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(C) 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 t

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$\sqrt{\frac{K_1}{K_2}}$

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(C) 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 maximum velocities are equal $(v_{max,A} = v_{max,B})$,we have $A_1 \omega_1 = A_2 \omega_2$. Since $\omega = \sqrt{\frac{K}{m}}$,we have $A_1 \sqrt{\frac{K_1}{m}} = A_2 \sqrt{\frac{K_2}{m}}$. Squaring both sides,we get $A_1^2 \frac{K_1}{m} = A_2^2 \frac{K_2}{m}$. Simplifying,$A_1^2 K_1 = A_2^2 K_2$. Therefore,the ratio of the amplitude of $B$ $(A_2)$ to that of $A$ $(A_1)$ is $\frac{A_2}{A_1} = \sqrt{\frac{K_1}{K_2}}$.

Two bodies $A$ and $B$ of equal mass are suspended from two separate massless springs of spring constants $K_1$ and $K_2$ respectively. The two bodies oscillate vertically such that their maximum velocities are equal. The ratio of the amplitude of $B$ to that of $A$ is

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