Question

A block \(B\) is attached to two unstretched springs \(S_1\) and \(S_2\) with spring constants \(k\) and \(4k\) respectively (see the figure I). The other ends are attached to the identical supports \(M_1\) and \(M_2\) which are not fixed to the walls. The springs and supports have negligible masses. There is no friction anywhere. The block \(B\) is displaced towards the wall \(1\) by a small distance \(x\) (see the figure II) and then released. The block returns and moves a maximum distance \(y\) towards the wall \(2\). The displacements \(x\) and \(y\) are measured with respect to the equilibrium position of the block \(B\). The ratio \(\dfrac{y}{x}\) is


\( \begin{array}{l} \underset{y}{*} \mid \\ \end{array} \) I

• A. $\frac{1}{4}$
• B. $4$
• C. $\frac{1}{2}$
• D. $2$

Answer 1

The correct option is C $\frac{1}{2}$

Given, Spring constant of spring $S_1=k$ and spring constant of spring $S_2=4k$

Case $1:$ When Block $B$ is moved towards wall $1$ by a distance $x.$

In this case, there will be compression in spring $S_1$ and spring $S_2$ will remain unchanged as it will move with support $M_2$, the same distance $x$ from the mean position of block $B.$
So, energy stored in spring $S_1$ will be $\frac{1}{2}k{x}^2$

Case $2:$ When Block $B$ returns and moves a maximum distance $y$ towards wall $2.$
In this case, there will be compression in spring $S_2$ and spring $S_1$ will move with support $M_1$, the same maximum distance $y$ from the mean position of block $B.$
So, energy stored in spring $S_2$ will be $\frac{1}{2}\left(4k\right)y^2$
By the law of conservation of mechanical energy,
$\frac{1}{2}k{x}^2=\frac{1}{2}\left(4k\right)y^2$
$\Rightarrow \frac{y}{x}=\frac{1}{2}$.
Hence, option $\left(C\right)$ is correct.