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(C) For a particle performing simple harmonic motion,the speed $v$ at displacement $x$ is given by the formula:$v^2 = \omega^2(A^2 - x^2)$Expanding th

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(C) For a particle performing simple harmonic motion,the speed $v$ at displacement $x$ is given by the formula:$v^2 = \omega^2(A^2 - x^2)$Expanding this equation,we get:$v^2 = \omega^2 A^2 - \omega^2 x^2$This equation is of the form $y = mx + c$,where $y = v^2$,$x = x^2$,$m = -\omega^2$ (slope),and $c = \omega^2 A^2$ (y-intercept).Since the slope is negative $(-\omega^2)$,the graph of $v^2$ versus $x^2$ is a straight line with a negative slope and a positive y-intercept.Therefore,the correct graph is a straight line starting from the y-axis and sloping downwards.

$A$ particle is performing simple harmonic motion with angular frequency $\omega$ and amplitude $A$. If $a$ is acceleration and $v$ is speed at any instant,then the graph showing the correct variation between $v^2$ and $x^2$ is (where $x$ is displacement):

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