Question

In the figure shown, a block $\text{A}$ of mass $m$ is rigidly attached to a light spring of stiffness $k$ and suspended from a fixed support. Another block $\text{B}$ of same mass is just placed on it and the blocks are in equilibrium. Suddenly, the block $\text{B}$ is removed. Choose the correct option(s) afterward.

• A. Block $\text{A}$ will start SHM.
• B. Amplitude of oscillation of the block $\text{A}$ is $\frac{mg}{k}$.
• C. Maximum speed acquired by the block $\text{A}$ is $\sqrt{\frac{m{g}^2}{k}}$.
• D. Period of oscillation is $2\pi \sqrt{\frac{m}{2k}}$.

Answer 1

Answer: (A), (B), (C)

Maximum speed acquired by the block $\text{A}$ is $\sqrt{\frac{m{g}^2}{k}}$.

At equilibrium, $k{x}_1=2mg........\left(1\right)$
From $\left(1\right)$, we get,
$x_1=\frac{2mg}{k}......\left(2\right)$

When block $\text{B}$ is removed, $k{x}_1>mg$

Therefore, block $\text{A}$ excutes SHM.

Let '$x_2$' be the extension for new equilibrium of block $\text{A}$.
$\Rightarrow x_2=\frac{mg}{k}......\left(3\right)$

From $\left(2\right)$ and $\left(3\right)$,

Amplitude of oscillation $\left(A\right)=x_1-x_2=\frac{mg}{k}$

As only block $\text{A}$ executes SHM, we can write that
Time period of oscillation $\left(T\right)=2\pi \sqrt{\frac{m}{k}}$

Maximum speed acquired by block $\text{A},\left(V_{max}\right)=A\omega$
$\Rightarrow V_{max}=\frac{mg}{k}\times \sqrt{\frac{k}{m}}=\sqrt{\frac{m{g}^2}{k}}$

Thus, options (a), (b) and (c) are the correct answers.