Question

A mass M, attached to a horizontal spring, executes S.H.M. with amplitude $A_1$. When the mass M passes through its mean position then a smaller m is placed it and both of them move together with amplitude $A_2$. The ratio of $\left(\frac{A_1}{A_2}\right)$ is :-

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

Answer 1

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

At mean position $F_{net}=0$

Therefore, By conservation of linear momentum,

$M{v}_1=(M+m)v_2$

$M{\omega }_1{A}_1=(M+m){\omega }_2{A}_2$

$\Rightarrow \left(\frac{M}{M+m}\right)=\frac{{\omega }_2{A}_2}{{\omega }^1{A}_1}$

$\Rightarrow$ But ${\omega }_1=\sqrt{\frac{k}{M}}$ and ${\omega }_2=\sqrt{\frac{k}{M+m}}$

on solving, we get $\frac{A_1}{A_2}=\sqrt{\frac{m+M}{M}}$

At mean position $F_{net}=0$

Therefore, By conservation of linear momentum,

$M{v}_1=(M+m)v_2$

$M{\omega }_1{A}_1=(M+m){\omega }_2{A}_2$

$\Rightarrow \left(\frac{M}{M+m}\right)=\frac{{\omega }_2{A}_2}{{\omega }_1{A}_1}$

$\Rightarrow$ But ${\omega }_1=\sqrt{\frac{k}{M}}$ and ${\omega }_2=\sqrt{\frac{k}{M+m}}$

on solving, we get $\frac{A_1}{A_2}=\sqrt{\frac{m+M}{M}}$