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Correct option is A)

$⇒x=Asin((T2π )t+ϕ)$ ; $T=6s$

so the equation becomes

$x=Asin(3πt +ϕ)$

We are given that at $t=1s,x=0$; so we have

$0=Asin(3π +ϕ)⇒(3π +ϕ)=0⇒ϕ=−3π ⇒x=Asin(3πt −3π )⇒v=dtdx =dtd (Asin(3πt −3π ))=3πA cos(3πt −3π )$

Now we are given that at $t=2s,∣v∣=0.25m/s$, so we have

$0.25=∣∣∣∣∣ 3πA cos(32π −3π )∣∣∣∣∣ ⇒0.25=∣∣∣∣∣ 3πA cos(3π )∣∣∣∣∣ ⇒0.25=6πA ⇒A=2π3 $

Solve any question of Oscillations with:-

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