A particle is executing linear S.H.M. starting from mean position. The ratio of the kinetic energy to the potential energy of the particle at a point of half the amplitude is

For a particle performing simple harmonic motion with amplitude $A$ and angular frequency $\omega$, whose displacement from mean position is $x$, the force constant is given by $k = m\omega^2$.

The potential energy at displacement $x$ is $PE = \frac{1}{2} k x^2$, while the kinetic energy is $KE = \frac{1}{2} m\omega^2 (A^2 - x^2) = \frac{1}{2} k (A^2 - x^2)$.

At the point where the displacement is half the amplitude, $x = A/2$, the potential energy becomes $PE = \frac{1}{2} k (A/2)^2 = \frac{1}{8} k A^2$
and the kinetic energy becomes $KE = \frac{1}{2} k (A^2 - (A/2)^2) = \frac{1}{2} k (3A^2/4) = \frac{3}{8} k A^2$.

The ratio of kinetic to potential energy is therefore $\frac{KE}{PE} = \frac{3/8}{1/8} = 3$.

Ratio: 3:1
The correct option is B.