1. Total mechanical energy in SHM

For a particle executing simple harmonic motion with amplitude $A$ and spring constant $k$:

$E = \frac{1}{2}kA^2$

This total energy remains constant.

2. Position midway between mean and extreme position

The extreme position is at $x = A$ and the mean position is at $x=0$.

So the midway position is:

$x = \frac{A}{2}$

3. Potential energy at this position

Potential energy in SHM is:

$U = \frac{1}{2}kx^2$

Substituting $x=\frac{A}{2}$:

$U = \frac{1}{2}k\left(\frac{A}{2}\right)^2 = \frac{1}{2}k\cdot \frac{A^2}{4} = \frac{1}{8}kA^2$

4. Kinetic energy at this position

Since

$K = E-U$

we get

$K = \frac{1}{2}kA^2 - \frac{1}{8}kA^2 = \frac{4}{8}kA^2 - \frac{1}{8}kA^2 = \frac{3}{8}kA^2$

5. Fraction of total energy that is kinetic

$\frac{K}{E} = \frac{\frac{3}{8}kA^2}{\frac{1}{2}kA^2} = \frac{3}{8}\cdot \frac{2}{1} = \frac{3}{4}$

6. Option check

• A: $\frac{1}{2}$ — incorrect
• B: $\frac{3}{4}$ — correct
• C: $\frac{1}{3}$ — incorrect
• D: $\frac{1}{4}$ — incorrect

Therefore, the correct answer is B.