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=O,K=E
B. V=E,K=O
C. V lt E, K=O
D. V=O, K lt E

Answer

B

Explanation

(b) 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
`E=PE+0=PE=V(x_(m))=(1)/(2)kx_(m)^(2)`