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.`

Explanation

As is known, the particle will turn back when whole of its energy is converted into potential energy,
i.e., `V(x)=(1)/(2)kx^(2)=1(jou l e )`
`(1)/(2)xx(1)/(2)x^(2)=1 or x^(2)=4`
`x=+-2m`