A particle oscillates along the $x$-axis according to the law, $x(t)=x_0\sin^2\left(\frac t2\right)$ where $x_0=1$ m. The kinetic energy (K) of the particle as a function of $x$ is correctly represented by the graph

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $x=\frac{x_0}2(1-\cos t)$ is SHM about $x=\frac12$ with amplitude $\frac12$. $K=\frac12m\omega^2\left[\frac14-\left(x-\frac12\right)^2\right]$, an inverted parabola that is zero at $x=0$ and $x=1$ with maximum at $x=\frac12$. This is option (4). Answer (4).

Q. A particle oscillates along the $x$-axis according to the law, $x(t)=x_0\sin^2\left(\frac t2\right)$ where $x_0=1$ m. The kinetic energy (K) of the particle as a function of $x$ is correctly represented by the graph

Explanation

Reviewed by Dr. Abubacker

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