Question

Two masses $m$ and $2m$ are suspended together by a massless spring of force constant $K$. When the masses are in equilibrium, mass $m$ is removed without disturbing the system. Then, the amplitude of vibration of the system is:

• A. $\frac{2mg}{K}$
• B. $\frac{mg}{K}$
• C. $\frac{3mg}{K}$
• D. $\frac{mg}{2K}$

Answer 1

The correct option is B $\frac{mg}{K}$


Let $l$ be the extension of the spring at equilibrium, when both the masses are hanging.
From Hooke's law,
$3mg=Kl\dots \dots \left(2\right)$
Let $l^{\prime }$ be the extension of the spring at equilibrium, when mass $2m$ alone is hanging.
From Hooke's law, we can write that
$2mg=K{l}^{\prime }\dots \dots \left(1\right)$
This will be the new mean position for oscillation when mass $m$ is removed.
Subtracting $\left(1\right)$ from $\left(2\right)$, we get
$l-l^{\prime }=\frac{mg}{K}$
i.e Amplitude of oscillation $A=(l^{\prime }-l)=\frac{mg}{K}$
Thus, option (b) is the correct answer.