Two particles $$A$$ and $$B$$ of equal masses are suspended from two massless springs of spring constants $$K_1$$ and $$K_2$$ respectively. If the maximum velocities during oscillations are equal, the ratio of the amplitude of $$A$$ and $$B$$ is:

For a mass-spring system undergoing simple harmonic motion, the angular frequency is $$\omega = \sqrt{\frac{K}{m}}$$ and the maximum velocity is $$v_{\max} = A\omega = A\sqrt{\frac{K}{m}}$$.

For particle $$A$$: $$v_{\max,A} = A_1 \sqrt{\frac{K_1}{m}}$$. For particle $$B$$: $$v_{\max,B} = A_2 \sqrt{\frac{K_2}{m}}$$. Since the masses are equal and the maximum velocities are equal, we set $$A_1 \sqrt{\frac{K_1}{m}} = A_2 \sqrt{\frac{K_2}{m}}$$.

This gives $$A_1 \sqrt{K_1} = A_2 \sqrt{K_2}$$, so $$\frac{A_1}{A_2} = \frac{\sqrt{K_2}}{\sqrt{K_1}} = \sqrt{\frac{K_2}{K_1}}$$.