Oscillations

JEE Main level, numerical answer, medium

25 of 194

Question 25 of 194

The potential energy of a particle of mass 4kg4kg in motion along the xx-axis is given by U=4(1−cos⁡4x)JU=4\left(1-\cos 4x\right)J. The time period of the particle for small oscillation (sin⁡θ≃θ)\left(\sin \theta ≃\theta \right) (πK)s\left(\dfrac{\pi }{K}\right)s. The value of KK is _____ .

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Answer: 2

Given here, potential energy, U=4(1−cos⁡4x)JU=4\left(1-\cos 4x\right)J
Using the relation between conservative force and potential energy, F=−dUdxF=-\dfrac{dU}{dx}.
We have, F=−4(+sin⁡4x)4=−16sin⁡(4x)F=-4\left(+\sin 4x\right)4=-16\sin \left(4x\right)
For small θ\theta, sin⁡θ≈θ\sin \theta \approx \theta.
Acceleration of particle is a=−64xm=−64x4=−16xa=-\dfrac{64x}{m}=\dfrac{-64x}{4}=-16x
As the oscillations are simple harmonic in nature, so a=−ω2x⇒ω2=16⇒ω=4rads−1a=-{\omega }^{2}x\Rightarrow {\omega }^{2}=16\Rightarrow \omega =4rad{s}^{-1}
Now, time period of oscillation isT=2πω=π2T=\dfrac{2\pi }{\omega }=\dfrac{\pi }{2}.
Thus, the value of K=2K=2.

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