The given graph illustrates a transverse wave | vedclass

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Question

(D) In a transverse wave,the elements of the string perform simple harmonic motion $(SHM)$.For an element in $SHM$,the displacement $y$ is given by $y

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The acceleration of the element at $S$ is maximum.

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(D) In a transverse wave,the elements of the string perform simple harmonic motion $(SHM)$.For an element in $SHM$,the displacement $y$ is given by $y = A \sin(\omega t + \phi)$.The velocity is $v = \frac{dy}{dt} = A\omega \cos(\omega t + \phi)$,which is maximum at the mean position $(y = 0)$.The acceleration is $a = \frac{d^2y}{dt^2} = -\omega^2 y$,which is maximum at the extreme positions $(y = \pm A)$.Points $P$ and $S$ are at the crests (extreme positions),so the acceleration of the elements at $P$ and $S$ is maximum.Point $Q$ is at the mean position,so its velocity is maximum,but its displacement is zero only at this instant,not always.Point $R$ is at the trough (extreme position),where its velocity is zero and its energy is entirely potential.Therefore,the correct statement is that the acceleration of the element at $S$ is maximum.

The given graph illustrates a transverse wave travelling on a string at a particular instant,and the points $P, Q, R$ and $S$ represent elements of the string. Which of the following statements about the motion of the elements is correct?

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