mJ B1 [2]

This question in 9702/42 Oct/Nov 2013

4 A student investigates the energy changes of a mass oscillating on a vertical spring, as shown in Fig. 4.1. spring mass Fig. 4.1 The student draws a graph of the variation with displacement x of energy E of the oscillation, as shown in Fig. 4.2. 2.5 2.0 E / mJ 1.5 1.0 0.5 0 – 1.5 – 1.0 – 0.5 0 0.5 1.0 1.5 x / cm Fig. 4.2 (a) State whether the energy E represents the total energy, the potential energy or the kinetic energy of the oscillations. … [1] (b) The student repeats the investigation but with a smaller amplitude. The maximum value of E is now found to be 1.8 mJ. Use Fig. 4.2 to determine the change in the amplitude. Explain your working. change in amplitude = … cm [3]

4 marks

Mark scheme: 4 (a) kinetic (energy)/KE/EK B1 [1] (b) either change in energy = 0.60 mJ or max E proportional to (amplitude)2/equivalent numerical working B1 new amplitude is 1.3 cm B1 change in amplitude = 0.2 cm B1 [3]

This question in 9702/41 May/June 2014

4 A student investigates the energy changes of a mass oscillating on a vertical spring, as shown in Fig. 4.1. spring mass Fig. 4.1 The student draws a graph of the variation with displacement x of energy E of the oscillation, as shown in Fig. 4.2. 2.5 2.0 E / mJ 1.5 1.0 0.5 0 – 1.5 – 1.0 – 0.5 0 0.5 1.0 1.5 x / cm Fig. 4.2 (a) State whether the energy E represents the total energy, the potential energy or the kinetic energy of the oscillations. … [1] (b) The student repeats the investigation but with a smaller amplitude. The maximum value of E is now found to be 1.8 mJ. Use Fig. 4.2 to determine the change in the amplitude. Explain your working. change in amplitude = … cm [3]

4 marks

Mark scheme: 4 (a) kinetic (energy)/KE/EK B1 [1] (b) either change in energy = 0.60 mJ or max E proportional to (amplitude)2/equivalent numerical working B1 new amplitude is 1.3 cm B1 change in amplitude = 0.2 cm B1 [3]

This question in 9702/43 May/June 2014

1 A light spring is suspended from a fixed point. A bar magnet is attached to the end of the spring, as shown in Fig. 1.1. spring bar magnet cardboard cup Fig. 1.1 In order to shield the magnet from draughts, a cardboard cup is placed around the magnet but does not touch it. The magnet is displaced vertically and then released. The variation with time t of the vertical displacement y of the magnet is shown in Fig. 1.2. 2.0 y / cm 1.0 0 0 0.1 0.2 0.3 0.4 0.5 0.6 t / s –1.0 –2.0 Fig. 1.2 The mass of the magnet is 130 g. (a) For the oscillations of the magnet, use Fig. 1.2 to (i) determine the angular frequency ω, ω = … rad s−1 [2] (ii) show that the maximum kinetic energy of the oscillating magnet is 6.4 mJ. [2] (b) The cardboard cup is now replaced with a cup made of aluminium foil. During 10 complete oscillations of the magnet, the amplitude of vibration is seen to decrease to 0.75 cm from that shown in Fig. 1.2. The change in angular frequency is negligible. (i) Use Faraday’s law of electromagnetic induction to explain why the amplitude of the oscillations decreases. … … … … … [3] (ii) Show that the loss in energy of the oscillating magnet is 4.8 mJ. [2] (c) The mass of the aluminium cup in (b) is 6.2 g. The specific heat capacity of aluminium is 910 J kg−1 K−1. The energy in (b)(ii) is transferred to the cup as thermal energy. Calculate the mean rise in temperature of the cup. temperature rise = … K [2] Please turn over for Question 2.

11 marks

Mark scheme: 1 (a) (i) either ω = 2π / T or ω = 2πf and f = 1 / T C1 ω = 2π / 0.30 = 20.9 rad s–1 (accept 2 s.f.) A1 [2] (ii) kinetic energy = ½mω 2x02 or v = ωx0 and ½mv2 C1 = ½ × 0.130 × 20.92 × (1.5 × 10–2)2 = 6.4 × 10–3 J A1 [2] (b) (i) as magnet moves, flux is cut by cup / aluminium giving rise to induced e.m.f. (in cup) B1 induced e.m.f. gives rise to currents and heating of the cup B1 thermal energy derived from oscillations of magnet so amplitude decreases B1 or induced e.m.f. gives rise to currents which generate a magnetic field (B1) the magnetic field opposes the motion of the magnet so amplitude decreases (B1) [3] (ii) either use of ½mω 2x02 and x0 = 0.75 cm or x0 is halved so ¼ energy C1 to give new energy = 1.6 mJ either loss in energy = 6.4 – 1.6 or loss = ¾ × 6.4 giving loss = 4.8 mJ A1 [2] (c) q = mc∆θ 4.8 × 10–3 = 6.2 × 10–3 × 910 × ∆θ C1 ∆θ = 8.5 × 10–4 K A1 [2]

This question in 9702/43 Oct/Nov 2014

4 (a) Distinguish between free oscillations and forced oscillations. free oscillations: … … forced oscillations: … … [2] (b) A trolley is held on a horizontal surface by means of two stretched springs, as shown in Fig. 4.1. spring trolley spring fixed point oscillator Fig. 4.1 One spring is attached to a fixed point. The other spring is attached to an oscillator that causes horizontal oscillations of the trolley. The oscillator vibrates with a constant amplitude of vibration. The frequency of vibration of the oscillator is gradually increased from a very low value. The variation with frequency f of the amplitude x0 of vibration of the trolley is shown in Fig. 4.2. x0 1.5 2.0 2.5 3.0 f / Hz Fig. 4.2 Use Fig. 4.2 to state and explain (i) the value of the natural frequency of vibration of the trolley, … … … … [3] (ii) whether there are any frictional forces acting on the trolley. … … … [1] (c) The oscillator in (b) is now stopped. The trolley is given a horizontal displacement of 4.7 cm along the line of the springs. The trolley is then released. Use information from Fig. 4.2 to estimate the maximum speed of the trolley. speed = … m s–1 [2]

8 marks

Mark scheme: 4 (a) free: (body oscillates) without any loss of energy/no resistive forces/no external forces applied B1 forced: continuous energy input (required)/body is made to vibrate by an (external) periodic force/driving oscillator B1 [2] (b) (i) idea of resonance B1 maximum amplitude at natural frequency B1 frequency = 2.1 Hz (allow 2.08 to 2.12 Hz) B1 [3] (ii) peak not very sharp/amplitude not infinite so frictional forces are present B1 [1] (c) v = ωx0 = 2π × 2.1 × 4.7 × 10–2 (allow ECF from (b)(i)) C1 = 0.62 m s–1 A1 [2]

This question in 9702/42 Oct/Nov 2015

4 An object of mass 80 g oscillates with simple harmonic motion. The variation with time t of the displacement x of the object is shown in Fig. 4.1. 2.0 x / cm 1.0 0 00 0.1 0.2 0.3 0.4 0.5 t / s –1.0 –2.0 Fig. 4.1 (a) Use Fig. 4.1 to determine the amplitude and the period of the oscillations. amplitude = … cm period = … s [1] (b) Use Fig. 4.1 and your answers in (a) to calculate the kinetic energy of the object at time t = 0.19 s. kinetic energy = … J [3] [Total: 4]

4 marks

Mark scheme: 4 (a) amplitude = 1.8 cm and period = 0.30 s A1 [1] (b) EK = ½m ω 2 (x02 – x2) or EK = ½mv2 and v = ± ω √(x02 – x2) C1 = ½ × 0.080 × (2π / 0.30)2 × [(1.8 × 10–2)2 – (1.2 × 10–2)2] C1 = 3.2 × 10–3 J A1 [3]

This question in 9702/42 Feb/March 2016

4 A metal block hangs vertically from one end of a spring. The other end of the spring is tied to a thread that passes over a pulley and is attached to a vibrator, as shown in Fig. 4.1. pulley vibrator spring block Fig. 4.1 (a) The vibrator is switched off. The metal block of mass 120 g is displaced vertically and then released. The variation with time t of the displacement y of the block from its equilibrium position is shown in Fig. 4.2. 3 y / cm 2 1 0 0 0.2 0.4 0.6 0.8 1.0 W / s –1 –2 –3 Fig. 4.2 For the vibrations of the block, calculate (i) the angular frequency ω, ω = … rad s−1 [2] (ii) the energy of the vibrations. energy = … J [2] (b) The vibrator is now switched on. The frequency of vibration is varied from 0.7f to 1.3f where f is the frequency of vibration of the block in (a). For the block, complete Fig. 4.3 to show the variation with frequency of the amplitude of vibration. Label this line A. [3] amplitude 0 0.7I I 1.3I frequency Fig. 4.3 (c) Some light feathers are now attached to the block in (b) to increase air resistance. The frequency of vibration is once again varied from 0.7f to 1.3f. The new amplitude of vibration is measured for each frequency. On Fig. 4.3, draw a line to show the variation with frequency of the amplitude of vibration. Label this line B. [2] [Total: 9]

9 marks

Mark scheme: 4 (a) (i) T = 0.60 s and ω = 2π / T C1 ω = 10 (10.47) rad s–1 A1 [2] (ii) energy = ½mω2x02 or ½mv2 and v = ωx0 C1 = ½ × 120 × 10–3 × (10.5)2 × (2.0 × 10–2)2 = 2.6 × 10–3 J A1 [2] (b) sketch: smooth curve in correct directions B1 peak at f M1 amplitude never zero and line extends from 0.7f to 1.3f A1 [3] (c) sketch: peaked line always below a peaked line A M1 peak not as sharp and at (or slightly less than) frequency of peak in line A A1 [2]

This question in 9702/42 May/June 2016

3 To demonstrate simple harmonic motion, a student attaches a trolley to two similar stretched springs, as shown in Fig. 3.1. spring trolley Fig. 3.1 The trolley has mass m of 810 g. The trolley is displaced along the line of the two springs and then released. The subsequent acceleration a of the trolley is given by the expression 2k x a = − m where the spring constant k for each of the springs is 64 N m−1 and x is the displacement of the trolley. (a) Show that the frequency of oscillation of the trolley is 2.0 Hz. [3] (b) The maximum displacement of the trolley is 1.6 cm. Calculate the maximum speed of the trolley. speed = … m s−1 [2] (c) The mass of the trolley is increased. The initial displacement of the trolley remains unchanged. Suggest the change, if any, that occurs in the frequency and in the maximum speed of the oscillations of the trolley. frequency: … maximum speed: … [2] [Total: 7]

7 marks

Mark scheme: 3 (a) 2k/m = ω2 M1 ω= 2πf M1 (2 × 64 / 0.810) = (2π × f)2 leading to f = 2.0 Hz A1 [3] (b) v0 = ωx0 or v0 = 2πfx0 or v = ω(x02 – x2)1/2 and x = 0 C1 v0 = 2π × 2.0 × 1.6 × 10–2 = 0.20 m s–1 A1 [2] (c) frequency: reduced/decreased B1 maximum speed: reduced/decreased B1 [2]

This question in 9702/43 Oct/Nov 2016

3 A uniform beam is clamped at one end. A metal block of mass m is fixed to the other end of the beam causing it to bend, as shown in Fig. 3.1. beam metal block mass m equilibrium position x clamp displaced position Fig. 3.1 The block is given a small vertical displacement and then released so that it oscillates with simple harmonic motion. The acceleration a of the block is given by the expression k a =- x m where k is a constant for the beam and x is the vertical displacement of the block from its equilibrium position. (a) Explain how it can be deduced from the expression that the block moves with simple harmonic motion. … … … [2] (b) For the beam, k = 4.0 kg s–2. Show that the angular frequency ω of the oscillations is given by the expression 2 .0 ω = . m [2] (c) The initial amplitude of the oscillation of the block is 3.0 cm. Use the expression in (b) to determine the maximum kinetic energy of the oscillations. maximum kinetic energy = … J [3] (d) Over a certain interval of time, the maximum kinetic energy of the oscillations in (c) is reduced by 50%. It may be assumed that there is negligible change in the angular frequency of the oscillations. Determine the amplitude of oscillation. amplitude = … m [2] (e) Permanent magnets are now positioned so that the metal block oscillates between the poles, as shown in Fig. 3.2. metal block beam permanent magnets Fig. 3.2 The block is made to oscillate with the same initial amplitude as in (c). Use energy conservation to explain why the energy of the oscillations decreases more rapidly than in (d). … … … … … [3] [Total: 12]

12 marks

Mark scheme: 3(a) m is constant or k / m is constant and so acceleration / a proportional to displacement / x B1 negative sign shows that acceleration / a is in opposite direction to displacement / x or negative sign shows acceleration / a is towards fixed point B1 3(b) evidence of comparison to expression to a = – ω2x B1 ω2 = k/m or ω2 = 4.0/m hence ω = 2.0/√m A1 3(c) EK = ½ m ω2x0 2 or EK = ½mv 2 and v = ωx0 C1 = ½m (4.0/m) (3.0 × 10–2)2 C1 = 1.8 × 10–3 J A1 Question Answer Marks 3(d) new x0 = –3 [( ) ( 1.8 10 / 2 2 / ( / 4.0))] m m × × × or (EK ∝ x0 2 so) new x0 = –2 2 [½ 3.0 10 ( ) ] × × C1 = 2.12 × 10–2 m A1 3(e) flux linked to block changes / flux is cut by block which induces an e.m.f. in block B1 (eddy) currents induced in block cause heating B1 thermal / heat energy comes from (kinetic / potential) energy of oscillations / block B1

This question in 9702/42 Feb/March 2017

2 A bar magnet of mass 180 g is suspended from the free end of a spring, as illustrated in Fig. 2.1. spring magnet coil Fig. 2.1 The magnet hangs so that one pole is near the centre of a coil of wire. The coil is connected in series with a resistor and a switch. The switch is open. The magnet is displaced vertically and then allowed to oscillate with one pole remaining inside the coil. The other pole remains outside the coil. At time t = 0, the magnet is oscillating freely as it passes through its equilibrium position. At time t = 3.0 s, the switch in the circuit is closed. The variation with time t of the vertical displacement y of the magnet is shown in Fig. 2.2. 2.0 1.5 y / cm 1.0 0.5 0 0 1 2 3 4 5 6 7 8 9 t / s –0.5 –1.0 –1.5 –2.0 Fig. 2.2 (a) Determine, to two significant figures, the frequency of oscillation of the magnet. frequency = … Hz [2] (b) State whether the closing of the switch gives rise to light, heavy or critical damping. … [1] (c) Calculate the change in the energy ΔE of oscillation of the magnet between time t = 2.7 s and time t = 7.5 s. Explain your working. ΔE = … J [6] [Total: 9]

9 marks

Mark scheme: 2(a) e.g. period = 3 / 2.5 C1 frequency = 0.83 Hz A1 2(b) light (damping) B1 2(c) at 2.7 s, A0 = 1.5 (cm) B1 energy = ½ m × 4π2f 2A0 2 B1 = ½ × 0.18 × 4π2 × 0.832 × (1.5 × 10–2)2 = 5.51 × 10–4 (J) C1 at 7.5 s, A0 = 0.75 (cm) B1 energy = ¼ × 5.51 × 10–4 or energy = ½ × 0.18 × 4π2 × 0.832 × (0.75 × 10–2)2 C1 energy = 1.38 × 10–4 (J) change = (5.51 × 10–4 – 1.38 × 10–4) = 4.13 J A1

This question in 9702/41 May/June 2017

3 A bar magnet of mass 250 g is suspended from the free end of a spring, as illustrated in Fig. 3.1. spring magnet coil Fig. 3.1 The magnet hangs so that one pole is near the centre of a coil of wire. The coil is connected in series with a resistor and a switch. The switch is open. The magnet is displaced vertically and then allowed to oscillate with one pole remaining inside the coil. The other pole remains outside the coil. At time t = 0, the magnet is oscillating freely as it passes through its equilibrium position. At time t = 6.0 s, the switch in the circuit is closed. The variation with time t of the vertical displacement y of the magnet is shown in Fig. 3.2. 2.0 1.5 y / cm 1.0 0.5 0 0 2 4 6 8 10 12 14 16 t / s –0.5 –1.0 –1.5 –2.0 Fig. 3.2 (a) For the oscillating magnet, use data from Fig. 3.2 to calculate, to two significant figures, (i) the frequency f, f = … Hz [2] (ii) the energy of the oscillations during the time t = 0 to time t = 6.0 s. energy = … J [3] (b) (i) State Faraday’s law of electromagnetic induction. … … … … [2] (ii) Use Faraday’s law and energy conservation to explain why the amplitude of the oscillations of the magnet reduces after time t = 6.0 s. … … … … … … [3] [Total: 10]

10 marks

Mark scheme: 3(a)(i) e.g. period = 6 / 2.5 C1 frequency = 0.42 Hz A1 3(a)(ii) energy = ½ m × 4π2f 2y0 2 C1 = ½ × 0.25 × 4π2 × 0.422 × (1.5 × 10–2)2 C1 = 2.0 × 10–4 J A1 3(b)(i) (induced) e.m.f. proportional to rate of M1 change of magnetic flux (linkage) or cutting of magnetic flux A1 3(b)(ii) coil cuts flux/field (of moving magnet) inducing e.m.f. in coil B1 (induced) current in resistor causes heating (effect) M1 thermal energy/heat derived from energy of oscillations (of magnet) A1

This question in 9702/42 May/June 2017

2 A bar magnet of mass 180 g is suspended from the free end of a spring, as illustrated in Fig. 2.1. spring magnet coil Fig. 2.1 The magnet hangs so that one pole is near the centre of a coil of wire. The coil is connected in series with a resistor and a switch. The switch is open. The magnet is displaced vertically and then allowed to oscillate with one pole remaining inside the coil. The other pole remains outside the coil. At time t = 0, the magnet is oscillating freely as it passes through its equilibrium position. At time t = 3.0 s, the switch in the circuit is closed. The variation with time t of the vertical displacement y of the magnet is shown in Fig. 2.2. 2.0 1.5 y / cm 1.0 0.5 0 0 1 2 3 4 5 6 7 8 9 t / s –0.5 –1.0 –1.5 –2.0 Fig. 2.2 (a) Determine, to two significant figures, the frequency of oscillation of the magnet. frequency = … Hz [2] (b) State whether the closing of the switch gives rise to light, heavy or critical damping. … [1] (c) Calculate the change in the energy ΔE of oscillation of the magnet between time t = 2.7 s and time t = 7.5 s. Explain your working. ΔE = … J [6] [Total: 9]

9 marks

Mark scheme: 2(a) e.g. period = 3 / 2.5 C1 frequency = 0.83 Hz A1 2(b) light (damping) B1 2(c) at 2.7 s, A0 = 1.5 (cm) B1 energy = ½ m × 4π2f 2A0 2 B1 = ½ × 0.18 × 4π2 × 0.832 × (1.5 × 10–2)2 = 5.51 × 10–4 (J) C1 at 7.5 s, A0 = 0.75 (cm) B1 energy = ¼ × 5.51 × 10–4 or energy = ½ × 0.18 × 4π2 × 0.832 × (0.75 × 10–2)2 C1 energy = 1.38 × 10–4 (J) change = (5.51 × 10–4 – 1.38 × 10–4) = 4.13 J A1

This question in 9702/43 May/June 2017

3 A spring is hung vertically from a fixed point. A mass M is hung from the other end of the spring, as illustrated in Fig. 3.1. spring L mass M Fig. 3.1 The mass is displaced downwards and then released. The subsequent motion of the mass is simple harmonic. The variation with time t of the length L of the spring is shown in Fig. 3.2. 16 L / cm 14 12 10 8 0 0.2 0.4 0.6 0.8 1.0 t / s Fig. 3.2 (a) State: (i) one time at which the mass is moving with maximum speed time = … s [1] (ii) one time at which the spring has maximum elastic potential energy. time = … s [1] (b) Use data from Fig. 3.2 to determine, for the motion of the mass: (i) the angular frequency ω ω = … rad s–1 [2] (ii) the maximum speed maximum speed = … m s–1 [2] (iii) the magnitude of the maximum acceleration. maximum acceleration = … m s–2 [2] (c) The mass M is now suspended from two springs, each identical to that in Fig. 3.1, as shown in Fig. 3.3. mass M Fig. 3.3 Suggest and explain the change, if any, in the period of oscillation of the mass. A numerical answer is not required. … … … [2] [Total: 10]

10 marks

Mark scheme: 3(a)(i) 0.10 s or 0.30 s or 0.50 s or 0.70 s or 0.90 s A1 3(a)(ii) 0 or 0.40 s or 0.80 s A1 3(b)(i) ω = 2π / T C1 = 2π / 0.40 = 16 rad s–1 A1 3(b)(ii) v0 = ωx0 C1 = 15.7 × 2.5 × 10–2 = 0.39 m s–1 A1 or tangent drawn at steepest part and working to show attempted calculation of gradient (C1) leading to v0 = 0.39 m s–1 (allow ± 0.15 m s–1) (A1) 3(b)(iii) a0 = ω 2x0 C1 a0 = (15.72 × 2.5 × 10–2) = 6.2 m s–2 A1 or a0 = ωv0 (C1) a0 = 15.7 × 0.39 = 6.2 m s–2 (A1) Question Answer Marks 3(c) period is shorter/lower B1 Any one from: • greater spring constant/stiffness • (restoring) force is greater (for any given extension) • acceleration is greater (for any given extension) • greater energy/maximum speed (for a given amplitude) B1

This question in 9702/42 May/June 2019

3 A simple pendulum consists of a metal sphere suspended from a fixed point by means of a thread, as illustrated in Fig. 3.1. thread L sphere mass 94.0 g 0.90 cm 12.7 cm Fig. 3.1 (not to scale) The sphere of mass 94.0 g is displaced to one side through a horizontal distance of 12.7 cm. The centre of gravity of the sphere rises vertically by 0.90 cm. The sphere is released so that it oscillates. The sphere may be assumed to oscillate with simple harmonic motion. (a) State what is meant by simple harmonic motion. … … … [2] (b) (i) State the kinetic energy of the sphere when the sphere returns to the displaced position shown in Fig. 3.1. kinetic energy = … J [1] (ii) Calculate the total energy ET of the oscillations. ET = … J [2] (iii) Use your answer in (ii) to show that the angular frequency ω of the oscillations of the pendulum is 3.3 rad s–1. [2] (c) The period T of oscillation of the pendulum is given by the expression L T = 2π  g where g is the acceleration of free fall and L is the length of the pendulum. Use data from (b) to determine L. L = … m [3] [Total: 10]

10 marks

Mark scheme: 3(a) acceleration (directly) proportional to displacement B1 acceleration is in opposite direction to displacement or acceleration is (directed) towards a fixed point B1 3(b)(i) zero B1 3(b)(ii) ET is maximum potential energy = mgh ET = 94 × 10–3 × 9.81 × 0.90 × 10–2 C1 = 8.3 × 10–3 J A1 3(b)(iii) EMAX = ½ mv02 and v0 = ωx0 or EMAX = ½m(ωx0)2 C1 8.3 × 10–3 = ½ × 94 × 10–3 × ω2 × (12.7 × 10–2)2 …leading to ω = 3.3 rad s–1 A1 3(c) T = 2π / ω C1 2π / 3.3 = 2π × (L / 9.81)½ C1 L = 0.90 m A1

This question in 9702/42 Oct/Nov 2020

3 A U-shaped tube contains some liquid. The liquid column in each half of the tube has length L, as shown in Fig. 3.1. x x L L Fig. 3.1 Fig. 3.2 The liquid columns are displaced vertically. The liquid then oscillates in the tube. The liquid levels are displaced from the equilibrium positions as shown in Fig. 3.2. The acceleration a of the liquid in the tube is related to the displacement x by the expression ⎛ g ⎞ a = − x ⎝ L ⎠ where g is the acceleration of free fall. (a) Explain how the expression shows that the liquid in the tube is undergoing simple harmonic motion. … … … … … [3] (b) The length L of each liquid column is 18 cm. Determine the period T of the oscillations. T = … s [3] (c) The oscillations of the liquid in the tube are damped. In any one complete cycle of the oscillations, the amplitude decreases by 6.0% of its value at the beginning of the oscillation. Determine the ratio energy of oscillations after 3 cycles . initial energy of oscillations ratio = … [3] [Total: 9]

9 marks

Mark scheme: 3(a) acceleration in opposite direction to displacement shown by – sign B1 g / L is constant M1 (so) acceleration is (directly) proportional to displacement A1 3(b) ω2 = g / L C1 ω = 2π / T or ω = 2πf and f = 1 / T C1 (2π / T)2 = 9.81 / 0.18 T = 0.85 s A1 3(c) energy ∝ x02 C1 (after 3 cycles,) amplitude = (0.94)3x0 = 0.83x0 C1 ratio final energy / initial energy = 0.832 = 0.69 A1

This question in 9702/42 May/June 2021

4 A trolley on a smooth surface is attached by springs to fixed blocks as shown in Fig. 4.1. springs trolley fixed block smooth surface fixed block Fig. 4.1 The trolley oscillates horizontally about its equilibrium position with an amplitude of 12 cm. Fig. 4.2 shows the variation of the acceleration a of the trolley with displacement x from its equilibrium position. Friction between the trolley and the surface can be assumed to be negligible. 0.8 a / m s–2 0.4 0 –12 –8 – 4 0 4 8 12 x / cm – 0.4 –0.8 Fig. 4.2 (a) Describe the features of the line in Fig. 4.2 that demonstrate that the motion of the trolley is simple harmonic. … … … [2] (b) Use Fig. 4.2 to determine the period T of the oscillations of the trolley. T = … s [3] (c) (i) On the line of the graph of Fig. 4.2, label with the letter P one point where the kinetic energy of the trolley is zero. [1] (ii) On the line of the graph of Fig. 4.2, label with the letter Q an approximate position of one point where the kinetic energy of the trolley is equal to the potential energy stored in the springs. [1] [Total: 7]

7 marks

Mark scheme: 4(a) straight line through the origin B1 negative gradient B1 4(b) a = (–)ω2x and T = 2π / ω C1 e.g. ω = √(0.80 / 0.12) (any correct pair of values of a and x) ( = 2.58 rad s–1) C1 T = 2π / 2.58 = 2.4 s A1 4(c)(i) Point labelled P at one end of the line B1 4(c)(ii) Point labelled Q at displacement with magnitude more than half but less than maximum B1

This question in 9702/42 Oct/Nov 2021

4 A pendulum consists of a bob (small metal sphere) attached to the end of a piece of string. The other end of the string is attached to a fixed point. The bob oscillates with small oscillations about its equilibrium position, as shown in Fig. 4.1. string L equilibrium position bob x oscillations Fig. 4.1 (not to scale) The length L of the pendulum, measured from the fixed point to the centre of the bob, is 1.24 m. The acceleration a of the bob varies with its displacement x from the equilibrium position as shown in Fig. 4.2. 0.4 a / m s–2 0.2 0 –0.06 –0.04 –0.02 0 0.02 0.04 0.06 x / m –0.2 –0.4 Fig. 4.2 (a) State how Fig. 4.2 shows that the motion of the pendulum is simple harmonic. … … … [2] (b) (i) Use Fig. 4.2 to determine the angular frequency ω of the oscillations. ω = … rad s–1 [2] (ii) The angular frequency ω is related to the length L of the pendulum by k ω = L where k is a constant. Use your answer in (b)(i) to determine k. Give a unit with your answer. k = … unit … [2] (c) While the pendulum is oscillating, the length of the string is increased in such a way that the total energy of the oscillations remains constant. Suggest and explain the qualitative effect of this change on the amplitude of the oscillations. … … … [2] [Total: 8]

8 marks

Mark scheme: 4(a) straight line through origin shows that a is proportional to x B1 negative gradient shows that a is in opposite direction to x B1 4(b)(i) a0 =  2x0 or a = – 2x or 2 = – gradient C1  = (0.40 / 0.050) = 2.8 rad s–1 A1 4(b)(ii) k =  2L = 2.82  1.24 C1 = 9.7 m s–2 A1 4(c) (increasing L causes)  to decrease or energy (= ½ m2x02) = ½ mkx02 / L (and L increases) M1 so amplitude increases A1

This question in 9702/41 May/June 2022

4 A pendulum consists of a bob (small metal sphere) attached to the end of a piece of string. The other end of the string is attached to a fixed point. The bob oscillates with small oscillations about its equilibrium position, as shown in Fig. 4.1. string L equilibrium position bob x oscillations Fig. 4.1 (not to scale) The length L of the pendulum, measured from the fixed point to the centre of the bob, is 1.24 m. The acceleration a of the bob varies with its displacement x from the equilibrium position as shown in Fig. 4.2. 0.4 a / m s–2 0.2 0 –0.06 –0.04 –0.02 0 0.02 0.04 0.06 x / m –0.2 –0.4 Fig. 4.2 (a) State how Fig. 4.2 shows that the motion of the pendulum is simple harmonic. … … … [2] (b) (i) Use Fig. 4.2 to determine the angular frequency ω of the oscillations. ω = … rad s–1 [2] (ii) The angular frequency ω is related to the length L of the pendulum by k ω = L where k is a constant. Use your answer in (b)(i) to determine k. Give a unit with your answer. k = … unit … [2] (c) While the pendulum is oscillating, the length of the string is increased in such a way that the total energy of the oscillations remains constant. Suggest and explain the qualitative effect of this change on the amplitude of the oscillations. … … … [2] [Total: 8]

8 marks

Mark scheme: 4(a) straight line through origin shows that a is proportional to x B1 negative gradient shows that a is in opposite direction to x B1 4(b)(i) a0 =  2x0 or a = – 2x or 2 = – gradient C1  = (0.40 / 0.050) = 2.8 rad s–1 A1 4(b)(ii) k =  2L = 2.82  1.24 C1 = 9.7 m s–2 A1 4(c) (increasing L causes)  to decrease or energy (= ½ m2x02) = ½ mkx02 / L (and L increases) M1 so amplitude increases A1

This question in 9702/43 May/June 2022

3 An object is suspended from a vertical spring as shown in Fig. 3.1. spring object oscillation Fig. 3.1 The object is displaced vertically and then released so that it oscillates, undergoing simple harmonic motion. Fig. 3.2 shows the variation with displacement x of the energy E of the oscillations. 7.0 P 6.0 5.0 4.0 Q E / mJ 3.0 R 2.0 1.0 0 –1.6 –1.2 –0.8 –0.4 0 0.4 0.8 1.2 1.6 x / cm Fig. 3.2 The kinetic energy, the potential energy and the total energy of the oscillations are each represented by one of the lines P, Q and R. (a) State the energy that is represented by each of the lines P, Q and R. P … Q … R … [2] (b) The object has a mass of 130 g. Determine the period of the oscillations. period = … s [4] (c) (i) State the cause of damping. … … [1] (ii) A light card is attached to the object. The object is displaced with the same initial amplitude and then released. During each complete oscillation the total energy of the system decreases by 8.0% of the total energy at the start of that oscillation. Determine the decrease in total energy, in mJ, of the system by the end of the first 6 complete oscillations. energy lost = … mJ [2] (iii) State, with a reason, the type of damping that the card introduces into the system. … … … [1] [Total: 10]

10 marks

Mark scheme: 3(a) P: total energy B2 Q: potential energy R: kinetic energy 3(b) E = ½m2x02 or E = ½mv02 and v0 = x0 C1 6.4  10 −3 = 1  0.130  2  0.0152 C1 2 (2 = 438) (= 20.9) T = 2 /  C1 = 2 / 20.9 A1 = 0.30 s 3(c)(i) resistive forces B1 3(c)(ii) 0.926 C1 decrease in energy = 6.4 – (6.4  0.926) A1 = 2.5 mJ 3(c)(iii) light damping because the amplitude of oscillations gradually reduces B1 or light damping because the system still oscillates

This question in 9702/42 Feb/March 2023

4 A small steel sphere is oscillating vertically on the end of a spring, as shown in Fig. 4.1. spring steel sphere oscillations Fig. 4.1 The velocity v of the sphere varies with displacement x from its equilibrium position according to v = ± 9.7 (11 .6 - x 2) where v is in cm s–1 and x is in cm. (a) (i) Calculate the frequency of the oscillations. frequency = … Hz [2] (ii) Show that the amplitude of the oscillations is 3.4 cm. [1] (iii) Calculate the maximum acceleration a0 of the sphere. a0 = … m s–2 [2] (b) On Fig. 4.2, sketch the variation with x of the acceleration a of the sphere. 2 a0 a a0 0 – 4 – 2 0 2 4 x / cm – a0 – 2a0 Fig. 4.2 [3] (c) Describe, without calculation, the interchange between the potential energy and the kinetic energy of the oscillations. … … … … … [3] [Total: 11]

11 marks

Mark scheme: 4(a)(i) C1 f = 9.7 / 2 = 1.5 Hz A1 4(a)(ii) amplitude = √(11.6) = 3.4 cm A1 4(a)(iii) a0 = 2x0 C1 = 9.72  3.4  10–2 = 3.2 m s–2 A1 4(b) sketch: straight line through the origin with negative gradient B1 line with negative gradient passing through (+3.4, –a0) and (–3.4, +a0) B1 line with ends at x =  3.4 cm and a =  a0 B1 4(c) sum of potential energy and kinetic energy is constant B1 at maximum displacement, kinetic energy is zero or at maximum displacement, potential energy is maximum B1 at zero displacement, kinetic energy is maximum or at zero displacement, potential energy is minimum B1

This question in 9702/42 May/June 2023

5 Fig. 5.1 shows a pendulum consisting of a metal sphere suspended by a thin string. thin string metal sphere oscillations Fig. 5.1 (not to scale) The sphere undergoes small oscillations about its equilibrium position. The oscillations may be considered to be simple harmonic. Fig. 5.2 shows the variation with time t of the displacement x of the sphere from its equilibrium position. 0.02 x / m 0.01 0 0 0.2 0.4 0.6 0.8 1.0 1.2 t / s –0.01 –0.02 Fig. 5.2 (a) On Fig. 5.1, draw an arrow, from the centre of the sphere, to represent the direction of the resultant force acting on the sphere when it is in the position shown. [1] (b) The mass of the sphere is 0.15 kg. (i) State the amplitude of the oscillations. amplitude = … m [1] (ii) Determine the angular frequency of the oscillations. angular frequency = … rad s–1 [2] (iii) Calculate the total energy of the oscillations. total energy = … J [2] (c) On Fig. 5.3, sketch the variation with x of the kinetic energy EK of the sphere. 6 EK / 10–3 J 4 2 0 –0.02 –0.01 0 0.01 0.02 x / m Fig. 5.3 [3] [Total: 9]

9 marks

Mark scheme: 5(a) arrow from sphere, perpendicular to string, pointing left and down B1 5(b)(i) amplitude = 0.016 m A1 5(b)(ii) angular frequency = 2 / T C1 = 2 / 0.40 A1 = 16 rad s–1 5(b)(iii) total energy = ½m2x02 C1 = ½  0.15  15.72  0.0162 A1 = 4.7  10–3 J 5(c) dome-shaped curve starting and ending on the x-axis, with peak at x = 0 B1 maximum EK shown as 4.7  10–3 J B1 minimum x shown as –0.016 m and maximum x shown as +0.016 m at the ends of the line B1

This question in 9702/42 Oct/Nov 2024

4 A small crystal is made to vibrate with simple harmonic motion. The variation with time t of the displacement x of one surface of the crystal from its equilibrium position is shown in Fig. 4.1. 50 x / 10−6 m t / 10−6 s 0 0 0.1 0.2 0.3 0.4 0.5 0.6 –50 Fig. 4.1 (a) Show that the angular frequency of the vibration of the surface is 4.2 × 107 rad s–1. [2] (b) Determine the maximum acceleration a0 of the vibration of the surface. a0 = … m s–2 [2] (c) The crystal may be modelled as a single mass of 2.4 × 10– 4 kg that vibrates as shown in Fig. 4.1. Calculate the total energy E of the vibrations. E = … J [3] (d) The crystal generates ultrasound waves that are used to obtain diagnostic information about internal structures. (i) The crystal is made from piezoelectric material. Explain how the crystal is made to vibrate. … … … … [2] (ii) A parallel beam of ultrasound waves is incident on a muscle‑bone boundary. Data for muscle and bone are given in Table 4.1. Table 4.1 material density / kg m–3 speed of sound / m s–1 muscle 1100 1600 bone 1900 4100 Calculate the percentage of the intensity of the ultrasound beam that is transmitted at this boundary. percentage transmitted = … % [3] [Total: 12]

12 marks

Mark scheme: 4(a)  = 2 / T C1 = 2 / (0.15  10–6) = 4.2  107 rad s–1 A1 4(b) a0 = 2x0 C1 = (4.2  107)2  40  10–6 A1 = 7.1  1010 m s–2 4(c) E = ½m2xo2 C1 = ½  2.4  10–4  (4.2  107)2  (40  10–6)2 C1 = 340 J A1 4(d)(i) apply alternating p.d. (to / across crystal) B1 applying p.d. to / across crystal causes it to distort B1 4(d)(ii) Z = c C1 Zm = 1100  1600 (= 1.76  106) Zb = 1900  4100 (= 7.79  106) intensity reflection co-efficient= [(7.79 – 1.76) / (7.79 + 1.76)]2 C1 = 0.40 or 40% percentage transmitted = 60% A1

This question in 9702/42 Feb/March 2025

5 A cuboidal block floats in a liquid with its base horizontal, as shown in Fig. 5.1. block liquid surface h Fig. 5.1 The base of the block is at a depth h below the surface of the liquid. The block is displaced downwards by a small distance and then released so that it oscillates. Fig. 5.2 shows the variation with h of the acceleration a of the block. 1.0 a / m s–2 0 0 0.4 0.8 1.2 1.6 2.0 2.4 h / m –1.0 Fig. 5.2 Fig. 5.3 shows the variation with h of the kinetic energy EK of the block. 10 EK / J 5 0 0 0.4 0.8 1.2 1.6 2.0 2.4 h / m Fig. 5.3 (a) (i) Determine the amplitude of the oscillations. amplitude = … m [1] (ii) State what the line in Fig. 5.2 shows about the nature of the oscillations. … [1] (b) State three other quantitative conclusions that can be drawn from Fig. 5.2 and Fig. 5.3 about the block and its oscillations. Use the space for any working. 1 … … 2 … … 3 … … [3] (c) On Fig. 5.4, sketch the variation with h of the potential energy EP of the oscillations. 10 EP / J 5 0 0 0.4 0.8 1.2 1.6 2.0 2.4 h / m Fig. 5.4 [3] [Total: 8]

8 marks

Mark scheme: 5(a)(i) amplitude = 0.60 m A1 5(a)(ii) oscillations are simple harmonic B1 5(b) Any three points from: B3 • mean / equilibrium position is at h = 1.4 m • total energy of oscillations = 9.0 J • angular frequency of oscillations = 1.2 rad s–1 or period of oscillations = 5.1 s or frequency of oscillation = 0.19 Hz • maximum speed of block = 0.73 m s–1 • mass of block = 33 kg 5(c) U-shaped curve resting on h axis (with minimum at EP = 0) B1 curve from h = 0.8 m to h = 2.0 m, with minimum EP shown at h = 1.4 m B1 both end-points of curve shown at EP = 9.0 J B1

This question in 9702/41 May/June 2025

5 A cuboidal block floats in a liquid with its base horizontal, as shown in Fig. 5.1. block liquid surface h Fig. 5.1 The base of the block is at a depth h below the surface of the liquid. The block is displaced downwards by a small distance and then released so that it oscillates. Fig. 5.2 shows the variation with h of the acceleration a of the block. 1.0 a / m s–2 0 0 0.4 0.8 1.2 1.6 2.0 2.4 h / m –1.0 Fig. 5.2 Fig. 5.3 shows the variation with h of the kinetic energy EK of the block. 10 EK / J 5 0 0 0.4 0.8 1.2 1.6 2.0 2.4 h / m Fig. 5.3 (a) (i) Determine the amplitude of the oscillations. amplitude = … m [1] (ii) State what the line in Fig. 5.2 shows about the nature of the oscillations. … [1] (b) State three other quantitative conclusions that can be drawn from Fig. 5.2 and Fig. 5.3 about the block and its oscillations. Use the space for any working. 1 … … 2 … … 3 … … [3] (c) On Fig. 5.4, sketch the variation with h of the potential energy EP of the oscillations. 10 EP / J 5 0 0 0.4 0.8 1.2 1.6 2.0 2.4 h / m Fig. 5.4 [3] [Total: 8]

8 marks

Mark scheme: 5(a)(i) amplitude = 0.60 m A1 5(a)(ii) oscillations are simple harmonic B1 5(b) Any three points from: B3 • mean / equilibrium position is at h = 1.4 m • total energy of oscillations = 9.0 J • angular frequency of oscillations = 1.2 rad s–1 or period of oscillations = 5.1 s or frequency of oscillation = 0.19 Hz • maximum speed of block = 0.73 m s–1 • mass of block = 33 kg 5(c) U-shaped curve resting on h axis (with minimum at EP = 0) B1 curve from h = 0.8 m to h = 2.0 m, with minimum EP shown at h = 1.4 m B1 both end-points of curve shown at EP = 9.0 J B1

This question in 9702/43 May/June 2025