A particle performing S.H.M is at its equilibrium when t=1s and it is found to have a speed of 0.25m/s at t=2s. If the period of oscillation is 6s. Calculate the amplitude of oscillation.
A
32πm
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B
34πm
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C
6πm
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D
38πm
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Solution
The correct option is A32πm Given that, T=6s⇒ω=2π6rad/s The general equation of SHM is x=Asin(ωt+ϕ) ∵ At t=1s;x=0 ⇒0=Asin[(ω×1)+ϕ]⇒ϕ=2nπ−ω (n is the integer) Now, v=dxdt=Aωcos(ωt+ϕ) ∵ at t=2s;v=14m/s ⇒14=Aωcos[(ω×2)+2nπ−ω] ⇒14=Aωcos(2nπ+ω)=Aωcos(ω) By putting values ⇒14=A×2π6cos2π6 ⇒A=32πm Hence, option (a) is the correct answer.