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 $\frac{A_1}{A}$ is $\left(m_2<m_1\right)$

• A. ${\left[\frac{m_1}{m_1+m_2}\right]}^{\frac{1}{2}}$
• B. ${\left[\frac{m_1+m_2}{m_1}\right]}^{\frac{1}{2}}$
• C. ${\left[\frac{m_2}{m_1+m_2}\right]}^{\frac{1}{2}}$
• D. ${\left[\frac{m_1+m_2}{m_2}\right]}^{\frac{1}{2}}$

Answer 1

Answer: (A)

Since amplitude of S.H.M. $=A$
Applying energy conservation
$\frac{1}{2}\times k\times A^2=\frac{1}{2}{m}_1{v}^2.........(1)$
where $v=$ velocity of block at mean position.
When $m_2$ will be placed over $m_1$,
then their common velocity can be calculated
using conservation of linear momentum as
$m_1\times v=\left(m_1+m_2\right)\times V$
$V_{\text{common }}=\frac{m_{1}v}{m_1+m_2}$
Now when both masses will execute S.H.M. Combinedly,
we can find their amplitude using energy conservation
$\frac{1}{2}\times k\times A^{\prime 2}=\frac{1}{2}\times \left(m_1+m_2\right){\left(V_{\text{common }}\right)}^2$
On solving we will get $\frac{A^{\prime }}{A}={\left[\frac{m_1}{m_1+m_2}\right]}^{\frac{1}{2}}$