IBDP Physics 4.1 – Oscillations: IB Style Question Bank SL Paper 2


This question is about simple harmonic motion (SHM).

The graph shows the variation with time \(t\) of the acceleration \(a\) of an object X undergoing simple harmonic motion (SHM).


Define simple harmonic motion (SHM).[2]


X has a mass of 0.28 kg. Calculate the maximum force acting on X.[1]


Determine the maximum displacement of X. Give your answer to an appropriate number of significant figures.[4]


A second object Y oscillates with the same frequency as X but with a phase difference of \(\frac{\pi }{4}\). Sketch, using the graph opposite, how the acceleration of object Y varies with \(t\).[2]




force/acceleration proportional to the displacement/distance from a (fixed/equilibrium) point/mean position;

directed towards this (equilibrium) point / in opposite direction to displacement/ distance;

Allow algebra only if symbols are fully explained.


0.73 (N); (allow answer in range of 0.71 to 0.75 (N))


use of \({a_0} =  – {\omega ^2}{x_0}\);

\(T = 7.9{\text{ (s)}}\) or \(\omega  = 0.795\) or \(\frac{\pi }{4}{\text{ (rad}}\,{{\text{s}}^{ – 1}})\); } (allow answers in the range of T = 7.8 to 8.0 (s) or \(\omega \) = 0.785 to 0.805 (rad s–1))

\({x_0} = 4.1(1){\text{ (m)}}\); (allow answers in the range of 4.0 to 4.25 (m))

two significant figures in final answer whatever the value;

Award [4] for a bald correct answer.


shape correct, constant amplitude for new curve, minimum of 10 s shown; } (there must be some consistent lead or lag and no change in T)

lead/lag of 1 s (to within half a square by eye);


Simple harmonic motion and forced oscillations

The graph shows the variation with time of the displacement of an object undergoing simple harmonic motion.


(i) State the amplitude of the oscillation.

(ii) Calculate the frequency of the oscillation.[3]


(i) Determine the maximum speed of the object.

(ii) Determine the acceleration of the object at 140 ms.[4]


The graph below shows how the displacement of the object varies with time. Sketch on the same axes a line indicating how the kinetic energy of the object varies with time.

You should ignore the actual values of the kinetic energy.





(i) 32 (mm);

(ii) period = 160 (ms);

frequency = 6.2/6.3 (Hz);

Allow ECF for incorrect period.


(i) ω=2π×6.25;
v(=39.3×32×10-3)=1.3(ms-1); (allow ECF from (a))


tangent drawn to graph at a point of zero displacement;

gradient calculated between 1.2 and 1.4;

(ii) displacement = 23–26 (mm);

35–40 (ms-2);

23 mm found by calculating displacement


double frequency;

always positive and constant amplitude;

correct phase ie cosine squared;

Ignore amplitude value.

A minimum of one complete, original oscillation needed to award [3].


This question is in two parts. Part 1 is about simple harmonic motion (SHM) and waves. Part 2 is about wind power and the greenhouse effect.

Part 1 Simple harmonic motion (SHM) and waves

A gas is contained in a horizontal cylinder by a freely moving piston P. Initially P is at rest at the equilibrium position E.


The piston P is displaced a small distance A from E and released. As a result, P executes simple harmonic motion (SHM).
Define simple harmonic motion as applied to P.


The graph shows how the displacement x of the piston P in (a) from equilibrium varies with time t.

(i) State the value of the displacement A as defined in (a).

(ii) On the graph identify, using the letter M, a point where the magnitude of the acceleration of P is a maximum.

(iii) Determine, using data from the graph and your answer to (b)(i), the magnitude of the maximum acceleration of P.

(iv) The mass of P is 0.32 kg. Determine the kinetic energy of P at t=0.052 s.[7]


The oscillations of P initially set up a longitudinal wave in the gas.

(i) Describe, with reference to the transfer of energy, what is meant by a longitudinal wave.

(ii) The speed of the wave in the gas is 340 m s–1. Calculate the wavelength of the wave in the gas.[4]




the acceleration of piston/P is proportional to its displacement from equilibrium;
and directed towards equilibrium;
There must be a clear indication what is accelerating otherwise award [1 max].


(i) 12(cm); (accept –12)

(ii) any maximum or minimum of the graph;

(iii) period= 0.04 (s); (allow clear substitution of this value)
\(\omega = \left( {\frac{{2\pi }}{T} = } \right)\frac{{2 \times 3.14}}{{0.04}} = 157\left( {{\rm{rad }}{{\rm{s}}^{ – 1}}} \right)\)

maximum acceleration=(2=)0.12×1572=3.0×103(ms-2); (watch for ECF from wrong period)

(iv) at t=0.052sx=(-)4(±1)cm;
\({\rm{KE = }}\left( {\frac{1}{2}m{\omega ^2}\left[ {{A^2} – {x^2}} \right] = } \right)0.5 \times 0.32 \times {157^2}\left[ {{{0.12}^2} – {{0.04}^2}} \right] = 50\left( { \pm 7} \right)\left( {\rm{J}} \right)\);

Watch for incorrect use of cm.
Allow ECF from calculations in (b)(iii).
Do not retrospectively credit a mark for ω to (b)(iii) if it was not gained there on original marking.
Allow use of sin ωt to obtain v.
Award [2] for a bald correct answer.


(i) the direction of the oscillations/vibrations/movements of the particles (in the medium/gas);

for a longitudinal wave are parallel to the direction of the propagation of the energy of the wave;

(ii) \(f = \left( {\frac{1}{T} = } \right)\frac{1}{{0.04}} = 25\left( {{\rm{Hz}}} \right)\);
\(\lambda = \left( {\frac{v}{f} = } \right)\frac{{340}}{{25}} = 14\left( {\rm{m}} \right)\);

Award [1 max] if frequency is not clearly stated.
Allow ECF from calculations in (b)(iii).