To potential energy function for a particle executing linear simple harmonic motion is given V(x) = `kx^(2) //2` , where k is the force constant of the oscillator . For k = `0.5 N m^(-1)` the graph of V(x) versus x is shown in Fig. Show that a particle of total energy 1 J moving under this potential must 'turn back' when it reaches `x = pm 2 m`

Explanation

At any instant, the total energy of an oscillator is the sum of K.E and P.E i.e.
`E = K.E + P.E, E = (1)/(2) m u^(2) + (1)/(2) kx^(2)`
The particles turns back at the instant, when its velocity becomes zero i.e. u = 0
`:. e = 0 + (1)/(2) kx6(2) ` as E =1 Joule and `K = (1)/(2) N//m`
`:. 1 = (1)/(2) xx (1)/(2) x^(2) (or) x^(2) = 4, x = pm 2m`