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JEE Main 2022 (28 Jul, Shift 2) · Periodic Motion & SHM Basics

Periodic Motion & SHM Basics question from JEE Main 2022 (28 Jul, Shift 2)

Q41JEE Main 2022 · 28 JUL · SHIFT 2Medium

Periodic Motion & SHM Basics

The potential energy of a particle of mass $4\text{ kg}$ in motion along the x-axis is given by $U = 4(1 - \cos 4x)$ J. The time period of the particle for small oscillation $(\sin\theta \simeq \theta)$ is $\left(\frac{\pi}{K}\right)\text{ s}$ . The value of $K$ is _________.

Numerical answer type. Work it out, then open the answer.

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Answer: 2

For small $x$ : $1 - \cos 4x \approx \frac{(4x)^2}{2} = 8x^2$ , so $U \approx 32x^2$ .
Compare with $U = \frac{1}{2}kx^2$ : $k = 64\text{ N/m}$ .
$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{64}{4}} = 4\text{ rad/s}$ , $T = \frac{2\pi}{4} = \frac{\pi}{2}\text{ s}$
So $K = 2$ .

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