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The potential energy of a particle of ma...

The potential energy of a particle of mass 0.1 kg, moving along the x-axis, is given by `U=5x(x-4)J`, where x is in meter. It can be concluded that

A

The speed of the particle is maximum at x=2 m

B

The particle executes simple harmonic motion

C

The period of oscillation of the particle is `(pi)/(5)` s

D

all are correct

Text Solution

Verified by Experts

The correct Answer is:
D
(d) `U=5x^(2)-20x,(dU)/(dx)=10x-20`
`F=(dU)/(dx)=(-10x+20)`
Let us suppose, x=(X+2). Then `F=-10X, F=0 " at "X=0`
or x=2 , i.e., x=2m is the mean position about which particle is in SHM, k=10
`therefore T=2pisqrt((m)/(k))=2pisqrt((0.1)/(10))=(pi)/(5)s`. At mean position ,kinetic energy is maximum and force acitng on particle is SHM is variable.
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