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 mass $m$ is placed over it and both of them move together with amplitude $A_2$. The ratio of $\left(\frac{A_1}{A_2}\right)$

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

Velocity of particle at mean position initially,

$V^{{}^{\prime }}=\omega A_1{V}^{{}^{\prime }}=\sqrt{\frac{K}{M}}{A}_1\to \left(1\right)$

After placing another mass,

$(m+M)V^{{}^{\prime \prime }}=M{V}^{{}^{\prime }}$

$V^{{}^{\prime \prime }}=\frac{M{V}^{{}^{\prime }}}{(m+M)}$

$V^{{}^{\prime \prime }}={\omega }^{{}^{\prime }}{A}_2$

$V^{{}^{\prime \prime }}=\sqrt{\frac{K}{m+M}}{A}_2\frac{M{V}^{{}^{\prime }}}{M+m}=\sqrt{\frac{K}{m+M}}{A}_2\to \left(2\right)ondividingboth,\frac{M+m}{M}=\sqrt{\frac{M+m}{M}}\frac{A_1}{A_2}\frac{A_1}{A_2}={\left(\frac{M+m}{M}\right)}^{\frac{1}{2}}$