A particle performing simple harmonic motion according to y = A sinωt. Then its kinetic energy (K.E.), potential energy (P.E.) and speed (V) at position y = A/2 are

The correct option is:
$K.E=\frac{3k{A}^2}{8}$
$P.E=\frac{k{A}^2}{8}$
$V=\frac{A}{2}\sqrt{\frac{3k}{m}}$

Step-by-Step Derivation:

The attached image shows a mass-spring system consisting of a block of mass m attached to a spring of spring constant K on a horizontal surface, executing simple harmonic motion (SHM).

1. Potential Energy (P.E.):
The potential energy of a particle performing SHM at any displacement y from the mean position is given by:
$P.E=\frac{1}{2}k{y}^2$
Given that the particle is at position:
$y=\frac{A}{2}$
Substituting y into the potential energy formula:
$P.E=\frac{1}{2}k{\left(\frac{A}{2}\right)}^2=\frac{1}{2}k\left(\frac{A^2}{4}\right)=\frac{k{A}^2}{8}$

2. Kinetic Energy (K.E.):
The total mechanical energy (E) of the particle performing SHM with amplitude A is:
$E=\frac{1}{2}k{A}^2$
Since total energy is the sum of kinetic energy and potential energy:
$E=K.E+P.E$
Therefore, the kinetic energy is:
$K.E=E-P.E$
Substituting the values of E and P.E.:
$K.E=\frac{1}{2}k{A}^2-\frac{1}{8}k{A}^2=\frac{3k{A}^2}{8}$

3. Speed (V):
The kinetic energy is also given by the formula:
$K.E=\frac{1}{2}m{V}^2$
Equating the two expressions for kinetic energy:
$\frac{1}{2}m{V}^2=\frac{3k{A}^2}{8}$
Solving for V²:
$V^2=\frac{3k{A}^2}{4m}$
Taking the square root on both sides:
$V=\frac{A}{2}\sqrt{\frac{3k}{m}}$