Question

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

• A. $5mg/2K$
• B. $4mg/K$
• C. $10mg/K$
• D. $mg/K$

Answer 1

The correct option is C $10mg/K$

Let the extension of spring 1 be $x_1$

Let the downward extension in spring 2 be $x_2$

If the extension in spring 1 is $x_1$, then pulley will displace by same $x_1$

Thus the string release will be double the displacement of pulley is ${2x}_1$

Similarly the extension in the spring 2 is $x_2$, then the displacement of mass is same as extension $x_2$.

Hence, the net displacement of mass is ${2x}_1+x_2$

First apply the force balance equation,

Force of spring 1 = Tension in string

${Kx}_2=T$ $\left(1\right)$

On pulley net force,

$2T={Kx}_1$ $\left(2\right)$

By simplifying equation $1$ and $2$, we get:

$x_1={2x}_2$ $\left(3\right)$

Applying law of conservation of energy:

$\frac{1}{2}K{\left(x_1\right)}^2+\frac{1}{2}K{\left(x_2\right)}^2=mg({2x}_1+x_2)$ $\left(4\right)$

From equation $3$ and $4,$ we get:

$x_2=\frac{2mg}{K}{x}_1=\frac{4mg}{K}$

Net displacement of mass is $={2x}_1+x_2$

Put the value of $x_1$ and $x_2$

Net displacement = ${2x}_1+x_2=2\frac{4mg}{K}+\frac{2mg}{K}=\frac{10mg}{K}$