Question

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 $(sin θ ≃ θ)$ is $(\frac{π}{K})$ s. The value of $K$ is ..........

Answer 1

Given here, potential energy, $U=4(1-cos4x) J$

Using the relation between conservative force and potential energy, $F=-\frac{dU}{dx}$.

We have, $F=-4+sin4x4=-16sin4x$

For small $θ, sinθ≈θ$.

Acceleration of particle is $α=-\frac{64x}{m}=-\frac{64x}{4}=-16x$

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

Now, time period of oscillation is $T=\frac{2π}{ω}=\frac{π}{2}$.

Thus, the value of $K=2$.