Question 32 of 194

Two massless springs with spring constants $2k$ and $9k$, carry $50g$ and $100g$ masses at their free ends. These two masses oscillate vertically such that their maximum velocities are equal. Then, the ratio of their respective amplitudes will be :

A$3:2$B$1:2$C$3:1$D$2:3$

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Answer: (A) $3:2$

Maximum velocity is given by ${v}_{\max }=\omega A$ $=\sqrt{\dfrac{k}{m}}A$

Here, $\sqrt{\dfrac{{k}_{1}}{{m}_{1}}}{A}_{1}=\sqrt{\dfrac{{k}_{2}}{{m}_{2}}}{A}_{2}$
$\Rightarrow \sqrt{\dfrac{2k}{50\times {10}^{-3}}}{A}_{1}=\sqrt{\dfrac{9k}{100\times {10}^{-3}}}{A}_{2}$
$\Rightarrow 2{A}_{1}=3{A}_{2}$
Thus, ratio of amplitude is $\dfrac{{A}_{1}}{{A}_{2}}=\dfrac{3}{2}=3:2$.