Two point masses m1 and m2 are connected by a spring of natural length l0. The spring is compressed

Q: Wait, why is the velocity of the center of mass constant even though the spring is pushing the blocks apart?

According to Newton's Second Law for a system of particles, the acceleration of the center of mass is determined solely by the net *external* force. Since the spring force is an *internal* force acting between the two masses, and there is no friction or other horizontal external force, the net external force is zero. Thus, the center of mass moves with a constant velocity, completely ignoring the internal oscillations.

Q: Why do we set the acceleration to zero to find the natural length? Shouldn't the velocity be zero?

No, that's a common trap! The natural length of a spring is its relaxed state, where it is neither stretched nor compressed. Hooke's Law states that the spring force is proportional to the extension or compression ($F = -kx$). Therefore, at the natural length, the spring force is zero. By Newton's Second Law ($F = ma$), if the force is zero, the acceleration must be zero. The velocity is actually at a *maximum* when passing through the natural length.

Q: Why did the $v_0 t$ terms cancel out when calculating the separation $x_2 - x_1$?

The term $v_0 t$ represents the steady, uniform translation of the entire system's center of mass. Both masses share this forward motion. The separation between the masses depends only on their relative motion (the oscillation caused by the spring). When you subtract their positions, the shared translational motion mathematically cancels out, leaving only the relative oscillatory terms.

Q: What does the term $A(1 - \cos \omega t)$ represent physically in the equation for $x_1$?

This term represents the oscillatory motion of the mass relative to the center of mass. The cosine function dictates the back-and-forth movement. The factor $A$ is related to the amplitude of this oscillation, and $\omega$ is the angular frequency. The $1 - \cos \omega t$ form ensures that at $t=0$, the oscillatory displacement is zero, matching the initial condition that the string just broke.

Q: Could we have found the natural length by analyzing mass $m_2$ instead of $m_1$?

Absolutely! The physics is perfectly symmetric. If you differentiate the expression we found for $x_2$ twice to find its acceleration $a_2$, and set $a_2 = 0$, you will arrive at the exact same condition: $\cos \omega t = 0$. Both masses experience zero acceleration at the exact same instant when the spring passes through its natural length.

Question

Two point masses $m_1$ and $m_2$ are connected by a spring of natural length $l_0$. The spring is compressed such that the two point masses touch each other and then they are fastened by a string. Then the system is moved with a velocity $v_0$ along positive x-axis. When the system reaches the origin the string breaks ($t = 0$). The position of the point mass $m_1$ is given by $x_1 = v_0 t - A (1 - \cos \omega t)$ where $A$ and $\omega$ are constants. Find the position of the second block as a function of time. Also, find the relation between $A$ and $l_0$.

Sigma Percentile · Accepted Answer

### The Setup: A System in Motion

Imagine a fascinating mechanical ballet unfolding on a frictionless horizontal plane. We have two point masses, $m_1$ and $m_2$, connected by a spring. Initially, this spring is compressed, and the two masses are held tightly together by a string. The entire assembly is gliding smoothly along the positive x-axis with a constant velocity $v_0$.

Suddenly, at the exact moment the system crosses the origin ($t=0$), the string snaps! The compressed spring is unleashed, and the masses begin to push against each other. However, despite this internal chaos, a profound physical principle governs the system as a whole.

### The Unshakable Center of Mass

To unravel this problem, we must look past the individual bouncing blocks and focus on the system's **Center of Mass (COM)**. According to Newton's laws, the acceleration of the center of mass is determined solely by the net *external* force acting on the system.

In our scenario, the spring force is purely *internal*—it acts between $m_1$ and $m_2$ but does not push or pull the system from the outside. Since the horizontal surface is frictionless, the net external force in the x-direction is strictly zero.

Because the external force is zero, the velocity of the center of mass remains perfectly constant. It continues to cruise at the initial velocity $v_0$. Therefore, the position of the center of mass at any time $t$ is beautifully simple:

$$x_{CM} = v_0 t$$

### Unraveling the Position of Mass 2

We are given the rather complex equation for the position of the first mass:

$$x_1 = v_0 t - A(1 - \cos \omega t)$$

Our goal is to find $x_2$, the position of the second mass. We can bridge the gap between the individual masses and the center of mass using the fundamental definition of the COM coordinate:

$$x_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$$

Now, we substitute our simple expression for $x_{CM}$ and the given expression for $x_1$ into this equation:

$$v_0 t = \frac{m_1 [v_0 t - A(1 - \cos \omega t)] + m_2 x_2}{m_1 + m_2}$$

To isolate $x_2$, we multiply both sides by the total mass $(m_1 + m_2)$:

$$(m_1 + m_2) v_0 t = m_1 v_0 t - m_1 A(1 - \cos \omega t) + m_2 x_2$$

Notice the elegance of the algebra here. When we expand the left side, we get $m_1 v_0 t + m_2 v_0 t$. The $m_1 v_0 t$ term appears on both sides of the equation and cancels out perfectly! This leaves us with:

$$m_2 v_0 t = - m_1 A(1 - \cos \omega t) + m_2 x_2$$

Dividing everything by $m_2$, we arrive at the position of the second block:

$$x_2 = v_0 t + \frac{m_1}{m_2} A(1 - \cos \omega t)$$

### The Secret of the Natural Length

The second part of the problem asks us to find the relationship between the amplitude constant $A$ and the spring's natural length $l_0$. To do this, we must translate the geometric concept of "natural length" into a dynamic physical condition.

When a spring is exactly at its natural length, it is neither stretched nor compressed. Consequently, it exerts **zero force** on the masses attached to it. By Newton's Second Law ($F = ma$), if the net force on a mass is zero, its acceleration must also be zero.

Therefore, the critical condition we are looking for is the moment when the acceleration of the masses is zero.

### The Calculus of Oscillation

Let's find the acceleration of the first mass, $m_1$. We start with its position equation and differentiate it with respect to time to find its velocity, $v_1$:

$$v_1 = \frac{dx_1}{dt} = v_0 - A \omega \sin \omega t$$

Differentiating a second time gives us the acceleration, $a_1$:

$$a_1 = \frac{dv_1}{dt} = - A \omega^2 \cos \omega t$$

Applying our physical condition, we set the acceleration to zero to find the moment the spring is at its natural length:

$$- A \omega^2 \cos \omega t = 0$$

This implies that at the exact moment the spring reaches its natural length, the condition **$\cos \omega t = 0$** must hold true.

### The Final Revelation

The natural length $l_0$ is simply the physical separation between the two masses at this specific instant. Let's calculate the general separation, $x_2 - x_1$:

$$x_2 - x_1 = \left[ v_0 t + \frac{m_1}{m_2} A(1 - \cos \omega t) \right] - \left[ v_0 t - A(1 - \cos \omega t) \right]$$

Notice how the $v_0 t$ terms cancel out. This makes perfect physical sense; the separation between the blocks depends only on their relative oscillation, not on the steady forward motion of the entire system. Factoring out the common terms, we get:

$$x_2 - x_1 = A(1 - \cos \omega t) \left( \frac{m_1}{m_2} + 1 \right)$$

Finally, we substitute our critical condition, $\cos \omega t = 0$, into this separation equation. The separation at this moment is, by definition, the natural length $l_0$:

$$l_0 = A(1 - 0) \left( \frac{m_1}{m_2} + 1 \right)$$

$$l_0 = A \left( \frac{m_1}{m_2} + 1 \right)$$

Through a beautiful synthesis of center-of-mass mechanics, kinematics, and calculus, we have unraveled the hidden dynamics of this oscillating system!

Visualized Solution

System Setup and Initial State

Motion of the Centre of Mass

COM Position Formula

Substituting Knowns into COM Equation

Isolating $x_{2}$

Position of the Second Block

Condition for Natural Length

Acceleration of Mass $m_{1}$

Applying Zero Acceleration Condition

Separation Between Blocks

Relation Between $A$ and $l_{0}$

The Sigma Insight: Motion of Centre of Mass

The Setup

The Unshakable Center of Mass

Unraveling the Position of Mass 2

The Secret of the Natural Length

The Calculus of Oscillation

The Final Revelation

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