SHM HL Paper 1&2

C.1 Simple Harmonic Motion Papers 1&2

Extended Question Bank: Multiple Choice and Extended Responses

These original IB-style practice questions cover restoring acceleration, spring and pendulum oscillations, graphs, phase, speed and energy. They are not official IB examination questions. Attempt each question before opening its answer.

Instructions and Formula Reference

Use $g=9.81\,\mathrm{m\,s^{-2}}$ unless stated otherwise. Give numerical answers to 2–3 significant figures. Retain extra digits during calculations and use radians for phase angles.

For ideal oscillators, assume negligible damping, a light spring within its Hooke’s-law range, and small angular amplitudes for pendulums. Measure displacement from equilibrium. Amplitude is positive; velocity and acceleration may be positive or negative.

IB formulas — Standard Level and Higher Level

Defining relation: $a=-\omega^2x$ .
Period, frequency and angular frequency: $T=\frac{1}{f}=\frac{2\pi}{\omega}$ .
Mass–spring oscillator: $T=2\pi\sqrt{\frac{m}{k}}$ .
Small-angle simple pendulum: $T=2\pi\sqrt{\frac{l}{g}}$ .

IB formulas — Additional Higher Level

Displacement: $x=x_0\sin(\omega t+\phi)$ .
Velocity: $v=\omega x_0\cos(\omega t+\phi)$ .
Velocity at a given displacement: $v=\pm\omega\sqrt{x_0^2-x^2}$ .
Total mechanical energy: $E_T=\frac{1}{2}m\omega^2x_0^2$ .
Oscillation potential energy: $E_p=\frac{1}{2}m\omega^2x^2$ .

Symbols: $x$ is signed displacement from equilibrium; $x_0$ is amplitude, the maximum magnitude of displacement; $l$ is pendulum length; and $\phi$ is the initial phase constant. Amplitude is not necessarily the displacement at $t=0$ : $x(0)=x_0\sin\phi$ .

Useful derived relations: $f=1/T$ , $\omega=2\pi f$ , $v_{\max}=\omega x_0$ , $a_{\max}=\omega^2x_0$ , and $E_k=E_T-E_p=\frac{1}{2}m\omega^2(x_0^2-x^2)$ . Here $v_{\max}$ denotes maximum speed. The sign in the IB velocity formula depends on the direction of motion.

Choosing a phase: For an equilibrium crossing in the positive direction at $t=0$ , choose $\phi=0$ . For an equilibrium crossing in the negative direction, choose $\phi=\pi$ . For release from rest at $x=+x_0$ , choose $\phi=\pi/2$ , giving $x=x_0\sin(\omega t+\pi/2)$ . The equivalent cosine form is a trigonometric rewrite of the same IB formula.

For a horizontal spring, $m\omega^2=k$ , so the IB energy equations become $E_T=\frac{1}{2}kx_0^2$ and $E_p=\frac{1}{2}kx^2$ . For a vertical spring measured from equilibrium, this potential energy is the combined spring-plus-gravitational oscillation energy relative to its equilibrium minimum; it is not the spring elastic energy alone.

Part A — Multiple-Choice Questions

A1 — Definition of SHM

Which acceleration–displacement relationship defines simple harmonic motion?

A. $a=+\omega^2x$ .
B. $a=-\omega x$ .
C. $a=-\omega^2x$ .
D. $a=-\omega^2x^2$ .

Show answer and explanation

Answer: C. Acceleration must be proportional to displacement and directed towards equilibrium. The negative sign gives the restoring direction.

A2 — Direction of acceleration

An oscillator undergoing SHM has positive displacement from equilibrium. What is the direction of its acceleration?

A. Negative, towards equilibrium.
B. Positive, away from equilibrium.
C. Always in the direction of its velocity.
D. Always zero.

Show answer and explanation

Answer: A. From $a=-\omega^2x$ , positive displacement gives negative acceleration. The velocity may be positive or negative.

A3 — At equilibrium

For a non-zero-amplitude ideal SHM oscillator, which statement is correct as it passes through equilibrium?

A. Speed is zero and acceleration magnitude is maximum.
B. Speed and acceleration magnitude are both maximum.
C. Speed and acceleration are both zero.
D. Speed is maximum and acceleration is zero.

Show answer and explanation

Answer: D. At $x=0$ , acceleration is zero. Energy conservation gives maximum kinetic energy and speed at equilibrium.

A4 — At a turning point

Which statement is correct at an extreme displacement in ideal SHM?

A. Velocity and acceleration are both zero.
B. Velocity is zero and acceleration magnitude is maximum.
C. Velocity is maximum and acceleration is zero.
D. Velocity and acceleration magnitude are both maximum.

Show answer and explanation

Answer: B. The particle momentarily stops before reversing. Since $|x|=x_0$ , its acceleration magnitude is $\omega^2x_0$ .

A5 — Period and frequency

An oscillator has period $0.250\,\mathrm{s}$ . What is its frequency?

A. 0.250 Hz.
B. 2.00 Hz.
C. 4.00 Hz.
D. 25.0 Hz.

Show answer and explanation

Answer: C. $f=1/T=1/0.250=4.00\,\mathrm{Hz}$ .

A6 — Angular frequency

An oscillator has frequency $3.00\,\mathrm{Hz}$ . What is its angular frequency?

A. $3.00\,\mathrm{rad\,s^{-1}}$ .
B. $3\pi\,\mathrm{rad\,s^{-1}}$ .
C. $\frac{1}{6\pi}\,\mathrm{rad\,s^{-1}}$ .
D. $6\pi\,\mathrm{rad\,s^{-1}}$ .

Show answer and explanation

Answer: D. $\omega=2\pi f=6\pi\,\mathrm{rad\,s^{-1}}$ , approximately $18.8\,\mathrm{rad\,s^{-1}}$ .

A7 — Acceleration graph

An acceleration–displacement graph is a straight line through the origin with gradient $-64\,\mathrm{s^{-2}}$ . What is the angular frequency?

A. 8.00 rad s⁻¹.
B. 64.0 rad s⁻¹.
C. 4.00 rad s⁻¹.
D. 4096 rad s⁻¹.

Show answer and explanation

Answer: A. The gradient is $-\omega^2$ . Therefore $\omega=\sqrt{64}=8.00\,\mathrm{rad\,s^{-1}}$ .

A8 — Changing spring mass

The mass of an ideal mass–spring oscillator is increased by a factor of four, while spring constant remains unchanged. What happens to the period?

A. It halves.
B. It doubles.
C. It becomes four times greater.
D. It remains unchanged.

Show answer and explanation

Answer: B. $T=2\pi\sqrt{m/k}$ , so multiplying mass by four multiplies period by $\sqrt{4}=2$ .

A9 — Changing spring constant

The spring constant is increased by a factor of four while the oscillating mass remains unchanged. What happens to the period?

A. It doubles.
B. It becomes four times greater.
C. It halves.
D. It remains unchanged.

Show answer and explanation

Answer: C. Period is proportional to $1/\sqrt{k}$ , so multiplying $k$ by four halves the period.

A10 — Changing amplitude

The amplitude of an ideal mass–spring oscillator is doubled while the spring remains within its Hooke’s-law range. What happens to its period?

A. It doubles.
B. It halves.
C. It becomes four times greater.
D. It remains unchanged.

Show answer and explanation

Answer: D. The ideal period $T=2\pi\sqrt{m/k}$ is independent of amplitude.

A11 — Pendulum length

The length of a simple pendulum is increased by a factor of four. For small oscillations at the same location, what happens to its period?

A. It doubles.
B. It halves.
C. It becomes four times greater.
D. It remains unchanged.

Show answer and explanation

Answer: A. $T=2\pi\sqrt{l/g}$ , so multiplying length by four doubles the period.

A12 — Pendulum bob mass

The bob mass of an ideal simple pendulum is doubled. Length, gravitational field strength and small angular amplitude remain unchanged. What happens to the period?

A. It doubles.
B. It remains unchanged.
C. It increases by a factor of the square root of two.
D. It halves.

Show answer and explanation

Answer: B. The ideal small-angle period depends on $l$ and $g$ , not bob mass.

A13 — Small-angle assumption

Which approximation is required when treating a simple pendulum as an SHM oscillator?

A. $\sin\theta\approx1$ .
B. $\cos\theta\approx\theta$ .
C. $\sin\theta\approx\theta$ , with the small angle in radians.
D. $\sin\theta\approx\theta^2$ .

Show answer and explanation

Answer: C. This approximation makes the tangential restoring acceleration approximately proportional to displacement from equilibrium.

A14 — Choosing an initial condition (HL)

An SHM oscillator is released from rest at $x=+x_0$ at $t=0$ . Which equation represents its displacement?

A. $x=x_0\sin(\omega t)$ .
B. $x=x_0\sin(\omega t+\pi)$ .
C. $x=x_0\sin(\omega t+3\pi/2)$ .
D. $x=x_0\sin(\omega t+\pi/2)$ .

Show answer and explanation

Answer: D. The IB form $x=x_0\sin(\omega t+\phi)$ requires $\phi=\pi/2$ . At $t=0$ , $x=x_0\sin(\pi/2)=x_0$ and $v=\omega x_0\cos(\pi/2)=0$ .

A15 — Initial phase (HL)

An oscillator crosses equilibrium at $t=0$ moving in the negative direction. In $x=x_0\sin(\omega t+\phi)$ , which initial phase gives this motion?

A. $\phi=\pi$ .
B. $\phi=0$ .
C. $\phi=\pi/2$ .
D. $\phi=3\pi/2$ .

Show answer and explanation

Answer: A. For $\phi=\pi$ , $x(0)=x_0\sin\pi=0$ and $v(0)=\omega x_0\cos\pi=-x_0\omega$ .

A16 — Maximum speed (HL)

An oscillator has amplitude $0.040\,\mathrm{m}$ and angular frequency $10.0\,\mathrm{rad\,s^{-1}}$ . What is its maximum speed?

A. 4.00 m s⁻¹.
B. 0.400 m s⁻¹.
C. 0.00400 m s⁻¹.
D. 0.0400 m s⁻¹.

Show answer and explanation

Answer: B. $v_{\max}=\omega x_0=(0.040)(10.0)=0.400\,\mathrm{m\,s^{-1}}$ .

A17 — Calculating acceleration

An oscillator has angular frequency $10.0\,\mathrm{rad\,s^{-1}}$ . What is its acceleration at $x=+0.030\,\mathrm{m}$ ?

A. +3.00 m s⁻².
B. −0.300 m s⁻².
C. −3.00 m s⁻².
D. +0.300 m s⁻².

Show answer and explanation

Answer: C. $a=-\omega^2x=-(10.0)^2(0.030)=-3.00\,\mathrm{m\,s^{-2}}$ .

A18 — Energy fraction (HL)

An ideal horizontal spring oscillator has displacement $x=x_0/2$ . What fraction of its total mechanical energy is kinetic?

A. $1/4$ .
B. $1/2$ .
C. $\sqrt{3}/2$ .
D. $3/4$ .

Show answer and explanation

Answer: D. $E_p/E_T=x^2/x_0^2=1/4$ , so $E_k/E_T=1-1/4=3/4$ .

A19 — Amplitude and energy (HL)

The amplitude of an ideal oscillator is tripled while its mass and angular frequency remain unchanged. By what factor does its total mechanical energy increase?

A. 9.
B. 3.
C. 6.
D. The square root of 3.

Show answer and explanation

Answer: A. $E_T=\frac{1}{2}m\omega^2x_0^2$ , so tripling amplitude multiplies energy by nine.

A20 — Phase and time (HL)

A time interval of $T/4$ corresponds to what increase in the phase angle of an SHM oscillator?

A. $\pi$ .
B. $\pi/2$ .
C. $2\pi$ .
D. $\pi/4$ .

Show answer and explanation

Answer: B. $\Delta\theta=\omega\Delta t=(2\pi/T)(T/4)=\pi/2$ radians.

Part B — Extended-Response Questions

Question 1 — Identifying SHM from a graph

A student measures the acceleration of an oscillator at different displacements from equilibrium. The acceleration–displacement graph is a straight line through the origin with gradient $-49\,\mathrm{s^{-2}}$ .

(a) State the defining acceleration–displacement equation for SHM.
(b) Determine the angular frequency.
(c) Calculate the period.

Show answer and explanation

(a) $a=-\omega^2x$ . Acceleration is proportional to displacement and directed towards equilibrium.

(b) The gradient is $-\omega^2$ , so $\omega=\sqrt{49}=7.0\,\mathrm{rad\,s^{-1}}$ .

(c) $\displaystyle T=\frac{2\pi}{\omega}=\frac{2\pi}{7.0}=0.898\,\mathrm{s}\approx0.90\,\mathrm{s}$

Question 2 — Mass–spring oscillator

A mass of $0.500\,\mathrm{kg}$ is attached to a horizontal spring with spring constant $80\,\mathrm{N\,m^{-1}}$ . The mass moves on a frictionless surface. Assume the spring has negligible mass and obeys Hooke’s law.

(a) Calculate the period.
(b) The amplitude is doubled. State and explain what happens to the period.

Show answer and explanation

(a) $\displaystyle T=2\pi\sqrt{\frac{m}{k}}=2\pi\sqrt{\frac{0.500}{80}}=0.497\,\mathrm{s}\approx0.50\,\mathrm{s}$

(b) The period remains unchanged. For an ideal mass–spring oscillator, the period depends on mass and spring constant, not amplitude. Doubling amplitude increases both the distance travelled and the characteristic speeds, leaving the period unchanged. The spring must remain within its Hooke’s-law range.

Question 3 — Simple pendulum

A simple pendulum of length $0.90\,\mathrm{m}$ oscillates with a small angular amplitude.

(a) Calculate its period on Earth.
(b) Calculate its period at a location on Mars where $g=3.70\,\mathrm{m\,s^{-2}}$ .
(c) State one assumption required for the equation used.

Show answer and explanation

(a) $\displaystyle T=2\pi\sqrt{\frac{l}{g}}=2\pi\sqrt{\frac{0.90}{9.81}}=1.90\,\mathrm{s}$

(b) $\displaystyle T_{\rm Mars}=2\pi\sqrt{\frac{0.90}{3.70}}=3.10\,\mathrm{s}$

The smaller gravitational field strength produces a weaker restoring acceleration for a given displacement, so the period is longer.

(c) The angular amplitude must be sufficiently small for $\sin\theta\approx\theta$ , with the angle in radians. Other assumptions include negligible damping, a light inextensible string and a bob treated as a point mass.

Question 4 — Interpreting a displacement–time graph

An oscillator completes six full oscillations in $3.0\,\mathrm{s}$ . Its maximum displacement from equilibrium is $0.040\,\mathrm{m}$ .

(a) Determine the frequency and period.
(b) Determine the angular frequency.
(c) State the amplitude.

Show answer and explanation

(a) $f=6/3.0=2.0\,\mathrm{Hz}$ and $T=1/f=0.50\,\mathrm{s}$ .

(b) $\omega=2\pi f=2\pi(2.0)=12.6\,\mathrm{rad\,s^{-1}}$ .

(c) $x_0=0.040\,\mathrm{m}$ . Amplitude is the maximum magnitude of displacement from equilibrium. The peak-to-peak displacement is $2x_0=0.080\,\mathrm{m}$ .

Question 5 — Position, velocity and phase (HL)

A particle undergoes SHM with amplitude $x_0=0.050\,\mathrm{m}$ and angular frequency $\omega=20\,\mathrm{rad\,s^{-1}}$ . It passes through equilibrium at $t=0$ , moving in the positive direction. Its motion is described by $x=x_0\sin(\omega t)$ .

(a) Determine the period.
(b) Calculate the displacement at $t=0.050\,\mathrm{s}$ .
(c) Calculate the velocity at the same time.
(d) State the phase angle at that time.

Show answer and explanation

(a) $T=2\pi/20=0.314\,\mathrm{s}$ .

(b) First calculate the phase: $\omega t=20(0.050)=1.0\,\mathrm{rad}$ . In radian mode, $x=0.050\sin(1.0)=0.0421\,\mathrm{m}$ .

(c) $\displaystyle v=\omega x_0\cos(\omega t)=0.050(20)\cos(1.0)=+0.540\,\mathrm{m\,s^{-1}}$

The positive sign indicates motion in the positive direction.

(d) The phase angle is $1.0\,\mathrm{rad}$ . The initial phase constant is zero for this sine equation; it is different from the later phase angle $\omega t$ .

Question 6 — Energy during an oscillation (HL)

An ideal horizontal mass–spring oscillator has spring constant $k=250\,\mathrm{N\,m^{-1}}$ and amplitude $x_0=0.080\,\mathrm{m}$ . Displacement is measured from equilibrium, where the spring is unstretched.

(a) Determine the total mechanical energy.
(b) Determine the elastic potential energy at $x=0.030\,\mathrm{m}$ .
(c) Determine the kinetic energy at this displacement.
(d) Explain why kinetic energy has two maxima per cycle.

Show answer and explanation

Use the IB formulas with $m\omega^2=k$ .

(a) $\displaystyle E_T=\frac{1}{2}m\omega^2x_0^2=\frac{1}{2}kx_0^2=\frac{1}{2}(250)(0.080)^2=0.800\,\mathrm{J}$

(b) $\displaystyle E_p=\frac{1}{2}m\omega^2x^2=\frac{1}{2}kx^2=\frac{1}{2}(250)(0.030)^2=0.1125\,\mathrm{J}\approx0.113\,\mathrm{J}$

(c) $E_k=E_T-E_p=0.800-0.1125=0.6875\,\mathrm{J}\approx0.688\,\mathrm{J}$ .

(d) The oscillator passes through equilibrium twice per cycle, once in each direction. Speed is greatest at both crossings. Since $E_k=\frac{1}{2}mv^2$ , kinetic energy is maximum regardless of the direction of motion.

Question 7 — Gradient of a displacement–time graph (HL)

A particle undergoes SHM with frequency $f=4.0\,\mathrm{Hz}$ and amplitude $x_0=0.030\,\mathrm{m}$ . Its motion is described by $x=x_0\sin(\omega t)$ .

(a) Determine the angular frequency.
(b) Explain what the gradient of the displacement–time graph represents.
(c) Determine the maximum magnitude of the gradient.

Show answer and explanation

(a) $\omega=2\pi f=2\pi(4.0)=25.1\,\mathrm{rad\,s^{-1}}$ .

(b) The gradient of the tangent is instantaneous velocity: $v=\frac{dx}{dt}$ . A positive gradient means positive velocity, and a negative gradient means negative velocity.

(c) $\displaystyle v_{\max}=\omega x_0=0.030(2\pi)(4.0)=0.754\,\mathrm{m\,s^{-1}}$

The steepest parts occur at equilibrium crossings. At maximum positive or negative displacement, the gradient is zero.

Question 8 — Linking acceleration and displacement

At one instant, a particle has displacement $x=+0.020\,\mathrm{m}$ and acceleration $a=-0.80\,\mathrm{m\,s^{-2}}$ .

(a) Explain why these measurements are consistent with SHM but do not establish that the motion is SHM.
(b) Assuming SHM, determine the angular frequency.
(c) Calculate the period.

Show answer and explanation

(a) Acceleration is opposite to displacement and points towards equilibrium. However, one measurement does not establish proportionality throughout the motion. Measurements at multiple displacements are needed to show a straight-line acceleration–displacement graph through the origin with a constant negative gradient.

(b) $\displaystyle \omega^2=-\frac{a}{x}=\frac{0.80}{0.020}=40\,\mathrm{s^{-2}}$

Therefore $\omega=\sqrt{40}=6.32\,\mathrm{rad\,s^{-1}}$ .

(c) $\displaystyle T=\frac{2\pi}{\sqrt{40}}=0.993\,\mathrm{s}\approx0.99\,\mathrm{s}$

Question 9 — Speed, acceleration and energy (HL)

A particle of mass $0.400\,\mathrm{kg}$ undergoes SHM with amplitude $0.060\,\mathrm{m}$ and angular frequency $15.0\,\mathrm{rad\,s^{-1}}$ .

(a) Calculate its maximum speed.
(b) Calculate its speed at $x=+0.040\,\mathrm{m}$ .
(c) Calculate its acceleration at this displacement.
(d) Calculate its total energy and its kinetic and oscillation potential energies at this displacement.
(e) Explain why displacement alone does not determine the sign of velocity.

Show answer and explanation

(a) $v_{\max}=\omega x_0=(0.060)(15.0)=0.900\,\mathrm{m\,s^{-1}}$ .

(b) $\displaystyle |v|=\omega\sqrt{x_0^2-x^2}=15.0\sqrt{0.060^2-0.040^2}=0.671\,\mathrm{m\,s^{-1}}$

(c) $a=-\omega^2x=-(15.0)^2(0.040)=-9.00\,\mathrm{m\,s^{-2}}$ .

(d) $\displaystyle E_T=\frac{1}{2}m\omega^2x_0^2=\frac{1}{2}(0.400)(15.0)^2(0.060)^2=0.162\,\mathrm{J}$

$E_p=\frac{1}{2}m\omega^2x^2=0.0720\,\mathrm{J}$ and $E_k=E_T-E_p=0.0900\,\mathrm{J}$ .

(e) At a displacement inside the turning points, the particle passes once in each direction per cycle. Its velocity can be $+0.671$ or $-0.671\,\mathrm{m\,s^{-1}}$ here. Its acceleration is the same on both passes.

Question 10 — Determining the initial phase (HL)

An oscillator has amplitude $x_0=0.080\,\mathrm{m}$ and period $T=0.400\,\mathrm{s}$ . At $t=0$ , its displacement is $+x_0/2$ and its velocity is negative. Use $x=x_0\sin(\omega t+\phi)$ and choose $\phi$ in the interval $[0,2\pi)$ .

(a) Determine $\omega$ and $\phi$ .
(b) Calculate its initial velocity.
(c) Calculate its displacement at $t=0.100\,\mathrm{s}$ .
(d) Write an equivalent cosine equation.

Show answer and explanation

(a) $\omega=2\pi/T=5\pi\,\mathrm{rad\,s^{-1}}$ . Since $\sin\phi=1/2$ , the possible phases are $\pi/6$ and $5\pi/6$ . Negative initial velocity requires $\cos\phi$ to be negative, so $\phi=5\pi/6$ .

(b) $\displaystyle v(0)=\omega x_0\cos\phi=(0.080)(5\pi)\cos(5\pi/6)=-1.09\,\mathrm{m\,s^{-1}}$

(c) $\displaystyle x(0.100)=0.080\sin\left(\frac{\pi}{2}+\frac{5\pi}{6}\right)=-0.0693\,\mathrm{m}$

(d) An equivalent equation is $x=0.080\cos(5\pi t+\pi/3)$ , with displacement in metres and time in seconds. This follows from $\sin u=\cos(u-\pi/2)$ . It is equivalent to the IB sine form; the physical motion is identical.

Question 11 — A vertical mass–spring oscillator

A $0.200\,\mathrm{kg}$ mass hangs from an ideal light spring. At equilibrium the spring extension is $0.0490\,\mathrm{m}$ . It then oscillates vertically with amplitude $0.0200\,\mathrm{m}$ . Measure displacement $x$ downwards from equilibrium.

(a) Determine the spring constant.
(b) Calculate the oscillation period.
(c) Calculate the net acceleration at the lowest turning point.
(d) Explain why gravity shifts equilibrium without changing the ideal period.
(e) Explain why $\frac{1}{2}kx^2$ is not the actual spring elastic energy when displacement is measured from vertical equilibrium.

Show answer and explanation

(a) At equilibrium, $ky_0=mg$ . Hence $k=mg/y_0=(0.200)(9.81)/0.0490=40.0\,\mathrm{N\,m^{-1}}$ .

(b) $\displaystyle T=2\pi\sqrt{\frac{m}{k}}=0.444\,\mathrm{s}$

(c) $\displaystyle a=-\frac{k}{m}x\approx-\frac{40.0}{0.200}(0.0200)=-4.00\,\mathrm{m\,s^{-2}}$

The negative sign means upwards, towards equilibrium. Although velocity is zero at the turning point, acceleration is not zero.

(d) The net force is $F=mg-k(y_0+x)=-kx$ , because $ky_0=mg$ . Therefore $a=-(k/m)x$ and the period remains $2\pi\sqrt{m/k}$ .

(e) Actual spring elastic energy is $\frac{1}{2}k(y_0+x)^2$ . The expression $\frac{1}{2}kx^2$ represents the change in combined spring-plus-gravitational potential energy relative to its equilibrium minimum, up to an additive constant.

Question 12 — Measuring gravitational field strength

A small-angle pendulum has length $l=0.800\pm0.005\,\mathrm{m}$ . A student measures the time for 20 complete oscillations as $36.0\pm0.2\,\mathrm{s}$ . Treat the counted number of cycles as exact. Estimate product and quotient uncertainties by adding fractional uncertainties; multiply the fractional uncertainty by the power for a squared quantity.

(a) Determine the period.
(b) Calculate gravitational field strength using $g=4\pi^2l/T^2$ .
(c) Estimate the percentage and absolute uncertainty in the result.
(d) Explain why timing many cycles helps.
(e) Predict the bias in calculated gravitational field strength if the angular amplitude is too large and the small-angle equation is still used.

Show answer and explanation

(a) $T=36.0/20=1.80\,\mathrm{s}$ , with estimated uncertainty $0.2/20=0.010\,\mathrm{s}$ .

(b) $\displaystyle g=\frac{4\pi^2(0.800)}{1.80^2}=9.75\,\mathrm{m\,s^{-2}}$

(c) $\displaystyle \frac{\delta g}{g}\approx\frac{0.005}{0.800}+2\frac{0.2}{36.0}=0.01736$

Percentage uncertainty is approximately $1.74\%$ . Absolute uncertainty is $\delta g\approx(9.7478)(0.01736)=0.169\,\mathrm{m\,s^{-2}}$ . Report $g=9.75\pm0.17\,\mathrm{m\,s^{-2}}$ .

(d) Timing many oscillations makes start/stop reaction-time uncertainty a smaller fraction of the total elapsed time. Repeat measurements and avoid miscounting cycles.

(e) A large-amplitude pendulum has a longer period than the small-angle prediction. Using that longer measured period in $g=4\pi^2l/T^2$ underestimates gravitational field strength.

Question 13 — Interpreting force and energy graphs (HL)

For an ideal horizontal spring oscillator, a graph of net force against displacement is a straight line through the origin with gradient $-12.0\,\mathrm{N\,m^{-1}}$ . The mass is $0.300\,\mathrm{kg}$ and amplitude is $0.0500\,\mathrm{m}$ .

(a) Determine the spring constant and explain why the force gives SHM.
(b) Calculate angular frequency and period.
(c) State the gradient of the acceleration–displacement graph.
(d) Determine the gradient of a graph of elastic potential energy against $x^2$ , including units.
(e) Calculate the maximum speed.

Show answer and explanation

(a) From $F=-kx$ , $k=12.0\,\mathrm{N\,m^{-1}}$ . For constant mass, $a=F/m=-(k/m)x$ , so acceleration is proportional to displacement and points towards equilibrium.

(b) $\displaystyle \omega=\sqrt{\frac{k}{m}}=\sqrt{40}=6.32\,\mathrm{rad\,s^{-1}}$

$T=2\pi/\omega=0.993\,\mathrm{s}$ .

(c) The gradient is $-\omega^2=-40.0\,\mathrm{s^{-2}}$ .

(d) Since $E_p=(k/2)x^2$ , the gradient is $k/2=6.00\,\mathrm{J\,m^{-2}}$ , equivalent to $6.00\,\mathrm{N\,m^{-1}}$ .

(e) $v_{\max}=\omega x_0=(0.0500)\sqrt{40}=0.316\,\mathrm{m\,s^{-1}}$ .

Question 14 — Equal kinetic and potential energies (HL)

An ideal horizontal spring oscillator has $k=50.0\,\mathrm{N\,m^{-1}}$ , amplitude $x_0=0.100\,\mathrm{m}$ and period $T=0.800\,\mathrm{s}$ . It is released from rest at $x=+x_0$ at $t=0$ .

(a) Calculate total mechanical energy.
(b) Determine the two displacements where kinetic and elastic potential energy are equal.
(c) Determine the first positive time when these energies are equal.
(d) Calculate the speed then.
(e) State how many times the energies are equal during one complete cycle.

Show answer and explanation

(a) $E_T=\frac{1}{2}m\omega^2x_0^2=\frac{1}{2}kx_0^2=0.250\,\mathrm{J}$ .

(b) Each energy is $E_T/2=0.125\,\mathrm{J}$ . Since $E_p=E_T/2$ , $\frac{1}{2}m\omega^2x^2=\frac{1}{4}m\omega^2x_0^2$ , the displacements are $x=\pm x_0/\sqrt{2}=\pm0.0707\,\mathrm{m}$ .

(c) Release from $x=+x_0$ gives $\phi=\pi/2$ , so $x=x_0\sin(\omega t+\pi/2)$ . The first equality occurs when $\omega t+\pi/2=3\pi/4$ , giving $\omega t=\pi/4$ , so $t=T/8=0.100\,\mathrm{s}$ .

(d) $\displaystyle |v|=\frac{\omega x_0}{\sqrt{2}}=\frac{(0.100)(2\pi/0.800)}{\sqrt{2}}=0.555\,\mathrm{m\,s^{-1}}$

(e) Four times per cycle, at $T/8$ , $3T/8$ , $5T/8$ and $7T/8$ for these initial conditions.

Question 15 — Damping and the limits of the ideal model (HL)

A horizontal mass–spring oscillator has spring constant $80.0\,\mathrm{N\,m^{-1}}$ . Weak damping reduces its turning-point amplitude from $0.0800\,\mathrm{m}$ to $0.0500\,\mathrm{m}$ over 20 complete oscillations. The spring obeys Hooke’s law throughout.

(a) Calculate the mechanical energy at the initial and final turning points.
(b) Calculate the total energy loss and the fraction of initial energy remaining.
(c) Calculate the average energy loss per cycle over this interval.
(d) Explain where the energy goes.
(e) Explain why the motion is not exact ideal SHM even if it remains approximately sinusoidal.

Show answer and explanation

(a) At a turning point, speed is zero, so $E_T=\frac{1}{2}kx_0^2$ . This is the instantaneous spring energy at that turning point. For the corresponding undamped oscillator, $k=m\omega_0^2$ , recovering the IB total-energy expression. With damping, total mechanical energy decreases between turning points. Initially $E_{T,i}=\frac{1}{2}(80.0)(0.0800)^2=0.256\,\mathrm{J}$ . Finally $E_{T,f}=\frac{1}{2}(80.0)(0.0500)^2=0.100\,\mathrm{J}$ .

(b) Energy loss is $E_{T,i}-E_{T,f}=0.156\,\mathrm{J}$ . The fraction remaining is $E_{T,f}/E_{T,i}=(0.0500/0.0800)^2\approx0.391$ , or $39.1\%$ .

(c) The average loss over the interval is $0.156/20=0.00780\,\mathrm{J}$ per cycle. This does not imply an identical loss in every cycle.

(d) Damping transfers mechanical energy to the surroundings, usually as thermal energy, and sometimes sound.

(e) A damping force depends on motion as well as displacement, so net acceleration is not simply $-\omega^2x$ with constant amplitude. For viscous damping, $ma=-kx-bv$ . The amplitude decreases; weak damping can leave the period close to its undamped value.

Exam Readiness Checklist

I can recognise SHM using $a=-\omega^2x$ .
I can distinguish displacement, speed, velocity and acceleration.
I can interpret acceleration–displacement and displacement–time graphs.
I can use $f=1/T$ and $\omega=2\pi f$ .
I can calculate spring and pendulum periods and state the assumptions.
I can use $x=x_0\sin(\omega t+\phi)$ and choose the phase to match initial conditions.
I can distinguish an initial phase constant from the phase at a later time.
I can use energy conservation to calculate speed and energy fractions.
I can distinguish spring elastic energy from combined potential energy in a vertical oscillator.
I can estimate uncertainties and explain the limitations of ideal SHM models.

Studying with Gratitude and Integrity

Memory verse — Psalm 111:2 (KJV):
“The works of the LORD are great, sought out of all them that have pleasure therein.”

Creation: Investigating God’s works

Studying oscillations invites us to investigate the order of the created world. We measure, compare, calculate and test our explanations. Psalm 111:2 encourages delight in the works of the Lord. Hebrews 1:2–3 teaches that God made the universe through His Son, who sustains all things by His powerful word. Our learning can become an expression of gratitude to Jesus Christ, the Creator and Sustainer.

The Fall: Our need for truth and grace

Sin affects how we use our abilities and how we treat others. Pride can make us unwilling to admit mistakes, while fear can tempt us to be dishonest about our work. First John 1:5–9 calls us to walk in God’s light, acknowledge our sin and receive His forgiveness.

In science, honesty includes recognising the limits of evidence. One measurement cannot establish an entire relationship. In our daily lives, we also need humility to acknowledge what we do not know and courage to speak truthfully.

Redemption: Jesus offers forgiveness and new life

Jesus died for our sins and rose again so that we could be reconciled to God. Our worth before Him is not measured by an examination grade. Through repentance and faith in Christ, we receive forgiveness and begin a new life shaped by His love.

Receive Jesus as your personal God, Saviour and Lord. Ask Him to guide your choices, strengthen your perseverance and help you learn with honesty and care.

Restoration: Growing together in faith

Revelation 21:3–5 gives believers hope of complete restoration in God’s presence. As we look forward to that future, we can grow today through reading and studying the Bible, spending time in prayer and connecting with other believers in a local church.

Use your learning to serve others. Encourage a classmate, explain a difficult concept patiently and celebrate progress without comparing people’s worth.

Reflection: What can you do this week to approach your studies with greater honesty, gratitude and love for others?

Prayer: Lord Jesus, thank You for Your creation and Your saving love. Forgive my sins and help me walk in Your truth. Give me understanding, patience and perseverance in my studies. Help me grow through Your Word, prayer and fellowship, and use what I learn to serve others. Amen.

IB Physics HL Book 2025; Papers 1 & 2 Practice 5 Sections65 Lessons54 Weeks

Curriculum

SHM HL Paper 1&2

C.1 Simple Harmonic Motion Papers 1&2

Instructions and Formula Reference

IB formulas — Standard Level and Higher Level

IB formulas — Additional Higher Level

Part A — Multiple-Choice Questions

A1 — Definition of SHM

A2 — Direction of acceleration

A3 — At equilibrium

A4 — At a turning point

A5 — Period and frequency

A6 — Angular frequency

A7 — Acceleration graph

A8 — Changing spring mass

A9 — Changing spring constant

A10 — Changing amplitude

A11 — Pendulum length

A12 — Pendulum bob mass

A13 — Small-angle assumption

A14 — Choosing an initial condition (HL)

A15 — Initial phase (HL)

A16 — Maximum speed (HL)

A17 — Calculating acceleration

A18 — Energy fraction (HL)

A19 — Amplitude and energy (HL)

A20 — Phase and time (HL)

Part B — Extended-Response Questions

Question 1 — Identifying SHM from a graph

Question 2 — Mass–spring oscillator

Question 3 — Simple pendulum

Question 4 — Interpreting a displacement–time graph

Question 5 — Position, velocity and phase (HL)

Question 6 — Energy during an oscillation (HL)

Question 7 — Gradient of a displacement–time graph (HL)

Question 8 — Linking acceleration and displacement

Question 9 — Speed, acceleration and energy (HL)

Question 10 — Determining the initial phase (HL)

Question 11 — A vertical mass–spring oscillator

Question 12 — Measuring gravitational field strength

Question 13 — Interpreting force and energy graphs (HL)

Question 14 — Equal kinetic and potential energies (HL)

Question 15 — Damping and the limits of the ideal model (HL)

Exam Readiness Checklist

Studying with Gratitude and Integrity

Creation: Investigating God’s works

The Fall: Our need for truth and grace

Redemption: Jesus offers forgiveness and new life

Restoration: Growing together in faith

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