C.1 Simple harmonic motion – IB Physics exam questions with mark scheme

1StandardHL only5 marks

A mass of 0.25 kg hangs on a spring of spring constant $40\ \mathrm{N\,m^{-1}}$ and oscillates vertically with simple harmonic motion of amplitude 6.0 cm.

(a) Calculate the time period of the oscillation. [2]

(b) Calculate the maximum speed of the mass. [1]

(d) Determine the total energy of the oscillation. [1]

$T=2\pi\sqrt{m/k}=2\pi\sqrt{0.25/40}$ seen (or $\omega=\sqrt{k/m}=\sqrt{40/0.25}=12.6\ \mathrm{rad\,s^{-1}}$)(1)
$T=0.50$ s (accept 0.497 s)(1)
$v_{max}=\omega A = 12.6\times0.060 = 0.76\ \mathrm{m\,s^{-1}}$ (0.76; allow 0.75 from rounded $T=0.50$ s; allow ECF from (a))(1)
$a_{max}=\omega^2A = 160\times0.060 = 9.6\ \mathrm{m\,s^{-2}}$ (9.6; allow 9.5 from rounded $T=0.50$ s; allow ECF)(1)
$E_T=\tfrac12 m\omega^2x_0^2 = \tfrac12\times0.25\times160\times0.060^2 = 0.072$ J (accept $\tfrac12 kx_0^2$ or $\tfrac12 mv_{max}^2$; allow 0.070–0.071 J from rounded values; allow ECF)(1)
Answer: (a) 0.50 s (b) 0.76 m s$^{-1}$ (c) 9.6 m s$^{-2}$ (d) 0.072 J
Examiner tip: Convert the amplitude from cm to m before substituting into $\omega A$ and $\omega^2 A$.

2HardHL only6 marks

The graph shows how the displacement $x$ of a $0.30\ \mathrm{kg}$ object undergoing simple harmonic motion varies with time $t$.

(a) Use the graph to determine the angular frequency of the motion. [2]

(b) Calculate the displacement of the object at $t=0.050\ \mathrm{s}$. [2]

Time period read from graph $T=0.40$ s (amplitude 8.0 cm)(1)
$\omega=2\pi/T = 15.7\ \mathrm{rad\,s^{-1}}$ (15.7; allow 16)(1)
Graph starts at maximum so $x=x_0\cos\omega t = 8.0\cos(15.7\times0.050)$ (calculator in radians)(1)
$x=5.7$ cm (5.7 cm; $5.7\times10^{-2}$ m; allow 5.6 cm if $\omega=16$ is used)(1)
$E_T=\tfrac12 m\omega^2x_0^2$ (amplitude in metres)(1)
$E_T=0.24$ J (0.24 J; allow 0.25 J if $\omega=16$ is used)(1)
Answer: (a) 15.7 rad s$^{-1}$ (b) 5.7 cm (c) 0.24 J
Examiner tip: Check that the calculator is in radian mode when evaluating $\cos\omega t$.

3Core5 marks

A buoy floating in a harbour bobs up and down. Its vertical displacement $x$ from its equilibrium position and its acceleration $a$ are related by

$$a=-2.5\,x$$

where $x$ is in m and $a$ is in m s⁻².

(a) State the two conditions that must be satisfied for an object to move with simple harmonic motion. [2]

(b) Outline the significance of the minus sign in the equation. [1]

Acceleration (or restoring force) is proportional to the displacement from the equilibrium position (allow $a\propto x$; do not accept 'proportional to distance' without reference to equilibrium/a fixed point)(1)
Acceleration (or restoring force) is directed towards the equilibrium position / is in the opposite direction to the displacement(1)
Acceleration is always opposite in direction to the displacement / always points back towards equilibrium (ignore 'it is decelerating'; do not accept 'the acceleration is negative' on its own)(1)
Compares with $a=-\omega^2x$: $\omega^2=2.5$ so $\omega=1.58\ \mathrm{rad\,s^{-1}}$(1)
$T=2\pi/\omega=4.0$ s (accept 3.97 s; award [2] for a bare correct answer; $\omega=2.5$ leading to 2.5 s scores [0])(1)
Answer: (a) acceleration proportional to displacement from equilibrium; directed towards equilibrium (b) acceleration is always opposite to displacement (c) $T=4.0$ s
Examiner tip: Write $\omega^2=2.5$, not $\omega=2.5$: the constant in front of $x$ is the square of the angular frequency.

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4Core5 marks

A playground swing can be modelled as a simple pendulum of length 2.4 m. A child on the swing is given a small push and then left to swing.

(a) Calculate the period of the swing. [2]

(b) The child is pushed harder so that the amplitude increases, but remains small. State and explain the effect, if any, on the period. [2]

Substitution into $T=2\pi\sqrt{l/g}=2\pi\sqrt{2.4/9.81}$(1)
$T=3.1$ s (accept 3.11 s; award [2] for a bare correct answer)(1)
Period is unchanged / stays the same (do not award if the period is said to increase)(1)
Period depends only on $l$ and $g$, not on amplitude (for small amplitudes) / the oscillation is isochronous (accept 'greater distance is covered at a greater average speed')(1)
$T\propto\sqrt{l}$ so length is $\tfrac14$ of 2.4 m $=0.60$ m (do not accept 1.2 m)(1)
Answer: (a) 3.1 s (b) unchanged: period is independent of amplitude (isochronous) (c) 0.60 m
Examiner tip: Halving the period needs a quarter of the length, because $T$ depends on $\sqrt{l}$, not on $l$.

5Standard6 marks

A trolley of mass 0.20 kg is attached to a spring and oscillates on a horizontal track. The graph shows how the acceleration $a$ of the trolley varies with its displacement $x$ from the equilibrium position.

(a) Explain how the graph shows that the trolley moves with simple harmonic motion. [2]

(b) Determine the period of the oscillation. [3]

Straight line through the origin, so $a$ is proportional to $x$(1)
Negative gradient, so $a$ is opposite in direction to $x$ / directed towards equilibrium(1)
Gradient $=-\omega^2$ with $x$ converted to metres: $\omega^2=3.6/0.040=90\ \mathrm{s^{-2}}$(1)
$\omega=\sqrt{90}=9.5\ \mathrm{rad\,s^{-1}}$ (allow ECF)(1)
$T=2\pi/\omega=0.66$ s (accept 0.662 s; using cm gives $T=6.6$ s: lose the conversion mark only, max [2] by ECF)(1)
$k=m\omega^2=0.20\times90=18\ \mathrm{N\,m^{-1}}$ (allow ECF from (b); accept use of $T=2\pi\sqrt{m/k}$)(1)
Answer: (a) straight line through origin with negative gradient: $a\propto -x$ (b) $T=0.66$ s (c) $k=18\ \mathrm{N\,m^{-1}}$
Examiner tip: Convert the displacement to metres before finding the gradient; most lost marks here come from using 4.0 cm as 4.0.

6Hard8 marks

A student investigates how the period $T$ of a simple pendulum depends on its length $l$. For each length the student times 20 oscillations. The student measures $l$ from the point of suspension to the top of the spherical bob. The graph shows $T^2$ against $l$ with the line of best fit.

(a) Explain why the student plots $T^2$ against $l$ rather than $T$ against $l$. [2]

(b) Determine, using the graph, a value for the acceleration of free fall $g$. [3]

(d) State the effect, if any, of the student's method of measuring $l$ on the value of $g$ found in (b). [1]

$T^2=\dfrac{4\pi^2}{g}\,l$, so $T^2$ is proportional to $l$ (for the true length)(1)
This gives a straight line, whose gradient $=4\pi^2/g$ (accept 'easier to identify a linear relationship / to find $g$ from the gradient')(1)
Gradient calculated from two well-separated points on the line, e.g. $(4.05-0.05)/1.00=4.0\ \mathrm{s^2\,m^{-1}}$ (accept 3.9 to 4.1; do not award if data points off the line are used)(1)
$g=4\pi^2/\text{gradient}$(1)
$g=9.9\ \mathrm{m\,s^{-2}}$ (accept 9.6 to 10.1 consistent with their gradient)(1)
$T^2=\dfrac{4\pi^2}{g}(l+r)$, so the intercept on the $l$ axis is $-r$ / $r=$ intercept on $T^2$ axis $\div$ gradient $=0.05/4.0$(1)
$r=0.0125$ m so diameter $=2.5$ cm (accept 2.4 to 2.6 cm; allow ECF from their gradient)(1)
No effect: the gradient is unchanged, only the intercept changes (do not accept 'g would be smaller/larger')(1)
Answer: (a) $T^2\propto l$ gives a straight line of gradient $4\pi^2/g$ (b) $g\approx9.9\ \mathrm{m\,s^{-2}}$ (c) about 2.5 cm (d) none: gradient unaffected
Examiner tip: A systematic error in $l$ shifts the line but not its gradient, so $g$ is still correct; use points on the line, not table data, for the gradient.

7Standard6 marks

A glider on a frictionless air track is attached between two identical stretched springs fixed to the ends of the track. The glider is pulled a short distance to one side and released.

Explain why the glider moves with simple harmonic motion, and describe the changes in the kinetic energy and elastic potential energy of the system during one complete oscillation. [6]

When the glider is displaced, one spring pulls harder than the other, so there is a resultant (restoring) force towards the equilibrium position(1)
The resultant force (from Hooke's law) is proportional to the displacement, so $a\propto x$(1)
The force and acceleration are always directed towards equilibrium / opposite to the displacement, so $a=-\omega^2x$(1)
At release (maximum displacement) the glider is momentarily at rest: kinetic energy is zero and elastic potential energy is maximum(1)
As it moves towards equilibrium, elastic potential energy is transferred to kinetic energy; the glider speeds up(1)
At the equilibrium position kinetic energy is maximum (maximum speed) and elastic potential energy is minimum(1)
Past equilibrium the glider slows down as kinetic energy is transferred back to elastic potential energy, reaching zero speed at the other extreme(1)
The sequence repeats on the return journey: kinetic energy reaches its maximum twice per oscillation(1)
With no friction, the total energy stays constant, so the amplitude stays constant (allow 'oscillation continues indefinitely')(1)

Level	Marks	What it looks like
Level 3	5–6	Clear explanation linking a restoring force that is proportional to displacement and directed towards equilibrium to SHM, together with a correct, ordered account of the energy changes through the whole oscillation, including the extremes and the equilibrium position, and that total energy is constant.
Level 2	3–4	Identifies a restoring force towards equilibrium or proportionality, but the link to SHM is incomplete; energy changes are mostly correct but may omit one key position or the conservation of total energy.
Level 1	1–2	Isolated correct statements, e.g. 'kinetic energy is greatest in the middle' or 'the springs pull it back', with little linking or sequence.
Level 0	0	No relevant content.

Answer: Restoring force from the springs is proportional to displacement and directed towards equilibrium, so $a=-\omega^2x$. Elastic PE is maximum and KE zero at the extremes; KE maximum and elastic PE minimum at equilibrium; total energy constant with no friction.
Examiner tip: Say both parts of the SHM condition (proportional to displacement AND towards equilibrium); 'it gets pulled back' alone is not enough.

8HardHL only6 marks

Two identical masses A and B oscillate with simple harmonic motion. The graph shows how the displacement $x$ of each mass varies with time $t$. A is shown by the solid line and B by the dashed line.

(a) Calculate the angular frequency of the oscillations. [1]

(b) State the phase difference between the two oscillations and identify which oscillation leads. [2]

(c) The displacement of B is given by $x_B=x_0\sin(\omega t+\phi)$. Using the appropriate equation, calculate the speed of B at $t=0.10$ s. [2]

(d) Calculate $\dfrac{\text{total energy of A}}{\text{total energy of B}}$. [1]

$T=0.80$ s so $\omega=2\pi/0.80=7.9\ \mathrm{rad\,s^{-1}}$ (accept 7.85)(1)
$\dfrac{\pi}{2}$ rad (accept 90° or a quarter of a cycle; accept $\tfrac{3\pi}{2}$ only if paired with 'B leads')(1)
A leads / B lags behind A (B reaches each maximum 0.20 s after A)(1)
$\phi=-\dfrac{\pi}{2}$ and $v=\omega x_0\cos(\omega t+\phi)=7.85\times2.0\times\cos(\tfrac{\pi}{4}-\tfrac{\pi}{2})$ (calculator in radians; allow use of $v=\omega\sqrt{x_0^2-x^2}$ with $x=-1.41$ cm)(1)
$v=11\ \mathrm{cm\,s^{-1}}$ / $0.11\ \mathrm{m\,s^{-1}}$ (accept 11.1 cm s⁻¹; allow ECF from (a))(1)
$E\propto x_0^2$ (same $m$, same $\omega$) so ratio $=(3.0/2.0)^2=2.25$ (accept 2.3; do not accept 1.5)(1)
Answer: (a) 7.9 rad s⁻¹ (b) $\pi/2$ rad, A leads (c) 0.11 m s⁻¹ (d) 2.25
Examiner tip: Read the phase difference from the time shift between matching points: $\Delta\phi=2\pi\,\Delta t/T$, and set the calculator to radians.

9HardHL only7 marks

A block of mass 1.2 kg rests on a frictionless horizontal surface and is attached to a horizontal spring. The block is set oscillating with simple harmonic motion of amplitude 0.15 m. The total energy of the oscillation is 0.54 J.

(a) Show that the angular frequency of the oscillation is about 6.3 rad s⁻¹. [2]

(b) Calculate the speed of the block when its displacement is 0.050 m. [2]

(d) Calculate the displacement at which the kinetic energy equals the elastic potential energy. [2]

$E_T=\tfrac12m\omega^2x_0^2$ rearranged: $\omega^2=\dfrac{2\times0.54}{1.2\times0.15^2}$ (accept finding $k=48\ \mathrm{N\,m^{-1}}$ first)(1)
$\omega^2=40$ so $\omega=6.32\ \mathrm{rad\,s^{-1}}$ (answer must be given to more s.f. than 6.3 or with the working shown)(1)
$v=\omega\sqrt{x_0^2-x^2}=6.32\sqrt{0.15^2-0.050^2}$(1)
$v=0.89\ \mathrm{m\,s^{-1}}$ (accept 0.894; allow energy method $\tfrac12mv^2=0.54-\tfrac12m\omega^2x^2$)(1)
$\dfrac{E_K}{E_T}=\dfrac{x_0^2-x^2}{x_0^2}=\dfrac{8}{9}$ / 0.89 (accept $0.48/0.54$; allow ECF from (b))(1)
$E_P=\tfrac12E_T$ so $\tfrac12m\omega^2x^2=\tfrac12\left(\tfrac12m\omega^2x_0^2\right)$ / $x=x_0/\sqrt2$(1)
$x=\pm0.11$ m (accept 0.106 m; do not accept 0.075 m)(1)
Answer: (a) $\omega=\sqrt{40}=6.32$ rad s⁻¹ (b) 0.89 m s⁻¹ (c) 8/9 (0.89) (d) $\pm0.11$ m
Examiner tip: Kinetic and potential energy are equal at $x_0/\sqrt2$, not at half the amplitude: energy depends on $x^2$.

10StretchHL only8 marks

A coin rests on a horizontal platform. The platform is made to oscillate vertically with simple harmonic motion of frequency 12 Hz. The amplitude of the oscillation can be adjusted.

(a) Explain why, if the amplitude is large enough, the coin loses contact with the platform, and state where in the oscillation this first happens. [3]

(b) Show that the maximum amplitude for which the coin stays in contact throughout the motion is about 1.7 mm. [2]

(c) The amplitude is set to 1.2 mm. Calculate the speed of the coin when it is 0.60 mm above the equilibrium position. [2]

(d) The frequency is now doubled with the amplitude kept at 1.2 mm. Deduce whether the coin remains in contact with the platform. [1]

Above the equilibrium position the acceleration of the platform is downwards, with magnitude $\omega^2x$, increasing with displacement(1)
The greatest downward acceleration the coin can have is $g$ (only its weight acts when the normal force becomes zero)(1)
Contact is lost when $\omega^2x$ exceeds $g$, which first happens at (or near) the highest point of the motion / above the equilibrium position (do not accept 'at the bottom')(1)
$\omega=2\pi\times12=75.4\ \mathrm{rad\,s^{-1}}$ and $\omega^2x_0=g$(1)
$x_0=9.81/75.4^2=1.73\times10^{-3}$ m (accept 1.73 mm; answer must be to more s.f. than 1.7 or with working shown)(1)
$v=\omega\sqrt{x_0^2-x^2}=75.4\sqrt{(1.2\times10^{-3})^2-(0.60\times10^{-3})^2}$(1)
$v=0.078\ \mathrm{m\,s^{-1}}$ (accept 7.8 cm s⁻¹; using mm without converting and quoting 78 m s⁻¹ scores [1] max)(1)
$a_{max}=(2\times75.4)^2\times1.2\times10^{-3}=27\ \mathrm{m\,s^{-2}}$, which is greater than $g$, so the coin loses contact (accept: new maximum amplitude is $1.73/4=0.43$ mm $<1.2$ mm; conclusion needs supporting calculation)(1)
Answer: (a) above equilibrium the platform accelerates downwards at $\omega^2x$; the coin can only accelerate down at $g$; contact first lost near the top (b) $x_0=g/\omega^2=1.73$ mm (c) 0.078 m s⁻¹ (d) no: $a_{max}=27\ \mathrm{m\,s^{-2}}>g$
Examiner tip: The coin leaves the platform at the top, not the bottom: at the bottom the platform pushes up harder, while at the top it must fall away faster than $g$.

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