Two bodies M and N of equal masses are suspended from two separate springs of the spring constant $k_{1}$ and $k_{2}$ respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of vibration of M to that on N is

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Solution

The correct option is B

$V_{m a x 1} = V_{m a x 2}$

$\Rightarrow ω_{1} A_{1} = ω_{2} A_{2}$

$\Rightarrow \sqrt{\frac{K_{1}}{m}} A_{1} = \sqrt{\frac{K_{2}}{m}} A_{2}$

$\Rightarrow \frac{A_{1}}{A_{2}} = \sqrt{\frac{K_{2}}{K_{1}}}$

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