Question

Two masses $M$ and $m$ are suspended together by a massless spring of force constant $k$. When the masses are in equilibrium, $M$ is removed without disturbing the system. The amplitude of oscillation is :

• A. $\frac{Mg}{k}$
• B. $\frac{mg}{k}$
• C. $\frac{(M+m)g}{k}$
• D. $\frac{(M-m)g}{k}$

Answer 1

The correct option is A $\frac{Mg}{k}$

With mass $m$ alone the extension of the spring $l$ is given as

$mg=kl$ -------------------$\left(1\right)$

With mass $(M+m)$ the extension $l^{\prime }$ is given by

$\left(M+m\right)g=k\left(l+\Delta l\right)$-----------------$\left(2\right)$

The increase in extension is $\Delta l$ which is amplitude of vibration$.$
Substituting $\left(1\right)$ from $\left(2\right),$ we get

$Mg=k\Delta l$

$or,\Delta l=\frac{Mg}{k}$

Hence,

option $\left(A\right)$ is correct answer.