Q 11-13-149JEE MainJEE Main 2017 (2 Apr)Easy
A particle is executing simple harmonic motion with a time period $T$. At time $t = 0$, it is at its position of equilibrium. The kinetic energy–time graph of the particle will look like:
Answer: (A) see figure
Starting from the mean position, $x = A\sin\omega t$ and $v = A\omega\cos\omega t$, so
$$K = \tfrac12 mA^2\omega^2\cos^2\omega t = \tfrac14 mA^2\omega^2(1 + \cos2\omega t)$$
- $K$ is maximum at $t = 0$ (mean position).
- $K$ becomes zero at the extreme position, $t = T/4$, and is maximum again at $t = T/2$.
- $K$ never becomes negative and repeats with period $T/2$.
Graph (1) shows exactly this.
Solution by Sreeraj P, M.Sc Physics