Two massless springs with spring constants $$2k$$ and $$9k$$, carry $$50$$ g and $$100$$ 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 :

Spring constants $$k_1 = 2k$$ and $$k_2 = 9k$$, masses $$m_1 = 50$$ g $$= 0.05$$ kg and $$m_2 = 100$$ g $$= 0.1$$ kg. The maximum velocities are equal.

For a mass-spring system, the maximum velocity is given by: $$v_{\max} = A\omega = A\sqrt{\frac{k}{m}}$$.

For the first system, $$v_1 = A_1\sqrt{\frac{2k}{0.05}} = A_1\sqrt{40k}$$, and for the second system, $$v_2 = A_2\sqrt{\frac{9k}{0.1}} = A_2\sqrt{90k}$$.

Setting $$v_1 = v_2$$ gives $$A_1\sqrt{40k} = A_2\sqrt{90k}$$.

Cancelling $$\sqrt{k}$$ from both sides yields $$A_1\sqrt{40} = A_2\sqrt{90}$$.

Hence, $$\frac{A_1}{A_2} = \frac{\sqrt{90}}{\sqrt{40}} = \sqrt{\frac{90}{40}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$$.

Therefore, the ratio of amplitudes is $$A_1 : A_2 = 3 : 2$$. The correct answer is Option A.