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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

AI Generated Solution

To solve the problem, we need to analyze the potential energy function and the total energy of the particle executing simple harmonic motion. ### Step-by-Step Solution: 1. **Identify the potential energy function**: The potential energy \( V(x) \) is given by: \[ V(x) = \frac{1}{2} k x^2 ...
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