Question

A particle of mass 10 kg is describing SHM along a straight line with time period 2s nd amplitude 10m .Its K.E when it is 5m away from mean position?

Answer 1

Given: Path length = 10 cm, amplitude = path length /2 = 10/2 = 5 cm, Initial displacement = x0 = 1 cm, Displacement = x = 2.5 cm.

To Find: Epoch = α =? and phase of S.H.M. = (ωt + α) = ?

Epoch = α = sin-1(xo/a) = sin-1(1/5) = sin-1(0.2) = 11°32’

Displacement of a particle performing S.H.M. is given by

x = a sin (ωt + α)

∴ 2.5 = 5 sin (ωt + α)

∴ sin (ωt + α) = 2.5/5 = 1/2

∴ (ωt + α) = sin-1(1/2) = π/6

Ans: Initial phase is 11°32’ and phase of S.H.M. is π/6 or 30o.