IB Physics HL C.1 Simple harmonic motion Questions

SLSL HLHL

Practise IB Physics HL C.1 by testing a = −ω²x, solving phase-dependent motion equations, relating period to frequency, and reading SHM energy graphs.

147 questions in this syllabus scope, with question parts counted together.

Exam points & syllabus

Exam points

Identify SHM by a restoring acceleration proportional to displacement from equilibrium and opposite in direction: a = −ω²x.
From a supplied restoring force, use F = ma to obtain a = −Cx, identify ω² = C, and state why the resulting motion is SHM.
Compare a force model with linear SHM only over its valid displacement range; a softer nonlinear outer region gives slower motion and lower stored energy.
Keep displacement and acceleration signs opposite, and distinguish the accelerating object from a fixed reference point in a diagram.
On an a–x graph, SHM is a straight line through the origin with gradient −ω²; convert displacement-axis units before finding ω.
Use amax = ω²A with displacement amplitude in metres; distinguish angular frequency from f before squaring.
Read amplitude from equilibrium to an extreme, and find a full period between equivalent states with the same direction of motion.
Find displacement amplitude from vmax = ωA, or from the area under a velocity graph over a suitable fraction of the cycle.
For a vertically loaded spring, distinguish extension from natural length from displacement about equilibrium when extracting oscillation amplitude.
Use KE = 0 at each extreme to obtain amplitude from a KE–x graph; energy is maximum at equilibrium.
Use f = 1/T and ω = 2πf consistently; obtain period from the square root of a supplied restoring-force or a–x coefficient.
For regular repetitions, use time = number of cycles/f and identify the correct event spacing before converting speed into repetition frequency.
For a given sampling rate, use Δt = 1/f with kHz converted to Hz; distinguish the interval per sample from the time per bit.
Relate a wave’s speed and wavelength to f, then use ω = 2πf for the particle; travelling-wave speed differs from particle oscillation speed.
For release from an extreme, the first arrival at equilibrium takes T/4; keep the restoring-force model within the stage where it remains valid.
For the same spring, mass scales as T²; at equal period, spring stiffness scales with mass, and identical stiffness and period imply equal masses.
Sum the restoring contributions when two opposing springs act on a displaced mass, then use T = 2π√(m/keffective).
A spring oscillator’s period is independent of g in the ideal model, whereas a simple pendulum has T ∝ 1/√g.
For a small-angle simple pendulum, use T ∝ √l and mass independence; reducing l to a quarter gives half the period even if bob mass changes.
When comparing pendulum and spring periods after a mass change, apply each model separately before forming the requested ratio.
A kinetic-energy graph repeats twice per mechanical oscillation; distinguish its repeat interval from the full motion period.
Read velocity direction from the sign of the displacement–time gradient; maximum speed occurs at equilibrium crossings.
On a velocity–time trace, displacement extrema occur at v = 0; a downward crossing identifies the positive displacement maximum.
Displacement and acceleration are opposite in phase, while velocity is a quarter-cycle from displacement; v and a share direction for half a cycle.
Sketch velocity after release from an extreme with v = 0 initially and sinusoidal variation; use the stated amplitude, period and damping.
Calculate maximum speed with vmax = ωA, or with √(2EK,max/m) when maximum kinetic energy is explicitly identified.
Select the phase from both initial position and subsequent direction: release at −A corresponds to x = −A cos ωt and v = +ωA sin ωt.
For x = A sin(ωt + φ), calculate v = ωA cos(ωt + φ) in radians with the same phase and retain its sign.
Convert a time shift to phase using Δφ = 2πΔt/T, identify the earlier matching trace as leading, and express phase calculations in radians.
Sketch a phase shift horizontally while preserving period and sinusoidal shape; an unsigned phase difference may allow either lead or lag.
During ideal SHM, kinetic and potential energy exchange while total mechanical energy stays constant; use the correct physical energy stores.
Use ET = ½mω²A² and Ep = ½mω²x² for SHM energy relative to equilibrium, with EK = ET − Ep.
At x = A/2, SHM potential energy is ET/4 and kinetic energy is 3ET/4; identify whether a stated E denotes total or kinetic energy.
Measure displacement from equilibrium before using Ep/ET = x²/A²; a point 0.20A back from an extreme is at x = 0.80A.
At fixed spring stiffness, total oscillation energy scales with A² and is independent of the mass; separate this from the period’s mass dependence.
At fixed mass and amplitude, SHM total energy scales with f² or 1/T²; multiplying period by five reduces energy to one twenty-fifth.
To double oscillation energy at fixed system parameters, increase amplitude by √2; maximum speed scales with amplitude.
When KE = Ep, |x| = A/√2; two positions are visited twice each, so equality occurs four times in one full oscillation.
Draw Ep–x as an upward parabola and EK–x as an inverted parabola, meeting total energy at the appropriate extrema.
Energy–time sketches stay nonnegative and repeat twice per motion cycle; set phase from the starting displacement or velocity.
For damped energy sketches, use the given envelope: potential-energy maxima touch total energy and minima reach zero at the stated half-cycle spacing.
Use the original amplitude and period to estimate initial energy of a damped oscillator; keep the graph-reading tolerance stated in the mark scheme.
For a stretched vertical spring, decreasing actual elastic energy means decreasing extension; infer upward motion from that change.
Keep actual spring elastic energy, gravitational energy and equilibrium-relative SHM potential distinct when using energy to find velocity.
Combine timing, amplitude and maximum kinetic energy to determine mass or period, retaining sufficient precision for a show-that result.
When the full extreme-to-extreme distance is b, use A = b/2; at x = A/2, speed is (√3/2)ωA.

Covered syllabus points

C.1.1—Conditions for SHM C.1.2—SHM defining equation C.1.3—SHM quantities C.1.4—Period, frequency and angular frequency C.1.5—Mass–spring period C.1.6—Simple pendulum period C.1.7—Qualitative energy changes in SHM C.1.8 (HL)—Phase angle C.1.9 (HL)—SHM equations

[Maximum number: 2]

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 $18 \mathrm{Nm}^{-1}$ .

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".

[Maximum number: 4]

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.

(a)

Show that $\omega^{2}=\frac{k}{m}$ .

[ 2 ]

m a=-k x;
$a=-\frac{k}{m} x ;($ condone lack of negative sign )

\left(\omega^{2}=\frac{k}{m}\right)

or
implied use of defining equation for simple harmonic motion $a=-\omega^{2} x$ ;
(so $\omega^{2}=\frac{k}{m}$ )
m a=-k x so $a=-\left(\frac{k}{m}\right) x$ ;

(b)

One cycle of the variation of displacement with time is shown for two separate mass-spring systems, A and B .

[ 2 ]

(i)

Outline how you would use the graph to confirm that A is performing simple harmonic motion.

[ 2 ]

defines simple harmonic motion as acceleration proportional to negative displacement $/ x=x_{0} \sin \omega t$ ;
correct method for showing how graph leads to the definition;

(c)

The graph shows the variation of the potential energy of A with displacement.

On the axes,

[ 5 ]

(i)

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

[ 2 ]

(ii)

sketch a graph to show the variation of kinetic energy with displacement for the mass in B. Label this B.

[ 3 ]

end displacements correct $\pm 0.01 \mathrm{~m}$ ;
maximum lower than 0.16 J ;
maximum equal to $0.04 \mathrm{~J} \pm$ half square;

[Maximum number: 4]

This question is in two parts.

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

amplitude.

Marking guidance:

max velocity $=0.74( \pm 0.02) \mathrm{ms}^{-1}$ ;
recognize max velocity $=\omega x_{0}$ ;
$\omega=\left(\frac{2 \pi}{T}=\frac{2 \pi}{0.30}=\right) 20.9 \mathrm{rads}^{-1} ;$ $x_{0}=\left(\frac{0.74}{20.9}=\right) 3.5( \pm 0.2) \times 10^{-2} \mathrm{~m} ;$
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 ;

[Maximum number: 3]

One end of a light spring is attached to a rigid horizontal support.

Determine the period of oscillation of the spring.

$\omega=\sqrt{\frac{a}{x}}$ ;
$=10.95 \mathrm{~rad} \mathrm{~s}^{-2}$ ;
$T=\left(\frac{2\pi}{\omega}=\right) \frac{6.28}{10.95}=0.57 \mathrm{~s}$ ;

[Maximum number: 4]

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.

(a)

Show that $\omega^{2}=\frac{k}{m}$ .

[ 2 ]

m a=-k x;
$a=-\frac{k}{m} x ;($ condone lack of negative sign )

\left(\omega^{2}=\frac{k}{m}\right)

or
implied use of defining equation for simple harmonic motion $a=-\omega^{2} x$ ;
(so $\omega^{2}=\frac{k}{m}$ )
m a=-k x so $a=-\left(\frac{k}{m}\right) x$ ;

(b)

One cycle of the variation of displacement with time is shown for two separate mass-spring systems, A and B .

[ 2 ]

(i)

Outline how you would use the graph to confirm that A is performing simple harmonic motion.

[ 2 ]

defines simple harmonic motion as acceleration proportional to negative displacement $/ x=x_{0} \sin \omega t$ ;
correct method for showing how graph leads to the definition;

(c)

The graph shows the variation of the potential energy of A with displacement.

On the axes,

[ 5 ]

(i)

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

[ 2 ]

(ii)

sketch a graph to show the variation of kinetic energy with displacement for the mass in B. Label this B.

[ 3 ]

end displacements correct $\pm 0.01 \mathrm{~m}$ ;
maximum lower than 0.16 J ;
maximum equal to $0.04 \mathrm{~J} \pm$ half square;

[Maximum number: 1]

An object oscillates at the free end of a vertical spring. The graph shows the variation of the object's position with time.

At which position does the object have zero velocity and a negative acceleration?

[Maximum number: 2]

A mass is attached to a vertical spring. The other end of the spring is attached to the driver of an oscillator.

The mass is performing very lightly damped harmonic oscillations. The frequency of the driver is higher than the natural frequency of the system. At one instant the driver is moving downwards.

State and explain the direction of motion of the mass at this instant.

because the mass and the driver are out of phase «by $\pi » \vee$ so upwards

Justification needed for MP2

Exam points

Calculate maximum SHM speed from vmax = ωA, converting displacement amplitude to metres and using ω = 2πf or 2π/T.
Use displacement from equilibrium as the SHM amplitude; a vertically loaded spring’s total extension includes a separate static extension.
Choose phase and sign from the initial position and motion: starting at −A gives x = −A cos ωt and v = +ωA sin ωt.
For x = A sin(ωt + φ), use v = ωA cos(ωt + φ) with the same phase, evaluating the angle in radians and retaining the velocity sign.
An energy method for maximum speed must use maximum kinetic energy explicitly; then apply vmax = √(2EK,max/m).

[Maximum number: 2]

A boat is moved from land to water by rolling it across a set of cylindrical airbags.

During the testing of the rod, it is rotated about its central axis. Point P is on the surface of the rod. At time t=0, P is at a point $45^{\circ}$ above the horizontal. The linear speed of P is $2.0 \mathrm{~ms}^{-1}$ .

The vertical displacement x of P can be modelled with an equation $x=x_{0} \sin (\omega t+\phi)$ .

Determine the vertical velocity of P at $t=3.0 \mathrm{~s}$ .

Use of $v=\omega x_{0} \cos (\omega t+\phi)$ and $\phi=\frac{\pi}{4}$ $v=\omega x_{0} \cos (\omega t+\phi)=(2.22)(0.9) \cos \left((2.22)(3)+\frac{\pi}{4}\right)=0.79 m s^{-1} \quad$

More syllabus points

C.1.7—Qualitative energy changes in SHM1 question

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