1. Analyze the Physical Setup

The problem describes a block B connected to two springs, S1 (spring constant k) and S2 (spring constant 4k). These springs are in turn connected to supports M1 and M2. The crucial information is that the supports are 'not attached to the walls' and there is no friction. This implies that a support can transmit a force to the block only if it is pushed against a wall. If a spring pulls a support away from a wall, the support will move freely along with the end of the spring, and the spring will not be stretched.

Let's define the equilibrium position of block B as z = 0. Displacements towards wall 1 are negative (-z), and towards wall 2 are positive (+z).

2. Energy Stored During Displacement to the Left

• The block is displaced towards wall 1 by a distance x. The position of the block is z = -x.
• As the block moves left, it pushes spring S1. Spring S1, in turn, pushes support M1 against wall 1. Since wall 1 prevents M1 from moving further left, M1 acts as a fixed support. Therefore, spring S1 is compressed by a distance x.
• Simultaneously, as the block moves left, it pulls on spring S2. Spring S2 pulls support M2. Since M2 is not attached to wall 2, there is nothing to hold it in place. It will simply move to the left along with the block. Consequently, spring S2 is not stretched or compressed.
• The total potential energy $U_initial$ stored in the system when the block is at z = -x is solely due to the compression of spring S1.

$U_{initial} = \frac{1}{2} k x^2 + 0 = \frac{1}{2} k x^2$

• The block is released from rest at this position, so its initial kinetic energy is zero. The total initial mechanical energy of the system is $E_initial = U_initial$.

3. Energy Stored During Displacement to the Right

• The block is released and moves towards the right. It passes the equilibrium position z = 0 and reaches a maximum displacement y towards wall 2. Its position is z = +y.
• As the block moves to the right of z = 0, it pushes spring S2. Spring S2 pushes support M2 against wall 2. Since wall 2 prevents M2 from moving further right, M2 acts as a fixed support. Therefore, spring S2 is compressed by a distance y.
• Simultaneously, as the block moves right, it pulls on spring S1. Spring S1 pulls support M1 away from wall 1. Since M1 is not attached to wall 1, it moves freely to the right. Consequently, spring S1 is not stretched.
• At the maximum displacement y, the block momentarily comes to rest. The total potential energy $U_final$ stored in the system is solely due to the compression of spring S2.

$U_{final} = 0 + \frac{1}{2} (4k) y^2 = 2 k y^2$

• Since the block is momentarily at rest at its maximum displacement, its final kinetic energy is zero. The total final mechanical energy of the system is $E_final = U_final$.

4. Apply Conservation of Energy

• The system is frictionless, so the total mechanical energy is conserved.

$E_{initial} = E_{final}$ $U_{initial} = U_{final}$

• Substituting the expressions for the potential energies:

$\frac{1}{2} k x^2 = 2 k y^2$

5. Solve for the Ratio y/x

• We can simplify the energy conservation equation:

$\frac{1}{2} x^2 = 2 y^2$ $x^2 = 4 y^2$

• Taking the square root of both sides (since x and y are distances, they are positive):

$x = 2y$

• The question asks for the ratio y/x:

$\frac{y}{x} = \frac{1}{2}$

This corresponds to option C.