JEE MAIN - Physics (2022 - 28th July Evening Shift - No. 22)
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
2
Explanation
$$U = 4(1 - \cos 4x)$$
$$ \Rightarrow F = - {{dU} \over {dx}} = - (4)(4\sin 4x)$$
$$ = - 16\sin 4x$$
as small x
$$F = - 16(4x) = - 64x \equiv - kx$$
$$T = 2\pi \sqrt {{m \over k}} = 2\pi \sqrt {{4 \over {64}}} = {\pi \over 2}$$
$$ \Rightarrow K = 2$$
Comments (0)
