The potential energy of a particle of mass $4 \mathrm{~kg}$ in motion along the x-axis is given by $\mathrm{U}=4(1-\cos 4 x)$ J. The time period of the particle for small oscillation $(\sin \theta \simeq \theta)$ is $\left(\frac{\pi}{K}\right) s$. The value of $\mathrm{K}$ is _________.
Answer (integer)
2
Solution
<p>$U = 4(1 - \cos 4x)$</p>
<p>$\Rightarrow F = - {{dU} \over {dx}} = - (4)(4\sin 4x)$</p>
<p>$= - 16\sin 4x$</p>
<p>as small x</p>
<p>$F = - 16(4x) = - 64x \equiv - kx$</p>
<p>$T = 2\pi \sqrt {{m \over k}} = 2\pi \sqrt {{4 \over {64}}} = {\pi \over 2}$</p>
<p>$\Rightarrow K = 2$</p>
About this question
Subject: Physics · Chapter: Oscillations · Topic: Simple Harmonic Motion
This question is part of PrepWiser's free JEE Main question bank. 88 more solved questions on Oscillations are available — start with the harder ones if your accuracy is >70%.