Given here, potential energy, U=4(1−cos4x)J
Using the relation between conservative force and potential energy, F=−dxdU.
We have, F=−4(+sin4x)4=−16sin(4x)
For small θ, sinθ≈θ.
Acceleration of particle is a=−m64x=4−64x=−16x
As the oscillations are simple harmonic in nature, so a=−ω2x⇒ω2=16⇒ω=4rads−1
Now, time period of oscillation isT=ω2π=2π.
Thus, the value of K=2.