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

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

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(D) The maximum velocity of a body performing simple harmonic motion is given by $v_{max} = A\omega$,where $A$ is the amplitude and $\omega$ is the angular frequency.For a spring-mass system,the angular frequency is $\omega = \sqrt{\frac{k}{m}}$.Thus,$v_{max} = A\sqrt{\frac{k}{m}}$.Given that the masses are equal $(m_M = m_N = m)$ and their maximum velocities are equal $(v_M = v_N)$:$A_M \sqrt{\frac{k_1}{m}} = A_N \sqrt{\frac{k_2}{m}}$.Simplifying this,we get $A_M \sqrt{k_1} = A_N \sqrt{k_2}$.Therefore,the ratio of the amplitude of $M$ to that of $N$ is $\frac{A_M}{A_N} = \sqrt{\frac{k_2}{k_1}}$.

Two bodies $M$ and $N$ of equal masses are suspended from two separate massless springs of force constants $k_1$ and $k_2$ respectively. If the two bodies oscillate vertically such that their maximum velocities are equal,the ratio of the amplitude of $M$ to that of $N$ is

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