Two bodies M and N of equal masses are suspended from two separate massless springs of spring 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-

\frac{k_{1}}{k_{2}}

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

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\frac{k_{2}}{k_{1}}

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

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Solution

The correct option is D $\sqrt{\frac{k_{2}}{k_{1}}}$
Given, $v_{1} (max) = v_{2} (max)$
$ω_{1} A_{1} = ω_{2} A_{2}$
$ω = \sqrt{\frac{k}{m}}$ spring constant are $k_{1}$ and $k_{2} .$
$\frac{A_{1}}{A_{2}} = \frac{ω_{2}}{ω_{1}} = \sqrt{\frac{k_{2}}{k_{1}}}$
$\frac{A_{1}}{A_{2}} = \sqrt{\frac{k_{2}}{k_{1}}}$
Here, m is same for both.

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