A particle is making simple harmonic motion along the X-axis. If at a distances x1 and x2 from the mean position the velocities of the particle are v1 and v2 respectively. The time period of its oscillation is given as :

T = 2\pi \sqrt x_2^2 - x_1^2 \over v_1^2 - v_2^2

T = 2\pi \sqrt x_2^2 + x_1^2 \over v_1^2 - v_2^2

T = 2\pi \sqrt x_2^2 + x_1^2 \over v_1^2 + v_2^2

T = 2\pi \sqrt x_2^2 - x_1^2 \over v_1^2 + v_2^2

Simple Harmonic Motion PYQ — JEE Main 2021 · 20 Jul · Shift 2 · Q54 with Solution

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Correct answer: D

For a particle in SHM, $v^2 = \omega^2(A^2 - x^2)$ where:

So at position $x_1$ , $v_1^2 = \omega^2(A^2 - x_1^2) \qquad (1)$

At position $x_2$ , $v_2^2 = \omega^2(A^2 - x_2^2) \qquad (2)$

Subtract (2) from (1): $v_1^2 - v_2^2 = \omega^2[(A^2 - x_1^2) - (A^2 - x_2^2)]$

$v_1^2 - v_2^2 = \omega^2(x_2^2 - x_1^2)$

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

We know, $T = \frac{2\pi}{\omega}$

Therefore, $T = 2\pi \sqrt{\frac{1}{\omega^2}}$

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

This matches: $\boxed{T = 2\pi \sqrt{\frac{x_2^2 - x_1^2}{v_1^2 - v_2^2}}}$

So the correct option is D.