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(D) The maximum velocity of an object in simple harmonic motion is given by $v_{max} = A \omega$,where $A$ is the amplitude and $\omega$ is the angula

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$\sqrt{\frac{k_2}{k_1}}$

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(D) The maximum velocity of an object in simple harmonic motion is given by $v_{max} = A \omega$,where $A$ is the amplitude and $\omega$ is the angular frequency.Given that the masses are equal $(m_1 = m_2 = m)$,the angular frequencies are $\omega_1 = \sqrt{\frac{k_1}{m}}$ and $\omega_2 = \sqrt{\frac{k_2}{m}}$.Since the maximum velocities are equal,we have $A_1 \omega_1 = A_2 \omega_2$.Rearranging for the ratio of amplitudes,we get $\frac{A_1}{A_2} = \frac{\omega_2}{\omega_1}$.Substituting the expressions for $\omega_1$ and $\omega_2$,we get $\frac{A_1}{A_2} = \frac{\sqrt{k_2/m}}{\sqrt{k_1/m}} = \sqrt{\frac{k_2}{k_1}}$.Thus,the ratio of the amplitudes is $\sqrt{\frac{k_2}{k_1}}$.

Two bodies $A$ and $B$ of equal masses are suspended from two separate springs of force constants $k_1$ and $k_2$ respectively. If the two bodies oscillate such that their maximum velocities are equal,the ratio of the amplitudes of oscillation of $A$ and $B$ will be

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