The potential energy function for a particle executing linear SHM is given by `V(x) =1/2kx^(2)` wher e k is the force constant of the oscillator (figure). For k = 0.5 N/m ,the graph ofV(x) versus x is shown in the figure . A particle of total energy E turns back when it reaches `x = pm x_(m)` . If V and K indicate the PE and KE , respectively of the particle at `x = +x_(m)` , then which of the following is correct ?

Options

A. V =0 ,K=E
B. V = E,K=0
C. `V lt E , K = 0 `
D. `V = 0 ,K=0`

Answer

B

Explanation

Total energy is E = PE +KE …(i)
At `x = x_(m), v = 0 ,KE = 0 `
From Eqn.(i)
`E = pE +0 = PE =V(x_(m)) =1/2 kx_(m)^(2)`