The potential energy function for a particle executing linear SHM is given by `V(x)= 1/2kx^2` where k is the force constant of the oscillator. For `k = 0.5 Nm^(-1)`, the graph of V(x) versus x is shown in the figure A particle of total energy E turns back when it reaches `x=pmx_m`.if V and K indicate the potential energy and kinetic energy 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 lt E`

Answer

B

Explanation

Total energy is `E = PE + KE " "…(i)`
When particle is at `x = x_m` i.e., at extreme position,
returns back. Hence, at `x = x_m, x = 0, KE = 0`
From Eq. (i) `E=PE +0 = PE = V(x_m) =1/2 kx_(m)^2`