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Question

(C) The displacement equation for a particle executing simple harmonic motion starting from the mean position is given by $y = a \sin(\frac{2\pi}{T}t)

Vedclass · Accepted Answer

$T/8$

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(C) The displacement equation for a particle executing simple harmonic motion starting from the mean position is given by $y = a \sin(\frac{2\pi}{T}t)$.Given that the displacement $y = \frac{a}{\sqrt{2}}$,we substitute this into the equation:$\frac{a}{\sqrt{2}} = a \sin(\frac{2\pi}{T}t)$Dividing both sides by $a$,we get:$\sin(\frac{2\pi}{T}t) = \frac{1}{\sqrt{2}}$Since $\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$,we equate the arguments:$\frac{2\pi}{T}t = \frac{\pi}{4}$Solving for $t$:$t = \frac{\pi}{4} \cdot \frac{T}{2\pi} = \frac{T}{8}$Thus,the shortest time taken is $\frac{T}{8}$ seconds.

$A$ particle is executing simple harmonic motion with a period of $T$ seconds and amplitude $a$ meters. The shortest time it takes to reach a point $\frac{a}{\sqrt{2}} \, m$ from its mean position in seconds is

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