A particle is making simple harmonic motion along the X-axis. If at distances $$x_1$$ 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:

For SHM, the velocity at displacement $$x$$ from the mean position is given by $$v^2 = \omega^2(A^2 - x^2)$$, where $$A$$ is the amplitude and $$\omega$$ is the angular frequency.

Applying this at the two positions: $$v_1^2 = \omega^2(A^2 - x_1^2)$$ and $$v_2^2 = \omega^2(A^2 - x_2^2)$$.

Subtracting the second from the first: $$v_1^2 - v_2^2 = \omega^2(x_2^2 - x_1^2)$$, so $$\omega^2 = \frac{v_1^2 - v_2^2}{x_2^2 - x_1^2}$$.

Since $$T = \frac{2\pi}{\omega}$$, we have $$T = 2\pi\sqrt{\frac{x_2^2 - x_1^2}{v_1^2 - v_2^2}}$$.