Question 1Multiple choice · 1 mark
A mass–spring system oscillates with period T. The mass is multiplied by 16.
What is the new period?
- A 0.25 T
- B 4.0 T
- C 16 T
- D 260 T
Show answer
Answer: B
Free C.1 Simple harmonic motion questions for IB Physics (SL & HL), part of C Wave behaviour. Each question has the answer and the marking points an examiner would look for.
Generate a full C.1 Simple Harmonic Motion worksheet →All C Wave behaviour questionsA mass–spring system oscillates with period T. The mass is multiplied by 16.
What is the new period?
Answer: B
An object performs simple harmonic motion. What is the phase difference between its displacement and its velocity, and between its displacement and its acceleration?
| displacement and velocity | displacement and acceleration | |
|---|---|---|
| A | π/2 | π |
| B | 0 | π |
| C | π | π/2 |
| D | π/2 | 0 |
Answer: A
A mass m hangs from a spring and oscillates vertically. The time for 20 oscillations is measured for different masses. Theory suggests T2 = 4π2k(m + ms), where ms is an effective mass of the spring.
| m / kg | 0.10 | 0.20 | 0.30 | 0.40 | 0.50 |
|---|---|---|---|---|---|
| time for 20 oscillations / s ± 0.2 | 8.7 | 11.8 | 14.2 | 16.2 | 18.0 |
Calculate T2 for each mass.
Use the graph to determine the spring constant k.
Use the graph to determine ms.
The student says that the gradient alone gives k, so the intercept does not matter. Comment on this claim.
Explain how the uncertainty in T compares with the uncertainty in the time for 20 oscillations.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 0.190, 0.350, 0.504, 0.660, 0.813 T = time ÷ 20 ✓ T2: 0.190, 0.350, 0.504, 0.660, 0.813 s2 ✓ | 2 |
| (b) | 25.4 N m−1 gradient = 4π2k = 1.56 s2 kg−1 ✓ k = 4π2gradient ✓ k = 25.4 N m−1 (accept ±5%) ✓ | 3 |
| (c) | 23.7 g extrapolate to T2 = 0: m = −ms ✓ ms = 0.0237 kg ✓ | 2 |
| (d) | true for k; intercept gives m_s correct: ms only shifts the line, not the gradient ✓ but the intercept is needed to find ms, and a large intercept shows the spring’s mass is not negligible ✓ | 2 |
| (e) | same absolute ÷ 20 ΔT = 0.2/20 = 0.01 s; the percentage uncertainty is the same ✓ | 1 |
A block of mass 0.25 kg on a frictionless horizontal surface is attached to a spring of spring constant 12 N m−1. It oscillates with simple harmonic motion of amplitude 0.05 m.
Show that the total energy ET can be written as ET = ½kx02.
Calculate the period of oscillation.
Calculate the speed of the block when its displacement is half the amplitude.
Determine the shortest time for the block to move from the equilibrium position to a displacement of half the amplitude.
Determine the fraction of the total energy that is kinetic at this displacement.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | E = ½mω²x₀² with ω² = k/m ET = ½mω2x02 ✓ ω2 = k/m so ET = ½kx02 ✓ | 2 |
| (b) | 0.907 s T = 2π√(m/k) ✓ = 0.907 s ✓ | 2 |
| (c) | 0.300 m s−1 v = ω√(x02 − x2) ✓ = 0.300 m s−1 ✓ | 2 |
| (d) | 0.0756 s x = x0sin ωt so ωt = π/6 ✓ t = T/12 = 0.0756 s ✓ | 2 |
| (e) | 0.75 ¾ ✓ | 1 |
A mass on a spring oscillates vertically with amplitude 3.0 cm and period 0.80 s.
Calculate the angular frequency.
Calculate the maximum speed and maximum acceleration.
State where in the oscillation the kinetic energy is a maximum.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 7.9 rad s−1 ω = 2π/T = 7.85 rad s−1 ✓ | 1 |
| (b) | 0.24 m s−1; 1.8 m s−2 vmax = ωx₀ = 0.24 m s−1 ✓ amax = ω2x₀ = 1.8 m s−2 ✓ | 2 |
| (c) | at equilibrium at the equilibrium position ✓ | 1 |
A particle of mass 0.2 kg oscillates in simple harmonic motion with frequency 2.5 Hz and amplitude 2 cm. At t = 0 it is at maximum positive displacement.
Write down an equation for the displacement x as a function of time.
Calculate the total energy of the oscillation.
Calculate the kinetic energy when the displacement is half the amplitude.
Determine the first time at which the kinetic energy equals the potential energy.
Sketch graphs of kinetic energy and potential energy against time for one period of the motion.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | x = 0.02cos(15.7t) x = 0.02 cos(15.7t) m ✓ | 1 |
| (b) | 9.87 × 10−3 J E = ½mω2A2 ✓ 9.87 × 10−3 J ✓ | 2 |
| (c) | 7.40 × 10−3 J Ek = ½mω2(A2 − x2) ✓ 7.40 × 10−3 J (three quarters of the total) ✓ | 2 |
| (d) | 0.0500 s Ek = Ep when x = A√2 ✓ cos ωt = 1√2, ωt = π4 ✓ t = 0.0500 s ✓ | 3 |
| (e) | sin² and cos² shapes both oscillate at twice the frequency of the motion ✓ Ep starts at maximum, Ek at zero; their sum is constant ✓ | 2 |
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