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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 π2
For a conservative force, we can say that
F=(Ux^i+Uy^j+Uz^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=16ω=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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