The potential energy function for a particle executing linear SHM is given by $V(x)=\frac{1}{2}k{x}^2$ where k is the force constant of the oscillator (Fig. 6.2). For k = 0.5 N/m, the graph of V(x) versus x is shown in the figure. A particle of total energy E turns back when it reaches x = ± x_m . If V and K indicate the P.E. and K.E., respectively of the particle at x = +x_m, then which of the following is correct?

EXPLANATION:

The total energy of the particle of mass m at any point, E = K.E + P.E 

→As there is no fictional force or air drag force associated with the motion of the particle, hence it is moving by conservative force,

(F = - kx) and hence its total energy remains constant. 

→When P.E increases K.E decreases and when K.E increases then P.E decreases, such that the total energy remains the same.

The particle's maximum displacement is ± x_m.

The potential at any point is $V(x)=\frac{1}{2}k{x}^2$

→At the extreme left and right ± xm 

potential energy is $V(x_m)=\frac{1}{2}k{x}_m^2$ .

→At this position the K.E = 0. 

∴ total energy E = K + V = 0 + V = $\frac{1}{2}k{x}_m^2$

We know that kinetic energy at any point,

= Total energy - potential energy 

= $\frac{1}{2}k{x}_m^2-\frac{1}{2}k{x}^2$

→At the mean position x = 0

potential energy = 0 and kinetic energy = $\frac{1}{2}k{x}_m^2$

∴ the potential energy and kinetic energy decrease and increase at the same rate to keep the total energy constant. 

∴ at ± xm V = E and K = 0 

⇒ total energy = E + 0 = E

So, the correct answer is option (2).