The potential energy of a particle of mass 4kg in motion along the...

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

Using the relation between conservative force and potential energy, $F = - \dfrac{d U}{d x}$.

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 $a = - \dfrac{64x}{m} = \dfrac{- 64x}{4} = - 16x$

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

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

Thus, the value of $K = 2$.