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(C) The maximum velocity of a particle in simple harmonic motion is given by $V_{\max} = \omega A$,where $\omega$ is the angular frequency and $A$ is

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$\sqrt{k_2/k_1}$

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(C) The maximum velocity of a particle in simple harmonic motion is given by $V_{\max} = \omega A$,where $\omega$ is the angular frequency and $A$ is the amplitude.Given that the masses $M$ are equal,the angular frequencies are $\omega_1 = \sqrt{k_1/M}$ and $\omega_2 = \sqrt{k_2/M}$.Since the maximum velocities are equal,we have $V_{\max, 1} = V_{\max, 2}$,which implies $\omega_1 A_1 = \omega_2 A_2$.Therefore,the ratio of the amplitudes is $\frac{A_1}{A_2} = \frac{\omega_2}{\omega_1}$.Substituting the expressions for $\omega_1$ and $\omega_2$,we get $\frac{A_1}{A_2} = \frac{\sqrt{k_2/M}}{\sqrt{k_1/M}} = \sqrt{\frac{k_2}{k_1}}$.

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 amplitudes of $A$ and $B$ is

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