The potential energy of a particle of mass 4 ~kg in motion along the x-axis is given by U=4(1-\cos 4 x) J. The time period of the particle for small oscillation (\sin \theta \simeq \theta) is (\frac\piK) s . The value of K is \underline\hspace2cm .

Simple Harmonic Motion PYQ — JEE Main 2022 · 28 Jul · Shift 2 · Q61 with Solution

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Correct answer: 2

\cos 4x)$$

For small oscillations about the equilibrium position, we use the small-angle approximation:

$\cos 4x \approx 1-\frac{(4x)^2}{2} = 1-8x^2$

Approximate the potential near equilibrium

Substitute into $U$ :

$U(x) \approx 4\left[1-(1-8x^2)\right] = 4(8x^2)=32x^2$
Compare with SHM potential energy form

For simple harmonic motion,

$U(x)=\frac{1}{2}kx^2$

So,

$\frac{1}{2}kx^2 = 32x^2$

Therefore,

$k=64\ \text{N/m}$
Find angular frequency

Given mass:

$m=4\ \text{kg}$

Angular frequency for SHM is

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{64}{4}}=\sqrt{16}=4\ \text{rad/s}$
Find time period

$T=\frac{2\pi}{\omega}=\frac{2\pi}{4}=\frac{\pi}{2}\ \text{s}$

Comparing with

$T=\frac{\pi}{K}$

we get

$K=2$
Comparison with stored answer

Stored correct answer = 2, which matches our result.