The displacement of a particle moving in S.H. | vedclass

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(D) The displacement of the particle is given by $y = a \sin \omega t$.The velocity $v$ is the derivative of displacement with respect to time: $v = \

Vedclass · Accepted Answer

$-a \omega^2$

Vedclass · Answer

(D) The displacement of the particle is given by $y = a \sin \omega t$.The velocity $v$ is the derivative of displacement with respect to time: $v = \frac{dy}{dt} = a \omega \cos \omega t$.The acceleration $A$ is the derivative of velocity with respect to time: $A = \frac{dv}{dt} = -a \omega^2 \sin \omega t$.Given $t = \frac{T}{4}$ and knowing $\omega = \frac{2\pi}{T}$,we have $\omega t = \frac{2\pi}{T} 	imes \frac{T}{4} = \frac{\pi}{2}$.Substituting this into the acceleration equation: $A = -a \omega^2 \sin(\frac{\pi}{2}) = -a \omega^2 (1) = -a \omega^2$.Thus,the correct option is $D$.

The displacement of a particle moving in $S.H.M.$ at any instant is given by $y = a \sin \omega t$. The acceleration after time $t = \frac{T}{4}$ is (where $T$ is the time period).

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