Question 75 of 194

A particle oscillates along the $x$-axis according to the law, $x(\mathrm{t})=x_0 \sin ^2\left(\frac{\mathrm{t}}{2}\right)$ where $x_0=1 \mathrm{~m}$. The kinetic energy $(\mathrm{K})$ of the particle as a function of $x$ is correctly represented by the graph

ABCD

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Answer: (D)

$x(t)=x_0 \sin ^2\left(\frac{t}{2}\right)=\frac{x_0}{2}(1-\cos t)$


Clearly $\frac{x_0}{2}$ is mean position.