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 $\frac{ A'}{ A }=\left[\frac{ m _{1}}{ m _{1}+ m _{2}}\right]^{\frac{1}{2}}$