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(C) When the block $B$ is displaced by a distance $x$ towards wall $1$,spring $S1$ is compressed by $x$,while spring $S2$ remains unstretched because

Vedclass · Accepted Answer

$\frac{1}{2}$

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(C) When the block $B$ is displaced by a distance $x$ towards wall $1$,spring $S1$ is compressed by $x$,while spring $S2$ remains unstretched because support $M2$ is not fixed to the wall and moves with the block. The potential energy stored in the system is $U_i = \frac{1}{2} k x^2$.When the block is released,it moves towards the equilibrium position and then towards wall $2$. When it moves towards wall $2$ by a distance $y$,spring $S2$ is compressed by $y$,while spring $S1$ remains unstretched. The potential energy stored in the system is $U_f = \frac{1}{2} (4k) y^2$.Since there is no friction and the supports have negligible mass,the total mechanical energy is conserved.Equating the initial and final potential energies: $\frac{1}{2} k x^2 = \frac{1}{2} (4k) y^2$.Simplifying this,we get $x^2 = 4y^2$,which implies $y^2 = \frac{x^2}{4}$.Taking the square root,we find $y = \frac{x}{2}$.Therefore,the ratio $\frac{y}{x} = \frac{1}{2}$.

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