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The potential energy function for a part...

The potential energy function for a particle executing simple harmonic motion is given by `V(x)=(1)/(2)kx^(2)`, where k is the force constant of the oscillatore. For `k=(1)/(2)Nm^(-1)`, show that a particle of total energy 1 joule moving under this potential must turn back when it reaches `x=+-2m.`

A

V=0, K=E

B

V=E, K=O

C

`V lt E`, K=O

D

V=O, `K lt E`

Text Solution

Verified by Experts

(b) Total energy is E=PE+KE
When particle is at `x=x_(m)` i.e. at extreme positron, returns back. Hence, at `x=x_(m) x=0 KE=0`
From Eq.(ii) `E=PE+O=PE=V((x_(m))=(1)/(2)kx_(m)^(2))`
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