Q 11-13-091JEE MainJEE Main 2022 (24 Jun, Shift 2)Medium
Two massless springs with spring constants $2k$ and $9k$ carry $50\ \text{g}$ and $100\ \text{g}$ masses at their free ends. These two masses oscillate vertically such that their maximum velocities are equal. Then, the ratio of their respective amplitudes will be:
Answer: (A) $3 : 2$
$v_{\max} = A\omega$, so $A_1\omega_1 = A_2\omega_2$ and $\dfrac{A_1}{A_2} = \dfrac{\omega_2}{\omega_1}$.
$$\omega_1 = \sqrt{\frac{2k}{0.05}} = \sqrt{40k},\qquad \omega_2 = \sqrt{\frac{9k}{0.1}} = \sqrt{90k}$$
$$\frac{A_1}{A_2} = \sqrt{\frac{90}{40}} = \frac32$$
Solution by Sreeraj P, M.Sc Physics