A block (B) is attached to two unstretched springs S1 and S2 with spring constants k and 4k, respect

Q: Wait, why doesn't the stretched spring store any energy?

For a spring to store energy, it must be stretched or compressed *between two points that resist the force*. Here, the supports $M_1$ and $M_2$ have negligible mass and are not fixed to the walls. If you pull the spring, the support simply glides along with it effortlessly. Without any resistance to stretch against, the spring remains in its natural length and stores zero energy!

Q: Why do we ignore kinetic energy in our final equation?

We are comparing the two states of *maximum displacement* ($x$ to the right, and $y$ to the left). At maximum displacement, the block momentarily comes to a complete stop before reversing direction. Since its velocity is exactly zero at these extreme points, the kinetic energy is zero, and all the mechanical energy is purely potential.

Q: What if the supports $M_1$ and $M_2$ had significant mass?

That would completely change the problem! If they had mass, pulling the spring would accelerate that mass. The spring *would* stretch to provide the necessary force ($F = ma$), and it would store potential energy. The system would become a complex coupled oscillator. The "negligible mass" assumption is what makes this problem elegantly simple.

Q: Are we sure $S_1$ has constant $k$ and $S_2$ has $4k$?

Yes! The problem states "$S_1$ and $S_2$ with spring constants $k$ and $4k$, respectively." The word "respectively" strictly maps the first item to the first value, and the second item to the second value.

Q: Could the block B lose energy when it hits the springs?

No, the problem explicitly states "There is no friction anywhere" and springs are considered ideal (perfectly elastic). Therefore, there are no non-conservative forces doing work, and the total mechanical energy of the system remains perfectly conserved throughout the entire motion.

Question

A block ($B$) is attached to two unstretched springs $S_1$ and $S_2$ with spring constants $k$ and $4k$, respectively. The other ends are attached to two supports $M_1$ and $M_2$ not attached to the walls. The springs and supports have negligible mass. There is no friction anywhere.

The block $B$ is displaced towards wall 1 by a small distance $x$ and released. The block returns and moves a maximum distance $y$ towards wall 2. Displacements $x$ and $y$ are measured with respect to the equilibrium position of the block $B$. The ratio $\frac{y}{x}$ is

Options:

(A) 4

(B) 2

(C) $\frac{1}{2}$

(D) $\frac{1}{4}$

Sigma Percentile · Accepted Answer

### The Illusion of the Two-Spring Oscillator: A Masterclass in Constraints

At first glance, this problem looks like a standard coupled oscillator. You have a block sandwiched between two springs, and you might immediately think about calculating an effective spring constant $k_{eff} = k_1 + k_2$. But physics is rarely about blindly applying formulas; it is about reading the physical constraints of the universe you are given.

The secret to unlocking this problem lies in a single, seemingly innocent sentence: **"The other ends are attached to two supports $M_1$ and $M_2$ not attached to the walls. The springs and supports have negligible mass."**

### The 'Unstretchable' Spring Paradox

Let's conduct a thought experiment. Imagine you grab spring $S_2$ and try to stretch it. For a spring to stretch and store potential energy, it must be pulled from both ends. It needs something to pull *against*.

However, the support $M_2$ is not bolted to the wall, and it has zero mass. According to Newton's Second Law ($F = ma$), if a mass is zero, it requires zero force to accelerate it. The moment you try to pull the spring, the support $M_2$ will instantly and effortlessly glide along with your pull. Because it offers no resistance, the spring never actually stretches! It remains at its natural length, completely relaxed, storing absolutely zero energy.

This brilliant constraint means our springs are **compression-only**. They only act like springs when they are pushed against the solid, immovable walls.

### The Rightward Journey

When we displace block $B$ to the right by a distance $x$ (towards Wall 1), spring $S_1$ is squeezed between the block and Wall 1. It compresses by $x$ and stores elastic potential energy.

Meanwhile, spring $S_2$ is being pulled to the right. But as we just discovered, its support $M_2$ simply slides to the right, keeping $S_2$ perfectly unstretched.

Therefore, the total initial mechanical energy of the system is entirely stored in $S_1$:

$$E_{initial} = \frac{1}{2} k_1 x^2 = \frac{1}{2} k x^2$$

### The Leftward Rebound

When we release the block, the compressed spring $S_1$ violently pushes it back towards the center. The block accelerates, converting all that potential energy into kinetic energy as it crosses the equilibrium point.

It then overshoots and travels to the left, reaching a maximum distance $y$ (towards Wall 2). Now the roles are reversed! Spring $S_2$ is crushed against Wall 2, compressing by $y$. Spring $S_1$ is pulled, but its support $M_1$ simply slides left, keeping $S_1$ unstretched.

At the point of maximum compression $y$, the block momentarily stops. Its kinetic energy is zero, and all the system's energy is now stored in $S_2$:

$$E_{final} = \frac{1}{2} k_2 y^2 = \frac{1}{2} (4k) y^2$$

### The Grand Equivalence

The problem assures us that "there is no friction anywhere." This means the universe of our problem is perfectly conservative. No energy is lost to heat or sound. The energy we put into the system at the start must equal the energy at the end.

By the Law of Conservation of Mechanical Energy, we equate the two states:

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

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

The beauty of physics is how complex physical realities collapse into elegant algebra. We can immediately cancel the $\frac{1}{2} k$ from both sides of the equation:

$$x^2 = 4 y^2$$

We are looking for the ratio $\frac{y}{x}$. Let's rearrange our equation:

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

Taking the square root of both sides, we arrive at our final, pristine answer:

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

This problem is a beautiful reminder that in physics, the boundary conditions and constraints are just as important as the equations themselves. Always read the fine print!

Select Answer:

Visualized Solution

Understanding the Setup

The 'Unstretchable' Springs

Displacement Towards Wall 1

Initial Energy Stored $E_{i}$

Displacement Towards Wall 2

Final Energy Stored $E_{f}$

Conservation of Energy

Calculating the Ratio $\frac{y}{x}$

The Sigma Insight: Conservation of Mechanical Energy

The Illusion of the Two-Spring Oscillator

The 'Unstretchable' Spring Paradox

The Rightward Journey

The Leftward Rebound

The Grand Equivalence

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