1. Given

A particle performs SHM with:

• mass $m$
• amplitude $a$
• frequency $\nu$

We need the average kinetic energy while the particle moves from the mean position to the extreme position.

2. Write kinetic energy in SHM

For SHM, $\omega = 2\pi \nu$

If displacement is $x$, then speed is $v_x = \omega\sqrt{a^2-x^2}$

So kinetic energy at displacement $x$ is $K = \frac12 m v_x^2 = \frac12 m\omega^2(a^2-x^2)$

3. Choose a convenient time expression

From mean position to extreme position, the motion takes time $\frac{T}{4}$

Let us take $x = a\sin(\omega t), \qquad 0 \le t \le \frac{T}{4}$

Then $v_x = a\omega \cos(\omega t)$

Hence kinetic energy is $K(t)=\frac12 m a^2 \omega^2 \cos^2(\omega t)$

4. Average kinetic energy over this interval

Average value over time interval $0$ to $T/4$ is $\overline{K} = \frac{1}{T/4}\int_0^{T/4} K(t)\,dt$

Substitute $K(t)$: $\overline{K} = \frac{4}{T}\int_0^{T/4} \frac12 m a^2 \omega^2 \cos^2(\omega t)\,dt$

$\overline{K} = \frac{2ma^2\omega^2}{T}\int_0^{T/4}\cos^2(\omega t)\,dt$

Now use $\int_0^{T/4}\cos^2(\omega t)\,dt = \frac{T}{8}$

because the average of $\cos^2$ over a quarter cycle is $1/2$.

Thus, $\overline{K} = \frac{2ma^2\omega^2}{T}\cdot \frac{T}{8} = \frac14 ma^2\omega^2$

Now put $\omega = 2\pi \nu$: $\overline{K} = \frac14 ma^2(2\pi\nu)^2$

$\overline{K} = \frac14 ma^2 \cdot 4\pi^2\nu^2$

$\overline{K} = \pi^2 m a^2 \nu^2$

5. Match with options

$\boxed{\pi^2 m a^2 \nu^2}$

So the correct option is B.