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A body is executing simple harmonic motion. At a displacement `x` its potential energy is `E_(1)` and at a displacement `y` its potential energy is `E_(2)` The potential energy `E` at displacement `(x+y)` is

A

`sqrtE=sqrt(E_1)-sqrt(E_2)`

B

`sqrtE=sqrt(E_1)+sqrt(E_2)`

C

`E=E_1-E_2`

D

`E=E_1+E_2`

Text Solution

Verified by Experts

The correct Answer is:
B

`E_1=1/2momega^2x^2`, `E_2=1/2momega^2y^2`
`E=1/2momega^2(x+y)^2`
`sqrtE=sqrt(1/2momega^2)(x+y)`
`=sqrt(1/2momega^2)((sqrtE_1)/(sqrt(1/2momega^2))+(sqrtE_2)/(sqrt1/2momega^2))`
`sqrtE=sqrtE_1+sqrtE_2`
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