The potential energy of a particle of mass 4 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 $$(\sin\theta \approx \theta)$$ is $$\frac{\pi}{K}$$ s. The value of $$K$$ is _____

We have a particle of mass $$m = 4 \text{ kg}$$ with potential energy $$U = 4(1 - \cos 4x)$$ J. We need to find the time period for small oscillations.

The force on the particle is $$F = -\frac{dU}{dx} = -4 \cdot 4\sin 4x = -16\sin 4x$$.

For small oscillations, we use the approximation $$\sin 4x \approx 4x$$ (since $$\sin\theta \approx \theta$$ for small $$\theta$$). So the force becomes $$F = -16 \cdot 4x = -64x$$.

This is of the form $$F = -kx$$ with effective spring constant $$k = 64 \text{ N/m}$$. The angular frequency of oscillation is $$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{64}{4}} = \sqrt{16} = 4 \text{ rad/s}$$.

The time period is $$T = \frac{2\pi}{\omega} = \frac{2\pi}{4} = \frac{\pi}{2} \text{ s}$$.

Comparing with the given form $$T = \frac{\pi}{K}$$, we get $$K = 2$$.

Hence, the value of $$K$$ is $$\textbf{2}$$.