JEE Main 2022 — Oscillations Question with Solution

Given here, potential energy, $U=4\left(1-\cos 4x\right) \mathrm{J}$

Using the relation between conservative force and potential energy, $F=-\frac{dU}{dx}$.

We have, $F=-4\left(+\sin 4x\right)4=-16\sin \left(4x\right)$

For small $\theta$, $\sin \theta \approx \theta$.

Acceleration of particle is $a=-\frac{64x}{m}=\frac{-64x}{4}=-16x$

As the oscillations are simple harmonic in nature, so  $a=-{\omega }^{2}x\Rightarrow {\omega }^2=16\Rightarrow \omega =4 \mathrm{rad} {\mathrm{s}}^{-1}$

Now, time period of oscillation is$T=\frac{2\pi }{\omega }=\frac{\pi }{2}$.

Thus, the value of $K=2$.