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Simple Harmonic Motion - Multiple Choice Questions

Q18. A particle of mass 0.20 kg moves with simple harmonic motion of amplitude 2.0 × 10–2m.

If the total energy of the particle is 4.0 × 10–5J, what is the time period of the motion?

A – π/4 seconds

B – π/2 seconds

C - π seconds

D - 2π seconds

Total energy = max kinetic energy = 1/2 mv 2 at the centre of the path where velocity = maximum

vmax = 2πfa - from data sheet

Total energy =1/2 m (2πfa)2 = 4.0 × 10–5

1/2 x 0.2 (2πf x2.0 × 10–2)2 = 4.0 × 10–5

4π2f2x 4.0 × 10–6 = 4.0 × 10–6

f2 = 1/(4π2)

T2 = 4π2

∴T = 2π - choice D