A mass m_1 connected to a horizontal spring performs S.H.M. with … - Vidyadip

MCQ

A mass $m_1$ connected to a horizontal spring performs $\text{S.H.M.}$ 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 $m_{2 <}m_1$

✓
$\left[\frac{m_1}{m_1+m_2}\right]^{\frac{1}{2}}$
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

Correct option: A.

$\left[\frac{m_1}{m_1+m_2}\right]^{\frac{1}{2}}$

Applying principle of momentum conservation at the mean position,
$p_i=p_f$
$ \Rightarrow m_1 v=\left(m_1+m_2\right) v_1$
$ \Rightarrow \frac{v}{v_1}=\frac{m_1+m_2}{m_1}...(i)$
Also$, \frac{v}{v_1}=\frac{\omega A}{\omega_1 A_1}$
$\therefore \frac{A_1}{A}=\frac{\omega}{\omega_1} \times \frac{v_1}{v}....(ii)$
And $\frac{\omega}{\omega_1}=\sqrt{\frac{k}{m_1}} / \sqrt{\frac{k}{m_1+m_2}}=\sqrt{\frac{m_1+m_2}{m_1}}...(iii)$
From eqns. $(i), (ii)$ and $(iii),$
$\frac{A_1}{A}=\sqrt{\frac{m_1+m_2}{m_1}} \times \frac{m_1}{m_1+m_2}=\sqrt{\frac{m_1}{m_1+m_2}}$

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