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

The potential energy of the aprticle
`= V = 1/2 kx^(2)`
`k = 1/2 N//m^(-1)`
`V = 1/2 xx1/2 x^(2) = 1/4x^(2)`

When total energy ` = P.E +K.E = 1 J `
As the oscillating particle turns back at the point where its velocity = 0
Therefore , K.E = 0
Therefore , total energy = P.E
So, `1/4 x^(2) = 1 `
`x^(2) = 4`
` x = pm 2 `
Hence , it can be said that particle must turn back when it reaches ` x = pm 2 ` m .