A particle oscillates along the $x$-axis according to the law, $x(\mathrm{t})=x_0 \sin…
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
Solution
$x(t)=x_0 \sin ^2\left(\frac{t}{2}\right)=\frac{x_0}{2}(1-\cos t)$ Clearly $\frac{x_0}{2}$ is mean position.
