IB Physics HL C.1 Simple Harmonic Motion Question Bank
Practise IB Physics HL C.1 by analysing oscillator equations, phase, energy and data-driven spring or pendulum motion at HL depth.
- Syllabus
- First assessment 2025
- Course
- Physics HL
- Level
- HL
Practise IB Physics HL C.1 by analysing oscillator equations, phase, energy and data-driven spring or pendulum motion at HL depth.
One end of a light spring is attached to a rigid horizontal support.

An object W of mass 0.15 kg is suspended from the other end of the spring. The extension x of the spring is proportional to the force F causing the extension. The force per unit extension of the spring k is 18Nm−1.
A student pulls W down such that the extension of the spring increases by 0.040 m . The student releases W and as a result W performs simple harmonic motion (SHM).
State what is meant by the expression "W performs SHM".
the acceleration of (force acting on) W is proportional to its displacement from equilibrium; and directed towards equilibrium;
Determine the period of oscillation of the spring.
ω=xa;
=10.95 rad s−2;
T=(ω2π=)10.956.28=0.57 s;
This question is in two parts. Part 1 is about simple harmonic motion (SHM). Part 2 is about gas in an engine.
Part 1 Simple harmonic motion (SHM)
An object is placed on a frictionless surface. The object is attached by a spring fixed at one end and oscillates at the end of the spring with simple harmonic motion (SHM).

The tension F in the spring is given by F=k x where x is the extension of the spring and k is a constant.
Show that ω2=mk.
m a=-k x;
a=−mkx;( condone lack of negative sign )
or
implied use of defining equation for simple harmonic motion a=−ω2x;
(so ω2=mk )
m a=-k x so a=−(mk)x;
One cycle of the variation of displacement with time is shown for two separate mass-spring systems, A and B .

Calculate the frequency of the oscillation of A.
0.833( Hz);
The springs used in A and B are identical. Show that the mass in A is equal to the mass in B.
frequency/period is the same so ω is the same;
k is the same (as springs are identical);
(so m is the same)
Outline how you would use the graph to confirm that A is performing simple harmonic motion.
defines simple harmonic motion as acceleration proportional to negative displacement /x=x0sinωt;
correct method for showing how graph leads to the definition;
The graph shows the variation of the potential energy of A with displacement.

On the axes,
draw a graph to show the variation of kinetic energy with displacement for the mass in A. Label this A.

correct shape;
maximum at 0.16 J ;
sketch a graph to show the variation of kinetic energy with displacement for the mass in B. Label this B.
end displacements correct ±0.01 m;
maximum lower than 0.16 J ;
maximum equal to 0.04 J± half square;
Using data from (b) and (c), calculate the mass in A.
maximum speed of oscillator (=ωx0=2π×0.833×0.2)=1.05( m s−1);}(N.B.ω=5.24rads−1)
maximum energy read-off =0.16( J);} (condone ECF for incorrectly drawn maximum kinetic energy value in (c)(i))
use to 21mv2 to give mass value ( =0.292 kg using correct data);
or
read-off from graph of kemax =0.16( J) and xmax =0.2( m); (both needed)
0.16=21m×(2πf)2×(0.2)2;
0.292 (kg); (use of 0.8 Hz gives 0.32)
Allow any alternative method if correct, eg: T from graph, k evaluated (8Nm−1).
Part 2 Gas in an engine
This question is in two parts.
A particle P moves with simple harmonic motion.
State, with reference to the motion of P , what is meant by simple harmonic motion.
the acceleration (of a particle/P) is (directly) proportional to displacement; and is directed towards equilibrium/in the opposite direction to displacement; Do not accept "directed towards the centre".
The graph shows how the velocity v of particle P varies with time t.

Use the graph opposite to determine for the motion of P the
period.
0.30 s ;
amplitude.
Marking guidance:
max velocity =0.74(±0.02)ms−1;
recognize max velocity =ωx0;
ω=(T2π=0.302π=)20.9rads−1;x0=(20.90.74=)3.5(±0.2)×10−2 m;
or
identifies displacement with area; uses one quarter of a cycle;
answer in the range of 30 to 40 mm ;
answer in the range of 33 to 37 mm ;
displacement of P from equilibrium at t=0.2 s.
v=0.64(±0.2)ms−1;
use v=ω(x02−x2) to get x=1.7(±0.2)×10−2 m;
or
recognition that x=x0cosωt;
x(=35cos[0.32π×0.2])=17.5 mm;