Q 11-13-111JEE MainJEE Main 2021 (20 Jul, Shift 2)Medium
A particle is making simple harmonic motion along the $X$-axis. If at a 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:
Answer: (D) $T = 2\pi\sqrt{\frac{x_2^2 - x_1^2}{v_1^2 - v_2^2}}$
$v_1^2 = \omega^2(A^2 - x_1^2)$ and $v_2^2 = \omega^2(A^2 - x_2^2)$. Subtracting: $v_1^2 - v_2^2 = \omega^2(x_2^2 - x_1^2)$.
$$T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{x_2^2 - x_1^2}{v_1^2 - v_2^2}}$$
Solution by Sreeraj P, M.Sc Physics