QJEEAdv08Ph8f23: JEE Advanced 2008 Physics - Class 11 JEEAdv08Ph8f23 Solution: A block (B) is attached to two unstretched springs S1 and S2…

When the block B is displaced towards wall-1 only spring S $_1$ is compressed and S $_2$ is in natural state.

If spring S $_2$ is not attached to wall but it is free but it gain momentum ( $\uparrow$ ) and moves towards S $_1$ .

The spring S $_1$ comes to natural length and spring S $_2$ gets compressed.

So if there is no frictional force i.e.,

$fr=0$

So potential energy of spring S $_1$ as the potential energy of spring.

Hence, ${(PE)_{{S_1}}} = {(PE)_{{S_2}}}$

$\Rightarrow {1 \over 2}{k_1}{x^2} = {1 \over 2}{k_2}y$

$\Rightarrow 1k{x^2} = (4k){y^2} \Rightarrow {x^2} = 4{y^2}$

Potential energy of spring.

Hence, ${(PE)_{{S_1}}} = {(PE)_{{S_2}}}$

$\Rightarrow {1 \over 2}{k_1}{x^2} = {1 \over 2}{k_2}{y^2}$

$\Rightarrow {1 \over 2}k{x^2} = {1 \over 2}(4k){y^2}$

$\Rightarrow {x^2} = 4{y^2}$

$\Rightarrow {y \over x} = {1 \over 2}$