A particle is making simple harmonic motion along the $x-\text{axis}$. If at a distance $x_1$ and $x_2$ from the mean position the velocities of the particle are $v_1$ and $v_2$ respectively. The time period of its oscillation is given as :

The correct option is B $T=2\pi \sqrt{\frac{x_2^2-x_1^2}{v_1^2-v_2^2}}$
We know that,
$v^2={\omega }^2(A^2-x^2)$

$\Rightarrow A^2=x^2+\frac{v^2}{{\omega }^2}$

Given:
At $x_1$, velocity $=v_1$ , and
at $x_2$, velocity $=v_2$

$A^2=x_1^2+\frac{v_1^2}{{\omega }^2}=x_2^2+\frac{v_2^2}{{\omega }^2}$

$\Rightarrow \frac{v_1^2-v_2^2}{{\omega }^2}=x_2^2-x_1^2$

$\Rightarrow {\omega }^2=\frac{v_1^2-v_2^2}{x_2^2-x_1^2}$

$\therefore T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{x_2^2-x_1^2}{v_1^2-v_2^2}}$

Hence, option $\left(\text{D}\right)$ is correct.