Question

A partical simple harmonic motion on x axis with angular velocity 400ead/sec when the parical x=25cm the velicoty of the partical 100c.M/sec...

Answer 1

A particle performs a linear S.H.M along a path 10 cm long. The particle starts from a distance of 1 cm from the mean position towards the positive extremity. Find the epoch and the phase of motion when the displacement is 2.5 cm.

Solution:

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.