Damped and forced oscillations, resonance — Paper 4 · A Level Physics 9702

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

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

4 (a) Explain what is meant by the natural frequency of vibration of a system. … … … [1] (b) A block of metal is fixed to one end of a vertical spring. The other end of the spring is attached to an oscillator, as shown in Fig. 4.1. oscillator spring metal block Fig. 4.1 The amplitude of oscillation of the oscillator is constant. The variation of the amplitude x0 of the oscillations of the block with frequency f of the oscillations is shown in Fig. 4.2. x0 0 f Fig. 4.2 (i) Name the effect shown in Fig. 4.2. … [1] (ii) State and explain whether the block is undergoing damped oscillations. … … … [2] (c) State one example in which the effect shown in Fig. 4.2 is useful. … … [1] [Total: 5]

5 marks

Mark scheme: 4(a) frequency at which body will vibrate when there is no (resultant external) resistive force acting on it OR frequency at which body will vibrate when there is no driving force / external force acting on it B1 4(b)(i) resonance B1 4(b)(ii) peak is not sharp / peak not infinite height M1 so damped A1 4(c) e.g. (quartz crystal) to produce ultrasound (quartz crystal) in watch to keep timing NMR / MRI microwave ovens tuning circuits B1

This question in 9702/42 Feb/March 2018

4 (a) State two conditions necessary for a mass to be undergoing simple harmonic motion. 1. … … 2. … … [2] (b) A trolley of mass 950 g is held on a horizontal surface by means of two springs attached to fixed points P and Q, as shown in Fig. 4.1. trolley mass 950 g spring P Q Fig. 4.1 The springs, each having a spring constant k of 230 N m–1, are always extended. The trolley is displaced along the line of the springs and then released. The variation with time t of the displacement x of the trolley is shown in Fig. 4.2. x 0 0 t1 t Fig. 4.2 (i) 1. State and explain whether the oscillations of the trolley are heavily damped, critically damped or lightly damped. … … 2. Suggest the cause of the damping. … … … [3] (ii) The acceleration a of the trolley of mass m may be assumed to be given by the expression 2 k a = – x . d m n 1. Calculate the angular frequency ω of the oscillations of the trolley. ω = … rad s–1 [3] 2. Determine the time t1 shown on Fig. 4.2. t1 = … s [2] [Total: 10]

10 marks

Mark scheme: 4(a) acceleration proportional to displacement B1 acceleration directed towards fixed point or displacement and acceleration in opposite directions B1 4(b)(i) 1. amplitude decreases gradually so light damping or oscillations continue so light damping B1 2. loss of energy B1 due to friction in wheels or due to friction between wheels and surface (during slipping) or due to air resistance (on trolley) B1 4(b)(ii)1. ω2 = 2k / m C1 = (2 × 230) / 0.950 C1 ω = 22 rad s–1 A1 4(b)(ii)2. T = 2π / ω C1 T = (2π / 22) = 0.286 s time = 1.5T = 0.43 s A1

This question in 9702/42 May/June 2018

3 A U-tube contains liquid, as shown in Fig. 3.1. x liquid x liquid L Fig. 3.1 Fig. 3.2 The total length of the column of liquid in the tube is L. The column of liquid is displaced so that the change in height of the liquid in each arm of the U-tube is x, as shown in Fig. 3.2. The liquid in the U-tube then oscillates with simple harmonic motion such that the acceleration a of the column is given by the expression 2 g a = – x e L o where g is the acceleration of free fall. (a) Calculate the period T of oscillation of the liquid column for a column length L of 19.0 cm. T = … s [3] (b) The variation with time t of the displacement x is shown in Fig. 3.3. +2.0 x / cm +1.0 0 0 T 2T 3T t –1.0 –2.0 Fig. 3.3 The period of oscillation of the liquid column of mass 18.0 g is T. The oscillations are damped. (i) Suggest one cause of the damping. … … [1] (ii) Calculate the loss in total energy of the oscillations during the first 2.5 periods of the oscillations. energy loss = … J [3] [Total: 7]

7 marks

Mark scheme: 3(a) C1 T = 2π / ω C1 ω2 = (2 × 9.81) / 0.19 ω = 10.2 (rad s–1) T = 2π / 10.2 = 0.62 s A1 3(b)(i) e.g. viscosity of liquid/friction within the liquid/viscous drag/friction between walls of tube and liquid B1 3(b)(ii) (maximum) KE = ½mv0 2 and v0 = ωx0 or energy = ½mω2x0 2 C1 change = ½ × 18 × 10–3 × 103 × [(2.0 × 10–2)2 – (0.95 ×10–2)2] C1 = 2.9 × 10–4 J A1

This question in 9702/41 Oct/Nov 2018

4 A U-tube contains liquid, as shown in Fig. 4.1. x x liquid L Fig. 4.1 Fig. 4.2 The total length of the liquid column is L. The column of liquid is displaced so that the change in height of the liquid level from the equilibrium position in each arm of the U-tube is x, as shown in Fig. 4.2. The liquid in the U-tube then oscillates such that its acceleration a is given by the expression 2 g a x =-d L n where g is the acceleration of free fall. (a) Show that the liquid column undergoes simple harmonic motion. [2] (b) The variation with time t of the displacement x is shown in Fig. 4.3. +2.0 x / cm +1.0 0 0 0.25 0.50 0.75 1.00 1.25 1.50 t / s –1.0 –2.0 Fig. 4.3 Use data from Fig. 4.3 to determine the length L of the liquid column. L = … m [3] (c) The oscillations shown in Fig. 4.3 are damped. (i) Suggest one cause of this damping. … … [1] (ii) Calculate the ratio total energy of oscillations after 1.5 complete oscillations total initial energy of oscillations ratio = … [2] [Total: 8]

8 marks

Mark scheme: 4(a) B1 g and L are constant (so a ∝ –x and hence s.h.m.) B1 4(b) T = 0.50 s and T = 2π / ω C1 ω2 = 2g / L C1 L = (2 × 9.81 × 0.502) / 4π2 = 0.12 m A1 4(c)(i) Any one from: • viscosity of liquid • friction within the liquid • viscous drag • friction/resistance between walls of tube and liquid B1 4(c)(ii) (maximum) KE = ½mv0 2 and v0 = ωx0 or energy = ½mω2x0 2 C1 ratio = (1.3 / 2.0)2 = 0.42 A1

This question in 9702/42 Oct/Nov 2018

3 A U-tube contains liquid, as shown in Fig. 3.1. x liquid x liquid L Fig. 3.1 Fig. 3.2 The total length of the column of liquid in the tube is L. The column of liquid is displaced so that the change in height of the liquid in each arm of the U-tube is x, as shown in Fig. 3.2. The liquid in the U-tube then oscillates with simple harmonic motion such that the acceleration a of the column is given by the expression 2 g a = – x e L o where g is the acceleration of free fall. (a) Calculate the period T of oscillation of the liquid column for a column length L of 19.0 cm. T = … s [3] (b) The variation with time t of the displacement x is shown in Fig. 3.3. +2.0 x / cm +1.0 0 0 T 2T 3T t –1.0 –2.0 Fig. 3.3 The period of oscillation of the liquid column of mass 18.0 g is T. The oscillations are damped. (i) Suggest one cause of the damping. … … [1] (ii) Calculate the loss in total energy of the oscillations during the first 2.5 periods of the oscillations. energy loss = … J [3] [Total: 7]

7 marks

Mark scheme: 3(a) C1 T = 2π / ω C1 ω2 = (2 × 9.81) / 0.19 ω = 10.2 (rad s–1) T = 2π / 10.2 = 0.62 s A1 3(b)(i) e.g. viscosity of liquid/friction within the liquid/viscous drag/friction between walls of tube and liquid B1 3(b)(ii) (maximum) KE = ½mv0 2 and v0 = ωx0 or energy = ½mω2x0 2 C1 change = ½ × 18 × 10–3 × 103 × [(2.0 × 10–2)2 – (0.95 ×10–2)2] C1 = 2.9 × 10–4 J A1

This question in 9702/43 Oct/Nov 2018

4 A ball of mass M is held on a horizontal surface by two identical extended springs, as illustrated in Fig. 4.1. ball mass M oscillator fixed point Fig. 4.1 One spring is attached to a fixed point. The other spring is attached to an oscillator. The oscillator is switched off. The ball is displaced sideways along the axis of the springs and is then released. The variation with time t of the displacement x of the ball is shown in Fig. 4.2. 1.5 x / cm 1.0 0.5 0 0 0.4 0.8 1.2 1.6 2.0 2.4 –0.5 t / s –1.0 –1.5 Fig. 4.2 (a) State: (i) what is meant by damping … … [1] (ii) the evidence provided by Fig. 4.2 that the motion of the ball is damped. … … [1] (b) The acceleration a and the displacement x of the ball are related by the expression 2 k a = – x c M m where k is the spring constant of one of the springs. The mass M of the ball is 1.2 kg. (i) Use data from Fig. 4.2 to determine the angular frequency ω of the oscillations of the ball. ω = … rad s–1 [2] (ii) Use your answer in (i) to determine the value of k. k = … N m–1 [2] (c) The oscillator is switched on. The amplitude of oscillation of the oscillator is constant. The angular frequency of the oscillations is gradually increased from 0.7ω to 1.3ω, where ω is the angular frequency calculated in (b)(i). (i) On the axes of Fig. 4.3, show the variation with angular frequency of the amplitude A of oscillation of the ball. A 0 0.7ω 1.0ω 1.3ω angular frequency Fig. 4.3 [2] (ii) Some sand is now sprinkled on the horizontal surface. The angular frequency of the oscillations is again gradually increased from 0.7ω to 1.3ω. State two changes that occur to the line you have drawn on Fig. 4.3. 1. … … 2. … … [2] [Total: 10]

10 marks

Mark scheme: 4(a)(i) loss of energy B1 4(a)(ii) amplitude (of oscillations) decreases (with time) B1 4(b)(i) ω = 2π / T C1 T = 0.80 s, so ω = 2π / 0.80 ω = 7.9 rad s–1 A1 4(b)(ii) ω2 = 2k / M C1 7.92 = 2k / 1.2 k = 37 N m–1 A1 4(c)(i) (one) smooth curve, not touching the f-axis, with two concave sides meeting at a peak in between them B1 (one) peak at 1.0ω B1 4(c)(ii) • lower peak/(whole) line is lower • flatter peak/peak is less sharp • peak at (slightly) lower angular frequency/peak moves to left any two points, one mark each B2

This question in 9702/42 Oct/Nov 2019

4 A dish is made from a section of a hollow glass sphere. The dish, fixed to a horizontal table, contains a small solid ball of mass 45 g, as shown in Fig. 4.1. ball surface mass 45 g of dish x C Fig. 4.1 The horizontal displacement of the ball from the centre C of the dish is x. Initially, the ball is held at rest with distance x = 3.0 cm. The ball is then released. The variation with time t of the horizontal displacement x of the ball from point C is shown in Fig. 4.2. 4 3 x / cm 2 1 0 0 1 2 3 4 5 6 7 t / s –1 –2 –3 –4 Fig. 4.2 The motion of the ball in the dish is simple harmonic with its acceleration a given by the expression a = x –(gR) where g is the acceleration of free fall and R is a constant that depends on the dimensions of the dish and the ball. (a) Use Fig. 4.2 to show that the angular frequency ω of oscillation of the ball in the dish is 2.9 rad s–1. [1] (b) Use the information in (a) to: (i) determine R R = … m [2] (ii) calculate the speed of the ball as it passes over the centre C of the dish. speed = … m s–1 [2] (c) Some moisture collects on the surface of the dish so that the motion of the ball becomes lightly damped. On the axes of Fig. 4.2, draw a line to show the lightly damped motion of the ball for the first 5.0 s after the release of the ball. [3] [Total: 8]

8 marks

Mark scheme: 4(a) (ω = 2π / T and T = 2.2 s so) ω = 2π / 2.2 = 2.9 rad s–1 4(b)(i) ω2 = g / R C1 R = 9.81 / 2.862 = 1.2 m A1 4(b)(ii) v0 = ωx0 C1 = 2.9 × 3.0 × 10–2 = 0.087 m s–1 A1 4(c) smooth wave starting at 3.0 cm when t = 0 B1 positions of peaks and troughs show same period (or slightly longer) B1 each peak and trough at lower amplitude than the previous one B1

This question in 9702/42 May/June 2020

3 (a) State what is meant by simple harmonic motion. … … … [2] (b) A trolley of mass m is held on a horizontal surface by means of two springs. One spring is attached to a fixed point P. The other spring is connected to an oscillator, as shown in Fig. 3.1. spring trolley spring oscillator P Fig. 3.1 The springs, each having spring constant k of 130 N m−1, are always extended. The oscillator is switched off. The trolley is displaced along the line of the springs and then released. The resulting oscillations of the trolley are simple harmonic. The acceleration a of the trolley is given by the expression ⎛ ⎞2k a = − x ⎝ ⎠m where x is the displacement of the trolley from its equilibrium position. The mass of the trolley is 840 g. Calculate the frequency f of oscillation of the trolley. f = … Hz [3] (c) The oscillator in (b) is switched on. The frequency of oscillation of the oscillator is varied, keeping its amplitude of oscillation constant. The amplitude of oscillation of the trolley is seen to vary. The amplitude is a maximum at the frequency calculated in (b). (i) State the name of the effect giving rise to this maximum. … [1] (ii) At any given frequency, the amplitude of oscillation of the trolley is constant. Explain how this indicates that there are resistive forces opposing the motion of the trolley. … … … [2] [Total: 8]

8 marks

Mark scheme: 3(a) acceleration (directly) proportional to displacement B1 acceleration is in opposite direction to displacement B1 3(b) ω2 = 2k / m and ω = 2πf C1 (2πf)2 = (2 × 130) / 0.84 C1 f = 2.8 Hz A1 3(c)(i) resonance B1 3(c)(ii) oscillator supplies energy (continuously) B1 energy of trolley constant so energy must be dissipated or without loss of energy the amplitude would continuously increase B1

This question in 9702/41 May/June 2021

3 (a) State what is meant by simple harmonic motion. … … … [2] (b) A trolley of mass m is held on a horizontal surface by means of two springs. One spring is attached to a fixed point P. The other spring is connected to an oscillator, as shown in Fig. 3.1. spring trolley spring oscillator P Fig. 3.1 The springs, each having spring constant k of 130 N m−1, are always extended. The oscillator is switched off. The trolley is displaced along the line of the springs and then released. The resulting oscillations of the trolley are simple harmonic. The acceleration a of the trolley is given by the expression ⎛ ⎞2k a = − x ⎝ ⎠m where x is the displacement of the trolley from its equilibrium position. The mass of the trolley is 840 g. Calculate the frequency f of oscillation of the trolley. f = … Hz [3] (c) The oscillator in (b) is switched on. The frequency of oscillation of the oscillator is varied, keeping its amplitude of oscillation constant. The amplitude of oscillation of the trolley is seen to vary. The amplitude is a maximum at the frequency calculated in (b). (i) State the name of the effect giving rise to this maximum. … [1] (ii) At any given frequency, the amplitude of oscillation of the trolley is constant. Explain how this indicates that there are resistive forces opposing the motion of the trolley. … … … [2] [Total: 8]

8 marks

Mark scheme: 3(a) acceleration (directly) proportional to displacement B1 acceleration is in opposite direction to displacement B1 3(b) ω2 = 2k / m and ω = 2πf C1 (2πf)2 = (2 × 130) / 0.84 C1 f = 2.8 Hz A1 3(c)(i) resonance B1 3(c)(ii) oscillator supplies energy (continuously) B1 energy of trolley constant so energy must be dissipated or without loss of energy the amplitude would continuously increase B1

This question in 9702/43 May/June 2021

4 (a) State what is meant by resonance. … … … [2] (b) Fig. 4.1 shows a heavy pendulum and a light pendulum, both suspended from the same piece of string. This string is secured at each end to fixed points. fixed points string heavy pendulum light pendulum Fig. 4.1 Both pendulums have the same natural frequency. The heavy pendulum is set oscillating perpendicular to the plane of the diagram. As it oscillates, it causes the light pendulum to oscillate. Fig. 4.2 shows the variation with time t of the displacements of the two pendulums for three oscillations. heavy displacement / cm 0 light 0 t / s Fig. 4.2 The variation with t of the displacement x of the light pendulum is given by x = 0.25 sin 5.0rt where x is in centimetres and t is in seconds. (i) Calculate the period T of the oscillations. T = … s [2] (ii) On Fig. 4.2, label both of the axes with the correct scales. Use the space below for any additional working that you need. [2] (iii) Determine the magnitude of the phase difference φ between the oscillations of the light and heavy pendulums. Give a unit with your answer. φ = … unit … [2] [Total: 8]

8 marks

Mark scheme: 4(a) oscillations (of object) at maximum amplitude B1 when driving frequency equals natural frequency (of object) B1 4(b)(i) T = 2 /  C1 = 2 / 5.0 = 0.40 s A1 4(b)(ii) displacement scale labelled –1.0, –0.5, (0), 0.5, 1.0 on the 2 cm tick marks B1 t scale labelled 0.2, 0.4, 0.6, 0.8, 1.0, 1.2 on the 2 cm tick marks B1 4(b)(iii) ϕ = 2t / T = 2  0.10 / 0.40 or 2  0.30 / 0.40 C1 = 1.6 rad or 4.7 rad A1

This question in 9702/42 May/June 2022

3 An object is suspended from a spring that is attached to a fixed point as shown in Fig. 3.1. fixed point spring object oscillations equilibrium position Fig. 3.1 The object oscillates vertically with simple harmonic motion about its equilibrium position. (a) State the defining equation for simple harmonic motion. Identify the meaning of each of the symbols used to represent physical quantities. … … … [2] (b) The variation with displacement x from the equilibrium position of the velocity v of the object is shown in Fig. 3.2. 0.2 v / m s–1 0.1 0 – 0.12 – 0.08 – 0.04 0 0.04 0.08 0.12 x / m –– 0.20.1 – 0.2 Fig. 3.2 The variation with x of the potential energy EP of the oscillations of the object is shown in Fig. 3.3. 0.050 EP / J 0.025 0 – 0.12 – 0.08 – 0.04 0 0.04 0.08 0.12 x / m Fig. 3.3 Use Fig. 3.2 and Fig. 3.3 to: (i) determine the amplitude x0 of the oscillations x0 = … m [1] (ii) show that the angular frequency of the oscillations is 1.7 rad s–1 [2] (iii) determine the mass M of the object. M = … kg [2] (c) The oscillations of the object are now lightly damped. (i) State what is meant by damping. … … … [2] (ii) Assume that the damping does not change the angular frequency of the oscillations. On Fig. 3.2, sketch the variation with x of v when the amplitude of the oscillations is 0.060 m. [2] [Total: 11]

11 marks

Mark scheme: 3(a) a = –  2x M1 a = acceleration, x = displacement from equilibrium position and = angular frequency A1 3(b)(i) x0 = 0.12 m A1 3(b)(ii) v = (x02 – x2) C1 two (x, v) pairs correctly read from Fig. 3.2 (one may be (x0, 0) or value of x0 from (i)) e.g. 0.20 = (0.122 – 0) leading to = 1.7 rad s–1 A1 3(b)(iii) E = ½M 2x02 C1 0.050 = ½  M  1.672  0.122 A1 M = 2.5 kg or (EK)max = ½Mv02 (C1) 0.050 = ½ M  0.202 (A1) M = 2.5 kg 3(c)(i) loss of (total) energy (of system) B1 due to resistive forces B1 3(c)(ii) closed loop surrounding the origin with maximum x at ± 0.060 m passing through v = 0 B1 maximum velocity shown as ± 0.10 m s–1 passing through x = 0 B1

This question in 9702/41 Oct/Nov 2022

3 An object is suspended from a spring that is attached to a fixed point as shown in Fig. 3.1. fixed point spring object oscillations equilibrium position Fig. 3.1 The object oscillates vertically with simple harmonic motion about its equilibrium position. (a) State the defining equation for simple harmonic motion. Identify the meaning of each of the symbols used to represent physical quantities. … … … [2] (b) The variation with displacement x from the equilibrium position of the velocity v of the object is shown in Fig. 3.2. 0.2 v / m s–1 0.1 0 – 0.12 – 0.08 – 0.04 0 0.04 0.08 0.12 x / m –– 0.20.1 – 0.2 Fig. 3.2 The variation with x of the potential energy EP of the oscillations of the object is shown in Fig. 3.3. 0.050 EP / J 0.025 0 – 0.12 – 0.08 – 0.04 0 0.04 0.08 0.12 x / m Fig. 3.3 Use Fig. 3.2 and Fig. 3.3 to: (i) determine the amplitude x0 of the oscillations x0 = … m [1] (ii) show that the angular frequency of the oscillations is 1.7 rad s–1 [2] (iii) determine the mass M of the object. M = … kg [2] (c) The oscillations of the object are now lightly damped. (i) State what is meant by damping. … … … [2] (ii) Assume that the damping does not change the angular frequency of the oscillations. On Fig. 3.2, sketch the variation with x of v when the amplitude of the oscillations is 0.060 m. [2] [Total: 11]

11 marks

Mark scheme: 3(a) a = –  2x M1 a = acceleration, x = displacement from equilibrium position and = angular frequency A1 3(b)(i) x0 = 0.12 m A1 3(b)(ii) v = (x02 – x2) C1 two (x, v) pairs correctly read from Fig. 3.2 (one may be (x0, 0) or value of x0 from (i)) e.g. 0.20 = (0.122 – 0) leading to = 1.7 rad s–1 A1 3(b)(iii) E = ½M 2x02 C1 0.050 = ½  M  1.672  0.122 A1 M = 2.5 kg or (EK)max = ½Mv02 (C1) 0.050 = ½ M  0.202 (A1) M = 2.5 kg 3(c)(i) loss of (total) energy (of system) B1 due to resistive forces B1 3(c)(ii) closed loop surrounding the origin with maximum x at ± 0.060 m passing through v = 0 B1 maximum velocity shown as ± 0.10 m s–1 passing through x = 0 B1

This question in 9702/43 Oct/Nov 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 heavy metal sphere of mass 0.81 kg is suspended from a string. The sphere is undergoing small oscillations from side to side, as shown in Fig. 4.1. string heavy sphere, mass 0.81 kg oscillations Fig. 4.1 The oscillations of the sphere may be considered to be simple harmonic with amplitude 0.036 m and period 3.0 s. (a) State what is meant by simple harmonic motion. … … … [2] (b) Calculate: (i) the angular frequency of the oscillations angular frequency = … rad s–1 [2] (ii) the total energy of the oscillations. total energy = … J [2] (c) The suspended sphere is now lowered into water. The sphere is given a sideways displacement of +0.036 m from its equilibrium position and is then released at time t = 0. The water causes the motion of the sphere to be critically damped. On Fig. 4.2, sketch the variation of the displacement x of the sphere from its equilibrium position with t from t = 0 to t = 6.0 s. 0.04 x / m 0.02 0 0 1 2 3 4 5 6 t / s – 0.02 – 0.04 Fig. 4.2 [3] [Total: 9]

9 marks

Mark scheme: 4(a) (motion in which) acceleration is (directly) proportional to displacement B1 (motion in which): B1 acceleration is (always) in the opposite direction to displacement or acceleration is (always) directed towards a fixed point 4(b)(i) = 2 / T C1  = 2 / 3.0 A1 = 2.1 rad s–1 4(b)(ii) E = ½m2x02 C1 = ½  0.81  2.12  0.0362 A1 = 2.3  10–3 J 4(c) sketch: line starting at (0, 0.036) and not reaching x =  0.036 m at any other time B1 smooth curve, with no sudden changes in gradient, showing continuously decreasing magnitude of x from B1 maximum displacement at t = 0 to final displacement of zero where the gradient is also zero displacement reaches final value of zero between t = 0.75 s and t = 3.0 s at the latest B1

This question in 9702/41 Oct/Nov 2023

4 A heavy metal sphere of mass 0.81 kg is suspended from a string. The sphere is undergoing small oscillations from side to side, as shown in Fig. 4.1. string heavy sphere, mass 0.81 kg oscillations Fig. 4.1 The oscillations of the sphere may be considered to be simple harmonic with amplitude 0.036 m and period 3.0 s. (a) State what is meant by simple harmonic motion. … … … [2] (b) Calculate: (i) the angular frequency of the oscillations angular frequency = … rad s–1 [2] (ii) the total energy of the oscillations. total energy = … J [2] (c) The suspended sphere is now lowered into water. The sphere is given a sideways displacement of +0.036 m from its equilibrium position and is then released at time t = 0. The water causes the motion of the sphere to be critically damped. On Fig. 4.2, sketch the variation of the displacement x of the sphere from its equilibrium position with t from t = 0 to t = 6.0 s. 0.04 x / m 0.02 0 0 1 2 3 4 5 6 t / s – 0.02 – 0.04 Fig. 4.2 [3] [Total: 9]

9 marks

Mark scheme: 4(a) (motion in which) acceleration is (directly) proportional to displacement B1 (motion in which): B1 acceleration is (always) in the opposite direction to displacement or acceleration is (always) directed towards a fixed point 4(b)(i) = 2 / T C1  = 2 / 3.0 A1 = 2.1 rad s–1 4(b)(ii) E = ½m2x02 C1 = ½  0.81  2.12  0.0362 A1 = 2.3  10–3 J 4(c) sketch: line starting at (0, 0.036) and not reaching x =  0.036 m at any other time B1 smooth curve, with no sudden changes in gradient, showing continuously decreasing magnitude of x from B1 maximum displacement at t = 0 to final displacement of zero where the gradient is also zero displacement reaches final value of zero between t = 0.75 s and t = 3.0 s at the latest B1

This question in 9702/43 Oct/Nov 2023

4 (a) State what is meant by resonance. … … … [2] (b) A small ball is held in place using a stretched string. One end of the string is fixed to a wall and the other end is attached to a vibration generator, as shown in Fig. 4.1. ball wall vibration generator string Fig. 4.1 Initially, the vibration generator is switched off. A student displaces the ball vertically and then releases it. Fig. 4.2 shows the variation of the displacement of the ball with time after it is released. displacement 0 0 0.1 0.2 0.3 0.4 0.5 0.6 time / s Fig. 4.2 (i) State the name of the phenomenon illustrated by the decrease in the amplitude of the oscillations in Fig. 4.2. … [1] (ii) Explain the decrease with time of the amplitude of the oscillations of the ball. … … … [2] (iii) Determine the frequency of the oscillations of the ball. frequency = … Hz [1] (c) The vibration generator in (b) is switched on and its frequency f of vibration is gradually increased from 0 to 10 Hz. On Fig. 4.3, sketch the variation with f of the amplitude of the oscillations of the ball. amplitude 0 0 2.5 5.0 7.5 10.0 f / Hz Fig. 4.3 [2] [Total: 8]

8 marks

Mark scheme: 4(a) oscillation (of object) at maximum amplitude B1 when driving frequency = natural frequency (of system) B1 4(b)(i) light damping B1 4(b)(ii) oscillations (of ball) lose energy B1 (due to) resistive forces (acting on ball) B1 4(b)(iii) frequency = 1 / 0.25 = 4.0 Hz A1 4(c) curve showing a maximum amplitude at a single non-zero frequency B1 single maximum amplitude shown at 4.0 Hz B1

This question in 9702/41 May/June 2024

4 (a) State what is meant by resonance. … … … [2] (b) A small ball is held in place using a stretched string. One end of the string is fixed to a wall and the other end is attached to a vibration generator, as shown in Fig. 4.1. ball wall vibration generator string Fig. 4.1 Initially, the vibration generator is switched off. A student displaces the ball vertically and then releases it. Fig. 4.2 shows the variation of the displacement of the ball with time after it is released. displacement 0 0 0.1 0.2 0.3 0.4 0.5 0.6 time / s Fig. 4.2 (i) State the name of the phenomenon illustrated by the decrease in the amplitude of the oscillations in Fig. 4.2. … [1] (ii) Explain the decrease with time of the amplitude of the oscillations of the ball. … … … [2] (iii) Determine the frequency of the oscillations of the ball. frequency = … Hz [1] (c) The vibration generator in (b) is switched on and its frequency f of vibration is gradually increased from 0 to 10 Hz. On Fig. 4.3, sketch the variation with f of the amplitude of the oscillations of the ball. amplitude 0 0 2.5 5.0 7.5 10.0 f / Hz Fig. 4.3 [2] [Total: 8]

8 marks

Mark scheme: 4(a) oscillation (of object) at maximum amplitude B1 when driving frequency = natural frequency (of system) B1 4(b)(i) light damping B1 4(b)(ii) oscillations (of ball) lose energy B1 (due to) resistive forces (acting on ball) B1 4(b)(iii) frequency = 1 / 0.25 = 4.0 Hz A1 4(c) curve showing a maximum amplitude at a single non-zero frequency B1 single maximum amplitude shown at 4.0 Hz B1

This question in 9702/43 May/June 2024

7 A bar magnet is suspended from a spring. One pole of the magnet oscillates freely in a coil of wire, as shown in Fig. 7.1. spring bar magnet resistor coil of wire S Fig. 7.1 The switch S is initially open. (a) The switch S is now closed. As a result, the oscillations of the magnet are lightly damped. (i) State what is meant by damping. … … … [2] (ii) Describe what is observed to indicate that the damping is light. … … [1] (iii) By reference to electromagnetic induction and to conservation of energy, explain why the oscillations are damped. … … … … [3] (b) The procedure in (a) is repeated after replacing the resistor with one of greater resistance. Suggest, with a reason, the effect of this change on the oscillations. … … … [2] [Total: 8]

8 marks

Mark scheme: 7(a)(i) loss of energy (of oscillations) B1 due to resistive forces B1 7(a)(ii) amplitude (of oscillations) decreases gradually B1 or oscillations continue (for several periods) 7(a)(iii) cutting of (magnetic) flux causes induced e.m.f. (in coil) B1 or induced e.m.f. causes current (in resistor / circuit) current causes dissipation of thermal energy in resistor B1 or current in resistor causes dissipation of thermal energy thermal energy comes from energy of oscillations B1 7(b) (for any given e.m.f.) current is lower so thermal energy dissipated at a lower rate B1 or (for any given e.m.f.) current is lower so the resistive force is lower oscillations are less damped B1 or smaller decrease in amplitude of oscillations (in each period) or oscillations continue for longer

This question in 9702/44 May/June 2025

5 A steel ball on the end of a thin string oscillates with small oscillations, as shown in Fig. 5.1. thin string equilibrium position steel ball x oscillations Fig. 5.1 (not to scale) The displacement of the centre of the ball from its equilibrium position is x. (a) Fig. 5.2 shows the variation with x of the acceleration a of the ball. 15 a / cm s–2 10 5 0 – 2 – 1 0 1 2 x / cm – 5 – 10 – 15 Fig. 5.2 (i) Explain how Fig. 5.2 shows that the oscillations of the ball are simple harmonic. … … … [2] (ii) Determine the period T of the oscillations. T = … s [3] (b) At time t = 0, when the displacement of the ball has its maximum value, the ball is immersed in a trough containing thick oil so that the ball is just below the surface of the oil. This results in the subsequent motion of the ball being heavily damped. (i) State what is meant by damping. … … … [2] (ii) On Fig. 5.3, sketch a possible variation of the displacement x of the ball with t between t = 0 and t = 2T. 1.5 1.0 x / cm 0.5 0 0 T 2T t – 0.5 – 1.0 – 1.5 Fig. 5.3 [3] [Total: 10]

10 marks

Mark scheme: 5(a)(i) straight line through the origin shows that a is proportional to x B1 negative gradient shows that a is always in the opposite direction to x B1 5(a)(ii) a0 = 2x0 C1 = 2 / T C1 T = 2 √(x0 / a0) A1 = 2 √ (1.2 / 13) = 1.9 s 5(b)(i) loss of energy of oscillations B1 due to resistive force(s) B1 5(b)(ii) line starting from x = 1.2 cm at t = 0 B1 line starting from non-zero value of x from t = 0 to t = 2T that is entirely either above or below the t-axis B1 curve from t = 0 starting from non-zero x value, with both magnitude of x value and magnitude of gradient continuously B1 decreasing

This question in 9702/42 Oct/Nov 2025