Question

A Particle Performing SHM is found at its equilibrium at t=1 sec.and it is found to have a speed of 0.25m/s at t=2 sec.If the period of oscillation is 6 sec.Calculate amplitude of oscillation.

Answer 1

The amplitude of oscillation is 3/2π m.

Explanation:

• ω=2π/T=2π/6

Displacement of particle is given by

x = A.sin[ωt+ϕ]

At t=1s, x=0

0 = A.sin[(2π/6)×1+ϕ]

Solving this we'll get

ϕ = -π/3

• Velocity of the particle is given by

v = A.ω.cos[ωt+ϕ]

0.25 = A.2π/6.cos[(2π/6)×1-π/3]

Solving this we'll get

A = 3/2π m

Hence the amplitude of oscillation is 3/2π m.

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