A mass m 1 = 1 kg connected to a horizontal spring performs S.H.M. with amplitude A . While mass m 1 is passing through mean position another mass m 2 = 3 kg is placed on it so that both the masses mo

Question

A mass m 1 = 1 kg connected to a horizontal spring performs S.H.M. with amplitude A . While mass m 1 is passing through mean position another mass m 2 = 3 kg is placed on it so that both the masses move together with amplitude A 1 . The ratio of A 1 A is p q 1 2 , where p and q are the smallest integers. Then what is the value of p + q ?

TestHub · Accepted Answer

P.E of the oscillating mass is given by

$E = \frac{1}{2} m ω x^{2} = \frac{1}{2} k x^{2}$ 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 $\frac{1}{2} k A_{1}^{2} = \frac{1}{2} (m_{1} + m_{2}) v_{2}^{2}$

$= \frac{1}{2} k A^{2} \times \frac{m_{1}}{m_{1} + m_{2}}$

$\Rightarrow \frac{A_{1}}{A} = \sqrt{\frac{m_{1}}{m_{1} + m_{2}}}$ $= \sqrt{\frac{1}{4}}$

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