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Question

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

A
32π m
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B
34π m
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C
6π m
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D
38π m
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Solution

The correct option is A 32π m
Given that, T=6 sω=2π6 rad/s
The general equation of SHM is
x=Asin(ωt+ϕ)
At t=1 s ; x=0
0=Asin[(ω×1)+ϕ] ϕ=2nπω (n is the integer)
Now, v=dxdt=Aωcos(ωt+ϕ)
at t=2 s ; v=14 m/s
14=Aωcos[(ω×2)+2nπω]
14=Aωcos(2nπ+ω)=Aωcos(ω)
By putting values
14=A×2π6cos2π6
A=32π m
Hence, option (a) is the correct answer.

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