The kinetic energy,K of a body performing sim | vedclass

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(A) The kinetic energy $K$ of a body performing simple harmonic motion is given by the formula:$K = \frac{1}{2} m v^2$where $v$ is the velocity of the

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(A) The kinetic energy $K$ of a body performing simple harmonic motion is given by the formula:$K = \frac{1}{2} m v^2$where $v$ is the velocity of the body.If the displacement is $y = a \sin \omega t$,then the velocity is:$v = \frac{dy}{dt} = a \omega \cos \omega t$Substituting this into the kinetic energy expression:$K = \frac{1}{2} m (a \omega \cos \omega t)^2 = \frac{1}{2} m a^2 \omega^2 \cos^2 \omega t$Using the trigonometric identity $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$,we get:$K = \frac{1}{2} m a^2 \omega^2 \left( \frac{1 + \cos 2\omega t}{2} ight) = \frac{1}{4} m a^2 \omega^2 (1 + \cos 2\omega t)$This expression shows that the kinetic energy $K$ is always non-negative and varies with a frequency double that of the simple harmonic motion (i.e.,frequency $2\omega$). The graph in option $A$ correctly represents this periodic variation,where $K$ oscillates between $0$ and a maximum value with a period of $T/2$.

The kinetic energy,$K$ of a body performing simple harmonic motion varies with time $t$,is indicated in graph

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