Two bodies M and W 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 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 $\sqrt{\frac{k_{2}}{k_{1}}}$
$\begin{matrix} max i m a v e l o c i t y i n t h e o s c i l l a t i o n i s g i v e n a s V_{max} = A w \Rightarrow V_{{max}_{1}} = V_{{max}_{2}} \Rightarrow A_{1} w_{1} = A_{2} w_{2} \Rightarrow \frac{A_{1}}{A_{2}} = \frac{w_{2}}{w_{1}} k_{1} = m w_{1}^{2}, k_{2} = m w_{1}^{2} \Rightarrow \frac{A_{1}}{A_{2}} = \sqrt{\frac{\frac{k_{2}}{m}}{\frac{k_{1}}{m}}} = \sqrt{\frac{k_{2}}{k_{1}}} H e n c e, t h e o p t i o n D i s t h e c o r r e c t a n s w e r . \end{matrix}$

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Two bodies M and W 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 M to that of N is