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Two particles A and B of equal masses are suspended from two massless springs of spring constants $k_{1}$ and $k_{2}$ , respectively. If the maximum velocities, during oscillation, are equal, the ratio of amplitudes of A and B is

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

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

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

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

answer is B.

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Detailed Solution

$v_{\max 1} = v_{\max 2} \Rightarrow ω_{1} a_{1} = ω_{2} a_{2}$

$\Rightarrow \frac{2 π}{T_{1}} a_{1} = \frac{2 π}{T_{2}} a_{2}$

$\Rightarrow \frac{a_{1}}{a_{2}} = \frac{T_{1}}{T_{2}} = \sqrt{\frac{k_{2}}{k_{1}}} (T = 2 π \sqrt{\frac{m}{k}})$

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