Question

The potential energy function for a particle executing linear SHM is given by $\frac{1}{2}k{x}^2$ where k is the force constant of the oscillator (Fig.). 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?

• A. V = O, K = E
• B. V = E, K = O
• C. V < E, K = O
• D. V = O, K < E.

Answer 1

The correct option is B V = E, K = O

If the particle executing S.H.M and $x_m$ is the maximum displacement i.e., amplitude.

As we know total energy from S.H.M is constant through out motion.

$E=\frac{1}{2}{kx}_m^2=\frac{1}{2}m{\omega }^2{x}_m^2$ $({\omega }^2=\frac{k}{m}=$ angular frequency at S.H.M)

We, have $k+v=E$

$k=$ kinetic energy

$v=$ potential energy

So, when $k=0,v=E$