Question 97 of 194

A particle is executing simple harmonic motion with a time period $T$. At time $t=0$, it is at its position of equilibrium. The kinetic energy - time graph of the particle will look like:

ABCD

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Answer: (A)

$x=A\sin ⁡\omega t$

$v=\dfrac{dx}{dt}=a\omega \cos ⁡\omega t$

$K.E.=\dfrac{1}{2}m{v}^{2}$

$=\dfrac{1}{2}m{A}^{2}{\omega }^{2}{\cos }^{2}⁡\omega t$

$=\dfrac{1}{2}m{A}^{2}{\omega }^{2}{\cos }^{2}⁡\dfrac{2\pi }{T}t$

At $t=0$, $K.E.$ is maximum.

The period of $K.E.$ graph is half of the period of displacement graph.

Comment:- The points on the graph of $x-$axis should be differentiable.

In all option mentioned, the graph has cusps at the points on $x-$axis.