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)

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