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(C) Given the displacement equation: $x = A \cos \omega t$.The velocity $v$ is the first derivative of displacement with respect to time:$v = \frac{dx

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(C) Given the displacement equation: $x = A \cos \omega t$.The velocity $v$ is the first derivative of displacement with respect to time:$v = \frac{dx}{dt} = \frac{d}{dt}(A \cos \omega t) = -A \omega \sin \omega t$.The acceleration $a$ is the derivative of velocity with respect to time:$a = \frac{dv}{dt} = \frac{d}{dt}(-A \omega \sin \omega t) = -A \omega^2 \cos \omega t$.Comparing this with the original displacement equation $x = A \cos \omega t$,we see that $a = -\omega^2 x$. At $t = 0$,$x = A$,so $a = -A \omega^2$. This means the acceleration starts at a negative maximum value and follows a negative cosine curve. Graph $C$ represents this variation correctly.

$A \ (mm \ s^{-2})$	$16$	$8$	$0$	$-8$	$-16$
$x \ (mm)$	$-4$	$-2$	$0$	$2$	$4$

Similar Questions

Values of the acceleration $A$ of a particle moving in simple harmonic motion as a function of its displacement $x$ are given in the table below:
$A \ (mm \ s^{-2})$ $16$ $8$ $0$ $-8$ $-16$
$x \ (mm)$ $-4$ $-2$ $0$ $2$ $4$

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