Two bodies M and N of equal masses are suspended from two separate massless springs of spring constants $k_{1} a n d k_{2}$ 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 of N is

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$k_{1} / k_{2}$

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

$k_{2} / k_{1}$

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

answer is B.

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Detailed Solution

The two blocks undergo SHM in the vertical di-reaction about their mean position.

Let the amplitudes of oscillations be $A_{1}$ and $A_{2}$ and the angular frequencies be $ω_{1}$ and $ω_{2}$ The angular frequencies of oscillations are given by
$ω_{1} = \sqrt{k_{1} / m,} ω_{2} = \sqrt{k_{2} / m .}$
The velocities are maximum at the mean positions and are given by
$υ_{1, \max} = A_{1} ω_{1}, υ_{2, \max} = A_{2} ω_{2} .$
The condition $υ_{1, \max} = υ_{2, \max}$ gives $A_{1} / A_{2} = \sqrt{k_{2.} / k_{1}} .$