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The potential energy of a particle of ma...

The potential energy of a particle of mass 1 kg in motin along the x-axis is given by
U = 4(1-cos2x)J
Here, x is in meter. The period of small osciallationis (in second) is

A

`2pi`

B

`pi`

C

`(pi)/(2)`

D

`sqrt2pi`

Text Solution

Verified by Experts

The correct Answer is:
C
`F=-(dU)/(dx)=-8sin2x`
For small oscillation, `sin2x=2x`
i.e.,
Since `amu-x`, the oscillation are simple harmonic in nature.
`T=pisqrt(|(x)/(a)|)=2pisqrt((1)/(16))=(pi)/(2)s`
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