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The potential energy of a particle oscil...

The potential energy of a particle oscillating on x-axis is given as `U =20 +(x-2)^(2)`. The mean position is at

A

x = 2m

B

x = 1m

C

x = 3m

D

x = 4m

Text Solution

Verified by Experts

The correct Answer is:
A
`U = 20 + (x - 2)^(2)`
For min `U, (du)/(dx) = 0 rArr 2 (x - 2) = 0`
`rArr x = 2`
U is min and K is max at mean position
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