Question

Both springs, string and pulley shown in the figure are light. Initially when both the springs were in their natural state, the system was released from rest. The maximum displacement of the block having mass $m$ is:

• A. $\frac{5mg}{k}$
• B. $\frac{3mg}{k}$
• C. $\frac{10mg}{k}$
• D. $\frac{4mg}{k}$

Answer 1

The correct option is C $\frac{10mg}{k}$

When block is at maximum displacement $(v=0)$, let the extension in upper spring be $x$. Then the pulley will be also be lowered by $x$. This will result in slackening of string by length $2x$, hence the block will be lowered by $2x$.

Balancing forces ( for massless pulley) gives tension $2T$ in upper spring,
$\text{Spring force}=2T$
$\Rightarrow kx=2T$
$\therefore T=\frac{kx}{2}$
In lower pulley, $T=k{x}^{\prime }$
$\frac{kx}{2}=k{x}^{\prime }$
Hence extension in lower spring,
$x^{\prime }=\frac{x}{2}$

Therefore, the block will be lowered by a total distance $=2x+\frac{x}{2}=\frac{5x}{2}$


Applying mechanical energy conservation for system:
$\text{Loss in Grav. PE of block}=\text{Gain in elastic PE of springs}$
$\Rightarrow mg\times \frac{5x}{2}=\frac{1}{2}k{x}^2+\frac{1}{2}k{\left(\frac{x}{2}\right)}^2$
$\Rightarrow \frac{5}{2}mgx=\frac{k{x}^2}{2}+\frac{k{x}^2}{8}$

$\Rightarrow \frac{5}{2}mgx=\frac{5}{8}k{x}^2$
$\therefore x=\frac{4mg}{k}$
The maximum displacment of block is,
$x_m=\frac{5x}{2}$
$\therefore x_m=\frac{5}{2}\times \frac{4mg}{k}=\frac{10mg}{k}$