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(A) For a body of mass $m$ oscillating with amplitude $A$ and angular frequency $\omega$,the maximum velocity is given by $v_{max} = A\omega$.Given th

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$\sqrt{\frac{k_{2}}{k_{1}}}$

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(A) For a body of mass $m$ oscillating with amplitude $A$ and angular frequency $\omega$,the maximum velocity is given by $v_{max} = A\omega$.Given that the masses are equal $(m_A = m_B = m)$ and the force constants are $k_1$ and $k_2$,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 $A_1 \omega_1 = A_2 \omega_2$.Substituting the expressions for $\omega_1$ and $\omega_2$:$A_1 \sqrt{\frac{k_1}{m}} = A_2 \sqrt{\frac{k_2}{m}}$.Squaring both sides and simplifying:$A_1^2 \frac{k_1}{m} = A_2^2 \frac{k_2}{m} \implies \frac{A_1^2}{A_2^2} = \frac{k_2}{k_1}$.Taking the square root,the ratio of the amplitudes is $\frac{A_1}{A_2} = \sqrt{\frac{k_2}{k_1}}$.

Two bodies $A$ and $B$ of equal mass are suspended from two separate massless springs of force constant $k_{1}$ and $k_{2}$ respectively. The bodies oscillate vertically such that their maximum velocities are equal. The ratio of the amplitudes of body $A$ to that of body $B$ is

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