Question

A particle performing shm is found at its equilibrium at t =1 s and it is found to have a speed of 0.25 m/s at t =2 .If the period of oscillation is 6 s calc

Answer 1

Hey dear,

Your question is incomplete.

# Complete 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 - 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

Amplitude of oscillation is 3/2π m.

Keep studying...

Answer 2

Explanation:

first put x=Asin ( wt+ Ø) calculate you got the right ans