A particle performing SHM is found at its equilibrium at t=1sec and it is found to have a speed of 0.25m/s at t=2s. If the period of oscillation is 6sec, calculate amplitude of oscillation.
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Correct option is A)
Since a SHM can be represented by x=Asin(ωt+ϕ)
⇒x=Asin((T2π)t+ϕ) ; T=6s
so the equation becomes x=Asin(3πt+ϕ)
We are given that at t=1s,x=0; so we have 0=Asin(3π+ϕ)⇒(3π+ϕ)=0⇒ϕ=−3π⇒x=Asin(3πt−3π)⇒v=dtdx=dtd(Asin(3πt−3π))=3πAcos(3πt−3π) Now we are given that at t=2s,∣v∣=0.25m/s, so we have 0.25=∣∣∣∣∣3πAcos(32π−3π)∣∣∣∣∣⇒0.25=∣∣∣∣∣3πAcos(3π)∣∣∣∣∣⇒0.25=6πA⇒A=2π3