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

k_{2} / k_{1}

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

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k_{1} / k_{2}

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

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Solution

The correct option is C $\sqrt{k_{2} / k_{1}}$

Since there maximum velocities are equal i.e.
$A_{1} ω_{1} = A_{2} ω_{2} \Rightarrow \frac{A_{1}}{A_{2}} = \frac{ω_{2}}{ω_{1}} = \frac{\sqrt{\frac{k_{2}}{m}}}{\sqrt{\frac{k_{1}}{m}}} = \sqrt{\frac{k_{2}}{k_{1}}}$

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Two bodies M and N of equal masses are suspended from two separate massless spring of spring constants k1 and k2 respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, then the ratio of the amplitude of M to that of N is

The correct option is C √k2/k1 Since there maximum velocities are equal i.e.A1ω1=A2ω2⇒A1A2=ω2ω1=√k2m√k1m=√k2k1

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

The correct option is C $\sqrt{k_{2} / k_{1}}$

Since there maximum velocities are equal i.e.
$A_{1} ω_{1} = A_{2} ω_{2} \Rightarrow \frac{A_{1}}{A_{2}} = \frac{ω_{2}}{ω_{1}} = \frac{\sqrt{\frac{k_{2}}{m}}}{\sqrt{\frac{k_{1}}{m}}} = \sqrt{\frac{k_{2}}{k_{1}}}$