Q 11-13-050JEE MainJEE Main 2025 (24 Jan, Shift 2)Medium
A particle oscillates along the $x$-axis according to the law $x(t) = x_0\sin^2\left(\dfrac{t}{2}\right)$, where $x_0 = 1\ \text{m}$. The kinetic energy $(K)$ of the particle as a function of $x$ is correctly represented by the graph
Answer: (D) see figure
$$x = \sin^2\frac{t}{2} = \frac{1}{2} - \frac{1}{2}\cos t$$
This is SHM about $x = \tfrac{1}{2}$ with amplitude $A = \tfrac{1}{2}$ and $\omega = 1$. So
$$K = \frac{1}{2}m\omega^2\left[A^2 - \left(x - \frac{1}{2}\right)^2\right]$$
an inverted parabola that is zero at $x = 0$ and $x = 1$ (the extremes) and maximum at $x = \tfrac{1}{2}$: graph (4).
Solution by Sreeraj P, M.Sc Physics