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(B) The maximum velocity of a body in simple harmonic motion is given by $v_{max} = \omega A$,where $\omega = \sqrt{\frac{k}{m}}$ is the angular frequ

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

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(B) The maximum velocity of a body in simple harmonic motion is given by $v_{max} = \omega A$,where $\omega = \sqrt{\frac{k}{m}}$ is the angular frequency and $A$ is the amplitude.Given that the masses are equal $(m_P = m_Q = m)$ and their maximum velocities are equal $(v_P = v_Q)$,we have $\omega_1 A_1 = \omega_2 A_2$,where $A_1$ and $A_2$ are the amplitudes of $P$ and $Q$ respectively.Substituting the expression for angular frequency: $\sqrt{\frac{k_1}{m}} A_1 = \sqrt{\frac{k_2}{m}} A_2$.Rearranging to find the ratio of amplitudes $\frac{A_1}{A_2}$:$\frac{A_1}{A_2} = \frac{\sqrt{\frac{k_2}{m}}}{\sqrt{\frac{k_1}{m}}} = \sqrt{\frac{k_2}{k_1}}$.

Two bodies $P$ and $Q$ of equal mass are suspended from two separate massless springs of force constants $k_1$ and $k_2$ respectively. If the maximum velocities of them are equal during their motion,the ratio of the amplitude of $P$ to $Q$ is:

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