Question

# The potential energy of a particle of mass 1 kg moving along the X-axis is given by U=4(1+cos2x) J, where x is in meters. Find the time period of small oscillation (in second).

A
2π
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B
π
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C
π2
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D
2x
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Solution

## The correct option is C π2For a conservative force, we can say that →F=−(∂U∂x^i+∂U∂y^j+∂U∂z^k) Since the given potential energy is only a function x we can write the above equation as F=−dUdx From the data given in the question, F=−ddx[4(1+cos2x)]=−8sin2x ....(1) Let m be the mass and a be the acceleration of the particle. From Newton's second law, F=ma Using(1) in the above equation we get, a=−8sin2x For small values of x, sin2x is approximately equal to 2x. Thus, a=−16x Comparing this with a=−ω2x we get, ω2=16⇒ω=4 We know that, The time period of oscillation is given by T=2πω ⇒T=2π4=π2 s Thus, option (c) is the correct answer.

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