A mass m1= 1 kg connected to a horizontal spring performs S.H.M. with amplitude A. While mass m1 is passing through mean position another mass m2 = 3 kg is placed on it so that both the masses move together with amplitude A1. The ratio of A1A is (pq)12, where p and q are the smallest integers. Then what is the value of p + q?

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A mass $m_{1}$ = 1 kg connected to a horizontal spring performs S.H.M. with amplitude A. While mass m₁ is passing through mean position another mass m₂ = 3 kg is placed on it so that both the masses move together with amplitude A₁. The ratio of $\frac{A_{1}}{A}$ is ${(\frac{p}{q})}^{\frac{1}{2}},$ where p and q are the smallest integers. Then what is the value of p + q?

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answer is 5.

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Detailed Solution

$\begin{array}{l} P . E of the oscillating mass is given by \\ E = \frac{1}{2} {mωx}^{2} = \frac{1}{2} {kx}^{2} where k = m ω^{2} \\ Let k = m ω^{2} be the velocity at mean position. when \\ {mass M}_{2} is attached \\ v_{2} = \frac{m_{1} v_{1}}{m_{1} + m_{2}} \\ {If A}_{1} be the new amplitude \\ \frac{1}{2} {kA}_{1}^{2} = \frac{1}{2} (m_{1} + m_{2}) v_{2}^{2} \\ = \frac{1}{2} {kA}^{2} \times \frac{m_{1}}{m_{1} + m_{2}} \\ \Rightarrow \frac{A_{1}}{A} = \sqrt{\frac{m_{1}}{m_{1} + m_{2}}} = \sqrt{\frac{1}{4}} \end{array}$