AS & A Level Physics (9702)
Oscillations Banker
27 questions · about 3% of the marks we indexed · examined in 9 of 9 sessions.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Simple harmonic motion · Core · Paper 4 style · 4 marks- (a) Define simple harmonic motion. [2]
- (b) Write down the defining equation of simple harmonic motion and state the meaning of each symbol. [2]
Show answer
- (a) Motion in which the acceleration is proportional to the displacement from a fixed point and is always directed towards that point (opposite to the displacement).
B1 a ∝ x; B1 directed towards the fixed point / opposite to x
- (b) a = −ω²x; a acceleration, ω angular frequency, x displacement from equilibrium.
B1 equation; B1 symbols
A mass hanging from a spring oscillates vertically. A student measures the time for 20 complete oscillations as 13.6 s.
- (a) Calculate the period of the oscillations. [1]
- (b) Calculate the frequency. [1]
- (c) Calculate the angular frequency. [1]
- (d) Explain why the student times 20 oscillations rather than one. [1]
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- (a) 0.68 s
13.6/20 = 0.68 s (B1)
- (b) 1.47 Hz
1/0.68 = 1.47 Hz (B1)
- (c) 9.24 rad s⁻¹
ω = 2π/T = 9.24 rad s⁻¹ (B1)
- (d) To reduce the percentage uncertainty in the period (reaction time is a smaller fraction of the total time).
B1
- (a) Define (i) the period and (ii) the angular frequency of an oscillation. [2]
- (b) The mains supply in Pakistan has a frequency of 50 Hz. Calculate its period and its angular frequency. [2]
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- (a) (i) the time for one complete oscillation; (ii) 2π × frequency (rate of change of phase angle), ω = 2π/T.
B1 each
- (b) T = 0.020 s; ω = 314 rad s⁻¹
T = 1/50 = 0.020 s (B1); ω = 2π × 50 = 314 rad s⁻¹ (B1)
- (a) For each motion, state whether it is simple harmonic and give a reason.(i) a ball bouncing vertically on a hard floor(ii) a mass oscillating on a vertical spring(iii) a child on a swing with small amplitude [3]
- (b) State the condition needed for a simple pendulum to move with approximately simple harmonic motion. [1]
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- (a) (i) not SHM – acceleration is constant (g) during flight, not proportional to displacement; (ii) SHM – restoring force ∝ displacement; (iii) approximately SHM for small angles.
B1 each with reason
- (b) The amplitude (angle) must be small.
B1
A particle moves with simple harmonic motion of angular frequency 15 rad s⁻¹.
- (a) Calculate the acceleration of the particle when its displacement is +2.0 cm. [2]
- (b) Calculate the frequency of the oscillation. [1]
Show answer
- (a) −4.5 m s⁻²
a = −ω²x = −15² × 0.020 (C1) = −4.5 m s⁻² (A1)
- (b) 2.4 Hz
f = 15/2π = 2.4 Hz (B1)
The needle of a sewing machine moves up and down with simple harmonic motion of amplitude 3.0 cm and frequency 2.0 Hz.
- (a) Calculate the maximum speed of the needle. [2]
- (b) Calculate the maximum acceleration of the needle. [2]
- (c) State where in its motion the needle has its maximum acceleration. [1]
Show answer
- (a) 0.38 m s⁻¹
ω = 2π × 2.0 = 12.6 rad s⁻¹ (C1); v₀ = ωx₀ = 12.6 × 0.030 = 0.38 m s⁻¹ (A1)
- (b) 4.7 m s⁻²
a₀ = ω²x₀ = 12.6² × 0.030 (C1) = 4.7 m s⁻² (A1)
- (c) At the extremes of its motion (maximum displacement).
B1
A trolley held between two springs oscillates with simple harmonic motion of amplitude 5.0 cm and angular frequency 8.0 rad s⁻¹.
- (a) Calculate the maximum speed of the trolley. [1]
- (b) Calculate the speed of the trolley when its displacement is 3.0 cm. [2]
- (c) Calculate the magnitude of its acceleration at this displacement. [1]
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- (a) 0.40 m s⁻¹
v₀ = ωx₀ = 8.0 × 0.050 (B1)
- (b) 0.32 m s⁻¹
v = ω√(x₀² − x²) = 8.0√(0.050² − 0.030²) (C1) = 0.32 m s⁻¹ (A1)
- (c) 1.9 m s⁻²
a = ω²x = 64 × 0.030 = 1.9 m s⁻² (B1)
A block of mass 0.40 kg oscillates with simple harmonic motion of angular frequency 6.0 rad s⁻¹ and amplitude 5.0 cm.
- (a) Calculate the total energy of the oscillations. [2]
- (b) Calculate the kinetic energy and the potential energy when the displacement is 3.0 cm. [2]
- (c) Describe how the kinetic energy varies with displacement x, and state the shape of the graph of kinetic energy against x. [2]
Show answer
- (a) 0.018 J
E = ½mω²x₀² = ½ × 0.40 × 6.0² × 0.050² (C1) = 0.018 J (A1)
- (b) KE = 0.0115 J; PE = 0.0065 J
KE = ½mω²(x₀² − x²) = 0.0115 J (B1); PE = 0.018 − 0.0115 = 0.0065 J (B1)
- (c) KE is maximum at x = 0 and zero at x = ±x₀; the graph is an inverted parabola.
B1 max at centre, zero at ends; B1 inverted parabola
A data-logger records the acceleration a of a pendulum bob at different displacements x.x / cm: −4.0, −2.0, 0, 2.0, 4.0a / m s⁻²: 1.58, 0.79, 0, −0.79, −1.58
- (a) Explain how the data show that the motion is simple harmonic. [2]
- (b) Determine the angular frequency and the period of the oscillation. [3]
Show answer
- (a) a is proportional to x (a/x constant = −39.5 s⁻²) and opposite in sign to x.
B1 proportional; B1 opposite sign/directed to equilibrium
- (b) ω = 6.3 rad s⁻¹; T = 1.0 s
ω² = 1.58/0.040 = 39.5 s⁻² (C1); ω = 6.28 rad s⁻¹ (A1); T = 2π/ω = 1.0 s (A1)
Two identical pendulums A and B each have period 2.0 s. B reaches its maximum displacement to the right 0.25 s after A does.
- (a) Calculate the phase difference between the two pendulums in degrees and in radians. [2]
- (b) State which pendulum leads. [1]
- (c) Pendulum B is adjusted so that it reaches maximum displacement on one side at the same instant as A reaches maximum displacement on the other side. State the new phase difference. [1]
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- (a) 45°; π/4 rad
0.25/2.0 × 360° = 45° (B1); = π/4 rad (B1)
- (b) A leads B.
B1
- (c) 180° (π rad), antiphase
B1
- (a) Describe the graph of acceleration a against displacement x for an object moving with simple harmonic motion. [2]
- (b) For one oscillator the gradient of this graph is −25 s⁻². Calculate the period. [2]
- (c) Describe the shape of the graph of velocity against displacement. [1]
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- (a) A straight line through the origin with negative gradient, extending from −x₀ to +x₀.
B1 straight line through origin; B1 negative gradient, between ±x₀
- (b) 1.26 s
ω² = 25, ω = 5.0 rad s⁻¹ (C1); T = 2π/5.0 = 1.26 s (A1)
- (c) An ellipse (a circle if axes are suitably scaled), centred on the origin.
B1
An oscillator moving with simple harmonic motion has total energy 2.4 mJ and amplitude 2.0 cm.
- (a) Calculate the potential energy of the oscillator when its displacement is 1.0 cm. [2]
- (b) Calculate the kinetic energy at this displacement. [1]
- (c) Calculate the displacement at which the kinetic energy equals the potential energy. [2]
Show answer
- (a) 0.60 mJ
PE ∝ x², so PE = 2.4 × (1.0/2.0)² (C1) = 0.60 mJ (A1)
- (b) 1.8 mJ
2.4 − 0.60 = 1.8 mJ (B1)
- (c) 1.4 cm
PE = 1.2 mJ: (x/2.0)² = 0.5 (C1); x = 1.4 cm (A1)
Water in a U-tube is displaced and then oscillates. The acceleration a of the water surface is related to its displacement x from equilibrium by a = −(2g/L)x, where L = 0.40 m is the total length of the water column and g = 9.81 m s⁻².
- (a) Explain why the motion of the liquid is simple harmonic. [1]
- (b) Calculate the period of oscillation of the liquid. [2]
Show answer
- (a) The acceleration is proportional to the displacement and in the opposite direction (a = −(2g/L)x has the form a = −ω²x).
B1
- (b) 0.90 s
ω² = 2 × 9.81/0.40 = 49.1, ω = 7.0 rad s⁻¹ (C1); T = 2π/7.0 = 0.90 s (A1)
A data-logger shows that the velocity of a swinging lantern varies sinusoidally with time, with maximum value 0.60 m s⁻¹ and period 1.5 s.
- (a) Calculate the amplitude of the oscillation. [2]
- (b) Calculate the maximum acceleration. [1]
Show answer
- (a) 0.14 m
ω = 2π/1.5 = 4.19 rad s⁻¹ (C1); x₀ = v₀/ω = 0.60/4.19 = 0.14 m (A1)
- (b) 2.5 m s⁻²
a₀ = ωv₀ = 4.19 × 0.60 = 2.5 m s⁻² (B1)
A buoy in the sea at Gwadar bobs up and down with simple harmonic motion. It makes 12 complete oscillations in 30 s and the vertical distance between its highest and lowest positions is 0.50 m.
- (a) Calculate the period and the maximum speed of the buoy. [2]
- (b) Calculate the displacement of the buoy 0.40 s after it passes upwards through its equilibrium position. [2]
Show answer
- (a) 2.5 s; 0.63 m s⁻¹
T = 30/12 = 2.5 s (B1); v₀ = (2π/2.5) × 0.25 = 0.63 m s⁻¹ (B1)
- (b) 0.21 m above equilibrium
x = 0.25 sin(2π × 0.40/2.5) (C1) = 0.21 m (A1)
- (a) State the phase difference between (i) the velocity and the displacement, (ii) the acceleration and the displacement. [2]
- (b) The maximum speed of an oscillator is 0.80 m s⁻¹. Calculate its speed when the displacement is half the amplitude. [2]
Show answer
- (a) (i) π/2 rad (velocity leads by 90°) (ii) π rad (antiphase)
B1 each
- (b) 0.69 m s⁻¹
v = ω√(x₀² − x₀²/4) = (√3/2)v₀ (C1) = 0.69 m s⁻¹ (A1)
The displacement x of an oscillating object is given by x = 0.12 sin(5.0t), where x is in metres and t is in seconds (angles in radians).
- (a) State the amplitude and the period of the oscillation. [2]
- (b) Calculate the displacement and the velocity of the object at t = 0.40 s. [3]
- (c) Calculate the first time after t = 0 at which the displacement is 0.060 m. [2]
Show answer
- (a) 0.12 m; 1.26 s
x₀ = 0.12 m (B1); T = 2π/5.0 = 1.26 s (B1)
- (b) x = 0.109 m; v = −0.25 m s⁻¹
x = 0.12 sin(2.0 rad) = 0.109 m (B1); v = 0.12 × 5.0 cos(2.0 rad) (C1) = −0.25 m s⁻¹, i.e. moving back towards equilibrium (A1)
- (c) 0.105 s
sin(5.0t) = 0.50 (C1); 5.0t = π/6, t = 0.105 s (A1)
The piston of a car engine moves with approximately simple harmonic motion. The total distance moved between its highest and lowest positions (the stroke) is 9.0 cm. The engine turns at 3000 revolutions per minute, with one oscillation of the piston per revolution. The mass of the piston is 0.35 kg.
- (a) Calculate the angular frequency of the piston's motion. [2]
- (b) Calculate the maximum acceleration of the piston. [2]
- (c) Calculate the maximum resultant force on the piston. [1]
Show answer
- (a) 314 rad s⁻¹
f = 3000/60 = 50 Hz (C1); ω = 2π × 50 = 314 rad s⁻¹ (A1)
- (b) 4.4 × 10³ m s⁻²
x₀ = 0.045 m (C1); a₀ = 314² × 0.045 = 4.4 × 10³ m s⁻² (A1)
- (c) 1.6 × 10³ N
F = 0.35 × 4.44 × 10³ = 1.6 × 10³ N (B1)
At a harbour near Karachi the height of the sea above its mean level varies approximately with simple harmonic motion, with amplitude 1.8 m and period 12.4 hours.
- (a) Write an equation for the height h of the water above mean level at time t, taking h = +1.8 m (high tide) at t = 0. [2]
- (b) A ship can only pass the harbour entrance when the water is at least 1.0 m above mean level. Calculate the length of time in each tidal cycle during which the ship can pass. [3]
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- (a) h = 1.8 cos(0.507t), t in hours
B1 cos form with amplitude 1.8; B1 ω = 2π/12.4 = 0.507 rad h⁻¹
- (b) 3.9 h
1.8 cos ωt = 1.0 → ωt = 0.982 rad (C1); time from 0.982/0.507 = 1.94 h before to 1.94 h after high tide (C1); window = 3.9 h (A1)
A trolley of mass 0.25 kg on a horizontal frictionless track is attached to a spring system that provides a restoring force F = −kx, where k = 40 N m⁻¹ and x is the displacement. The trolley is pulled 4.0 cm from equilibrium and released.
- (a) By considering the resultant force, show that the trolley moves with simple harmonic motion and that ω = √(k/m). [2]
- (b) Calculate the period of the oscillations. [2]
- (c) Calculate the maximum speed and the maximum kinetic energy of the trolley. [2]
Show answer
- (a) F = −kx so a = −(k/m)x, which has the form a = −ω²x with ω² = k/m.
M1 F = −kx and a = F/m; A1 compare with a = −ω²x
- (b) 0.50 s
ω = √(40/0.25) = 12.6 rad s⁻¹ (C1); T = 2π/ω = 0.50 s (A1)
- (c) 0.51 m s⁻¹; 0.032 J
v₀ = 12.6 × 0.040 = 0.51 m s⁻¹ (B1); KE = ½ × 0.25 × 0.506² = 0.032 J (B1)
An object moves with simple harmonic motion of amplitude 2.5 cm. At t = 0 it passes through its equilibrium position moving in the positive direction. It first reaches its maximum positive displacement at t = 0.15 s.
- (a) State the period and the angular frequency. [2]
- (b) Calculate the velocity of the object at t = 0.10 s. [2]
- (c) Calculate the acceleration of the object at t = 0.10 s. [2]
Show answer
- (a) T = 0.60 s; ω = 10.5 rad s⁻¹
max displacement reached at a quarter period, so T = 4 × 0.15 = 0.60 s (B1); ω = 2π/0.60 = 10.5 rad s⁻¹ (B1)
- (b) 0.13 m s⁻¹
v = x₀ω cos ωt = 0.025 × 10.5 × cos(1.05 rad) (C1) = 0.13 m s⁻¹ (A1)
- (c) −2.4 m s⁻²
a = −ω²x₀ sin ωt = −10.5² × 0.025 × sin(1.05 rad) (C1) = −2.4 m s⁻² (A1)
A coin rests on a horizontal platform that oscillates vertically with simple harmonic motion of frequency 4.0 Hz. The amplitude is slowly increased. g = 9.81 m s⁻².
- (a) Explain why the coin may lose contact with the platform, and state at which point in the oscillation this first happens. [2]
- (b) Calculate the minimum amplitude at which the coin loses contact. [2]
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- (a) At the top of the motion the platform's downward acceleration is greatest; if this exceeds g, the coin (whose maximum downward acceleration is g) cannot follow and leaves the surface.
B1 needed downward acceleration > g; B1 at the highest point
- (b) 1.6 cm
ω²x₀ = g, ω = 2π × 4.0 = 25.1 rad s⁻¹ (C1); x₀ = 9.81/25.1² = 0.016 m (A1)
The tip of each prong of a tuning fork vibrates with simple harmonic motion of frequency 440 Hz and amplitude 0.50 mm.
- (a) Calculate the maximum speed of the tip of the prong. [2]
- (b) Calculate the maximum acceleration of the tip and express it as a multiple of g (9.81 m s⁻²). [2]
Show answer
- (a) 1.4 m s⁻¹
ω = 2π × 440 = 2760 rad s⁻¹ (C1); v₀ = 2760 × 0.50 × 10⁻³ = 1.4 m s⁻¹ (A1)
- (b) 3.8 × 10³ m s⁻², about 390g
a₀ = ω²x₀ = 3.8 × 10³ m s⁻² (B1); ÷ 9.81 = 390 (B1)
When four passengers of total mass 300 kg get into a car, the car body moves down by 4.9 cm. g = 9.81 m s⁻².
- (a) Calculate the effective spring constant of the suspension. [2]
- (b) The loaded car (total mass 1500 kg) bounces vertically after going over a bump. Assuming the motion is simple harmonic with ω = √(k/m), calculate the period. [2]
- (c) State and explain the effect on the period of the passengers leaving the car. [1]
Show answer
- (a) 6.0 × 10⁴ N m⁻¹
k = F/x = 300 × 9.81/0.049 (C1) = 6.0 × 10⁴ N m⁻¹ (A1)
- (b) 0.99 s
ω = √(6.0 × 10⁴/1500) = 6.3 rad s⁻¹ (C1); T = 2π/ω = 0.99 s (A1)
- (c) The period decreases because the mass decreases (T ∝ √m).
B1
- (a) State what is meant by damping. [1]
- (b) Describe the motion of a system that is (i) lightly damped, (ii) critically damped, (iii) heavily damped, after being displaced and released. [3]
- (c) State which type of damping is used in the suspension of a car and explain why. [1]
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- (a) The loss of energy (and so a decrease in amplitude) of an oscillating system caused by resistive forces.
B1
- (b) (i) oscillates with amplitude decreasing gradually (period nearly unchanged); (ii) returns to equilibrium in the shortest time without oscillating; (iii) returns to equilibrium slowly without oscillating.
B1 each
- (c) Critical damping, so the car returns to its rest position quickly without bouncing.
B1
- (a) Explain what is meant by resonance. [2]
- (b) Describe the effect of increasing the damping on the graph of amplitude against driving frequency. [3]
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- (a) When the driving frequency of a forced oscillation equals the natural frequency of the system, the amplitude is a maximum (maximum energy transfer).
B1 driving frequency = natural frequency; B1 maximum amplitude
- (b) The peak amplitude is lower at all frequencies, the peak becomes broader (flatter), and the resonant frequency shifts slightly lower.
B1 lower amplitude; B1 broader peak; B1 peak moves to lower frequency
A lightly damped pendulum is released with amplitude 10.0 cm. After 5 complete oscillations its amplitude is 8.0 cm.
- (a) Calculate the fraction of the original energy that remains after 5 oscillations. [2]
- (b) Assume the amplitude decreases by the same fraction in every 5 oscillations. Calculate the amplitude after 10 oscillations. [1]
- (c) State one cause of the damping. [1]
Show answer
- (a) 0.64
E ∝ A² (C1); (8.0/10.0)² = 0.64 (A1)
- (b) 6.4 cm
10.0 × 0.80 × 0.80 = 6.4 cm (B1)
- (c) Air resistance (or friction at the pivot).
B1
A mass on a spring is driven by a vibration generator. The amplitude A of the forced oscillations is measured at different driving frequencies f.f / Hz: 1.0, 1.5, 2.0, 2.5, 3.0A / cm: 0.8, 1.6, 4.9, 1.9, 0.9
- (a) State the natural frequency of the system and explain how you identified it. [2]
- (b) The oscillating mass is 0.30 kg. Using ω² = k/m, estimate the spring constant of the spring. [2]
- (c) The mass is replaced by one of 1.20 kg. State the new natural frequency. [1]
Show answer
- (a) About 2.0 Hz, where the amplitude is a maximum (resonance).
B1 2.0 Hz; B1 maximum amplitude at resonance
- (b) 47 N m⁻¹
ω = 2π × 2.0 = 12.6 rad s⁻¹ (C1); k = 0.30 × 12.6² = 47 N m⁻¹ (A1)
- (c) 1.0 Hz
f ∝ 1/√m: mass ×4 → f halved (B1)
A car of mass 1200 kg has a suspension of effective spring constant 6.0 × 10⁴ N m⁻¹. It drives over a series of speed bumps spaced 9.0 m apart along a road in Islamabad.
- (a) Calculate the natural frequency of vertical oscillation of the car on its suspension, using f = (1/2π)√(k/m). [2]
- (b) Calculate the speed at which the car would bounce with the largest amplitude, and explain your answer. [3]
- (c) Suggest why the shock absorbers are designed to give a large amount of damping (close to critical damping). [1]
Show answer
- (a) 1.1 Hz
f = (1/2π)√(6.0 × 10⁴/1200) (C1) = 1.13 Hz (A1)
- (b) 10 m s⁻¹ (about 36 km h⁻¹)
bumps act as a periodic driving force (B1); resonance when bump frequency v/9.0 = 1.13 Hz (C1); v = 10 m s⁻¹ (A1)
- (c) The damping reduces the amplitude at resonance and makes the car return to its rest position quickly, without continuing to bounce.
B1 smaller amplitude at resonance / oscillations die away quickly
A footbridge across a canal has a natural frequency of vertical oscillation of 2.0 Hz.
- (a) A group of joggers, each with a stride of 1.5 m, crosses the bridge in step. Calculate the jogging speed at which the bridge would oscillate with the largest amplitude. [2]
- (b) Explain why soldiers are often told to break step when marching across a bridge. [2]
- (c) Suggest one way engineers can reduce the risk without changing the bridge's natural frequency. [1]
Show answer
- (a) 3.0 m s⁻¹
resonance when step frequency = 2.0 Hz (C1); v = 2.0 × 1.5 = 3.0 m s⁻¹ (A1)
- (b) Marching in step gives a periodic driving force; if its frequency matches the natural frequency of the bridge, resonance causes large, possibly damaging, oscillations.
B1 periodic force at one frequency; B1 resonance → large amplitude
- (c) Fit dampers (e.g. tuned mass dampers or viscous dampers) to increase damping.
B1
