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In SHM , potential energy of a particle ...

In SHM , potential energy of a particle at mean position is `E_(1)` and kinetic enregy is `E_(2)` , then

A

`E_(1)=E_(2)`

B

total potential energy at `x=(sqrt(3)A)/(2)` is `E_(1)+(3E_(2))/(4)`

C

total kinetic energy at `x=(sqrt(3)A)/(2)` is `(3E_(2))/(4)`

D

total kinetic energy at `x=(A)/(sqrt(2))` is `(E_(2))/(4)`

Text Solution

Verified by Experts

The correct Answer is:
B
(b) `U=U_("mean")+(1)/(2)kX^(2)`
Given , `U_("mean")=E_(1)` and `(1)/(2)kA^(2)=E_(2)` = maximum kinetic energy at mean position
`therefore X=(sqrt(3)A)/(2), U=E_(1)+(1)/(2)k((sqrt(3)A)/(2))^(2)=E_(1)+(3)/(4)E_(2)`
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