Questions on simple harmonic motion

Question 1

State the defining condition for simple harmonic motion.

Accepted Answer

The acceleration is directly proportional to the displacement from the equilibrium position (1) and is always directed towards it: a = −ω²x (a ∝ −x) (1).

Question 2

A body oscillates with a frequency of 2.0 Hz. Calculate its angular frequency.

Accepted Answer

ω = 2πf = 2π × 2.0 (1) ω = 12.6 rad s⁻¹ (1)

Question 3

An oscillator has an angular frequency of 10 rad s⁻¹ and an amplitude of 0.050 m. Calculate its maximum speed.

Accepted Answer

v_max = ωA = 10 × 0.050 (1) v_max = 0.50 m s⁻¹ (1)

Question 4

An oscillator has an angular frequency of 7.0 rad s⁻¹. Calculate the magnitude of its acceleration when its displacement is +0.030 m, and state the direction of this acceleration.

Accepted Answer

a = ω²x = 7.0² × 0.030 = 1.5 m s⁻² (1). It is directed towards the equilibrium position, opposite to the displacement (1).

Question 5

For a body in simple harmonic motion, state the displacement at which (a) the acceleration is greatest and (b) the speed is greatest.

Accepted Answer

(a) At maximum displacement, x = ±A, the extremes of the motion (1). (b) At zero displacement, the equilibrium position (1).

Question 6

A vibrating machine panel completes 50 oscillations in 40 s. Calculate the frequency and the period of the oscillation.

Accepted Answer

f = 50/40 = 1.25 Hz (1) T = 1/f = 0.80 s (1)

Question 7

A body performs SHM at 3.0 Hz with an amplitude of 0.020 m. Calculate its maximum acceleration.

Accepted Answer

ω = 2πf = 2π × 3.0 (1) ω = 18.85 rad s⁻¹ (1) a_max = ω²A = 7.11 m s⁻² (1)

Question 8

A glider on an air track performs SHM with amplitude 0.10 m and angular frequency 8.0 rad s⁻¹. Calculate its speed when its displacement is 0.060 m.

Accepted Answer

v = ±ω√(A² − x²) (1) = 8.0 × √(0.10² − 0.060²) (1) v = 0.64 m s⁻¹ (1)

Question 9

An oscillator moves as x = A cos ωt with A = 0.15 m and ω = 4.0 rad s⁻¹. Calculate its displacement at t = 0.20 s.

Accepted Answer

x = A cos ωt = 0.15 × cos(4.0 × 0.20) (1) = 0.15 × cos(0.80), with the calculator in radians (1) x = 0.105 m (1)

Question 10

An oscillator has an angular frequency of 6.28 rad s⁻¹. Calculate its frequency and its period.

Accepted Answer

f = ω/2π = 6.28/(2π) = 1.00 Hz (1) T = 1/f = 1.00 s (1)

Question 11

A buoy bobbing on water performs SHM with amplitude 0.12 m and frequency 0.50 Hz. Timing starts as the buoy passes upwards through its equilibrium position, so its displacement is x = A sin ωt. Calculate its displacement at t = 0.40 s.

Accepted Answer

ω = 2πf = 2π × 0.50 = 3.14 rad s⁻¹ (1) x = 0.12 × sin(3.14 × 0.40), with the calculator in radians (1) x = 0.114 m (1)

Question 12

A loudspeaker cone reproducing a test tone vibrates in SHM at 250 Hz with amplitude 0.35 mm. The manufacturer claims the cone's maximum acceleration exceeds 50g (g = 9.81 m s⁻²). Deduce whether the claim is correct.

Accepted Answer

ω = 2πf = 2π × 250 = 1571 rad s⁻¹ (1) a_max = ω²A = 1571² × 3.5 × 10⁻⁴ (1) a_max = 864 m s⁻² (1) 50g = 490 m s⁻², and 864 m s⁻² is larger, so the claim is correct (1)

Question 13

An oscillator is released from rest at its amplitude of 0.080 m and has a period of 1.2 s, so x = A cos ωt. Calculate the time at which its displacement first falls to 0.020 m.

Accepted Answer

cos ωt = 0.020/0.080 = 0.25 (1) ωt = cos⁻¹(0.25) = 1.32 rad, with ω = 2π/1.2 = 5.24 rad s⁻¹ (1) t = 1.32/5.24 = 0.25 s (1)

Question 14

A body performs SHM with amplitude 0.080 m and period 0.50 s. Calculate (a) its angular frequency, (b) its maximum speed and (c) its maximum acceleration.

Accepted Answer

(a) ω = 2π/T = 2π/0.50 (1) ω = 12.57 rad s⁻¹ (1) (b) v_max = ωA = 1.01 m s⁻¹ (1) (c) a_max = ω²A = 12.6 m s⁻² (1)

Question 15

An object performs SHM of amplitude 0.10 m. Calculate the displacement at which its speed is half of its maximum speed.

Accepted Answer

v = ½v_max: ½ωA = ω√(A² − x²) (1) ¼A² = A² − x² (1) x = (√3/2)A (1) x = 0.087 m (1)

Question 16

Describe the phase relationship between the displacement, velocity and acceleration in SHM.

Accepted Answer

Velocity is a quarter of a cycle (90°) ahead of displacement (1), and is greatest when displacement is zero (1). Acceleration is exactly out of phase (180°) with displacement, being greatest and opposite when displacement is at a maximum (1).

Question 17

A student measures the acceleration of an oscillating trolley at three displacements. At x = 0.020 m, a = −0.50 m s⁻²; at x = 0.040 m, a = −1.00 m s⁻²; at x = 0.060 m, a = −1.50 m s⁻². Deduce whether the motion is simple harmonic and, if it is, determine the angular frequency.

Accepted Answer

a/x is the same for every pair: −0.50/0.020 = −1.00/0.040 = −1.50/0.060 = −25 s⁻², so a ∝ x (1)
The acceleration is opposite in sign to the displacement, that is directed towards equilibrium, so the motion is SHM (1)
Comparing with a = −ω²x gives ω² = 25 s⁻² (1)
ω = 5.0 rad s⁻¹ (1)

Question 18

An oscillator of period T passes through its equilibrium position at t = 0, so x = A sin ωt. Show that the time it takes to travel from the equilibrium position to a displacement of A/2 is T/12.

Accepted Answer

A/2 = A sin ωt gives sin ωt = ½ (1) The first solution is ωt = π/6 (1) With ω = 2π/T, t = (π/6)/(2π/T) = T/12 (1)

Question 19

A component in a machine oscillates with SHM. Sensors record a maximum speed of 0.90 m s⁻¹ and a maximum acceleration of 14 m s⁻². Determine the angular frequency, the amplitude and the frequency of the oscillation.

Accepted Answer

a_max/v_max = ω²A/(ωA) = ω (1) ω = 14/0.90 = 15.6 rad s⁻¹ (1) A = v_max/ω = 0.90/15.6 = 0.058 m (1) f = ω/2π = 2.5 Hz (1)