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.`

Answer

B

Explanation

At `x=+-x_(m)` the particle tuns back. Therefore, its stone II reaches the bottom earlier than stone I. `K=0`. The total energy `E` is in the form of potential energy `i.e.V=E`.