A mass m 1 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 is placed on it so that both the masses move together wi

Question

A mass m 1 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 is placed on it so that both the masses move together with amplitude A 1 . The ratio of A 1 A is ( m 2 < m 1 )

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 $\frac{1}{2} k A_{1}^{2} = \frac{1}{2} (m_{1} + m_{2}) v_{2}^{2}$
$\frac{1}{2} (m_{1} + m_{2}) \frac{m_{1}^{2} v_{1}^{2}}{{(m_{1} + m_{2})}^{2}}$
$= \frac{1}{2} m_{1} v_{1}^{2} \times \frac{m_{1}}{m_{1} + m_{2}}$
$= \frac{1}{2} k A^{2} \times \frac{m_{1}}{m_{1} + m_{2}}$
$\Rightarrow \frac{A_{1}}{A} = \sqrt{\frac{m_{1}}{m_{1} + m_{2}}}$

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