Question

A mass $M$ attached to a horizontal spring, executes SHM with amplitude $A_1$ on a smooth horizontal surface. When the mass $M$ passes through its mean position, then a small mass $m$ is placed over it and both of them move together with amplitude $A_2$. The ratio of $\left(\frac{A_1}{A_2}\right)$ will be

• A. $\frac{M+m}{M}$
• B. ${\left(\frac{M}{M+m}\right)}^{1/2}$
• C. ${\left(\frac{M+m}{M}\right)}^{1/2}$
• D. $\frac{M}{M+m}$

Answer 1

The correct option is C ${\left(\frac{M+m}{M}\right)}^{1/2}$


When block $M$ crosses its mean position, its velocity along direction shown will be,
$v=v_{max}=A_1{\omega }_1$ ..........$\left(1\right)$
After the mass $m$ is placed on it, let the system moves with velocity $v^{{}^{\prime }}$.
Applying momentum conservation along horizontal direction (positive $x-$axis)
$(M+m)v^{{}^{\prime }}=Mv$
$\Rightarrow (M+m)A_2{\omega }_2=M{A}_1{\omega }_1$ ..........$\left(2\right)$
$[$from $\left(1\right)]$
The angular frequency of the system before and after placing mass $m$ will be,
${\omega }_1=\sqrt{\frac{k}{M}}$
${\omega }_2=\sqrt{\frac{k}{M+m}}$

On substituting in eq. $\left(2\right)$ we get,
$(M+m)\times A_2\times \sqrt{\frac{k}{M+m}}=M\times A_1\times \sqrt{\frac{k}{M}}$
$\Rightarrow \left(\sqrt{M+m}\right)\times A_2=\sqrt{M}\times A_1$
$\Rightarrow \frac{A_1}{A_2}=\sqrt{\frac{M+m}{M}}={\left(\frac{M+m}{M}\right)}^{1/2}$

Why this question? Tips: When a small block is placed on $M$, the interaction between them give rise to internal forces of the system $(M+m)$ whose sum is zero. Net force on the system $(M+m)$ at mean position is zero as friction is absent at the horizontal surface and $F_{\text{spring}}=0$. Bottom line: For the system $(M+m)$ linear momentum conservation is applicable at the mean position.