SHMHard

Question

A particle performing S.H.M. is found at its equilibrium at t = 1 sand it is found to have a speed of 0.25 m/s at t = 2s. If the period of osculation is 6s Calculate amplitude of oscillation

Options

A. m
B. m
C. m
D. m

Solution

x = a sin (ωt + φ)
at t = 1s, x = 0 = a sin (ωt + φ) ⇒ φ = - ω
∵ v = a  cos(ωt + φ)
∴ At t = 2s, = aωcos (2ω + φ)

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