Solution

Statement

$2.4.21.$ Two identical blocks, connected by a spring of stiffness $k$ and length $l_0$, lie on a smooth horizontal table. A constant force $F$, directed along the spring, suddenly begins to act on the left block. Find the minimum and maximum distances between the blocks.

Solution

Let us direct the X-axis along the table in the direction of the force $\mathbf{F}$.
Since no external forces other than $\mathbf{F}$ act on the system along the X-axis, the motion of the center of mass of the system can be determined by Newton's second law:

$$Ma = F$$

where $M = 2m$ is the mass of the system. Hence, the acceleration of the center of mass is:

$$a = \frac{F}{2m}$$

Now let us consider the relative motion of one block with respect to the other. An elastic force $k(x - l_0)$ acts within the system, where $x$ is the current distance between the blocks. Let us shift to a frame of reference moving with the center of mass. We write Newton's second law projected onto the X-axis for the left block:

$$m\ddot{x} = -k(x - l_0) + ma$$

Substituting the acceleration, we obtain:

$$m\ddot{x} = -k(x - l_0) + \frac{F}{2}$$

Let us introduce the relative coordinate $\xi = x - l_0$ and rewrite our equation of motion:

$$m\ddot{\xi} = -k\xi + \frac{F}{2}$$

This is a linear non-homogeneous second-order differential equation, whose solution is the sum of the general solution of the homogeneous equation and a particular solution of the non-homogeneous equation. The solution of the homogeneous equation is known:

$$\xi(t) = A \cos(\omega t + \varphi), \quad \omega = \sqrt{\frac{k}{m}}$$

We will look for a particular solution in the form $\xi_{part}(t) = D$, where $D$ is a constant. Substituting this into the equation:

$$0 = -kD + \frac{F}{2}$$

From which

$$D = \frac{F}{2k}$$

Thus, the general solution of the differential equation is:

$$\xi(t) = A \cos(\omega t + \varphi) + \frac{F}{2k}$$

The amplitude and initial phase of the oscillations are determined by the initial conditions. At the initial moment, the spring is unstretched $\xi(0) = 0$, and the relative velocity is zero $\dot{\xi}(0) = 0$:

$$\dot{\xi}(t) = -\omega A \sin(\omega t + \varphi)$$

Substituting $t=0$, we get:

$$0 = -\omega A \sin(\varphi)$$

Hence $\varphi = 0$. Now let's find the amplitude of the oscillations:

$$0 = A + \frac{F}{2k}$$

Which yields

$$A = -\frac{F}{2k}$$

So, the solution to the differential equation with the initial conditions is:

$$\xi(t) = -\frac{F}{2k} \cos(\omega t) + \frac{F}{2k}$$

Returning to the variable $x$:

$$x(t) = l_0 - \frac{F}{2k} \cos(\omega t) + \frac{F}{2k}$$

The minimum distance between the blocks occurs when $\cos(\omega t) = 1$:

$$\boxed{l_{min} = l_0 - \frac{F}{2k} + \frac{F}{2k} = l_0}$$

The maximum distance between the blocks occurs when $\cos(\omega t) = -1$:

$$\boxed{l_{max} = l_0 + \frac{F}{2k} + \frac{F}{2k} = l_0 + \frac{F}{k}}$$

Alternative Solution (via CM frame)

Let us shift to a non-inertial frame of reference associated with the center of mass (CM) of the system.
The acceleration of the center of mass of the entire system is:
$$a_{cm} = \frac{F}{2m}$$

In this frame of reference, an additional inertial force $F_{in} = -ma_{cm} = -\frac{F}{2}$ acts on each block, directed opposite to the acceleration.

Then, in the CM frame, the following effective forces act on the blocks:

• On the left block: the external force $F$ and the inertial force $F/2$ in the opposite direction. The net force is $F - \frac{F}{2} = \frac{F}{2}$.
• On the right block: only the inertial force $\frac{F}{2}$, directed in the opposite direction (away from the center of mass).

We get a perfectly symmetric picture: in the CM frame, the blocks are pulled apart by equal effective forces $F_{eff} = \frac{F}{2}$.

Let's find the displacement of the equilibrium position in this frame. In the new equilibrium position, the elastic force balances this pulling force:
$$k \Delta x_{eq} = \frac{F}{2} \implies \Delta x_{eq} = \frac{F}{2k}$$

Since the blocks were initially at rest (their relative velocity is zero), they begin their motion from the extreme point. The oscillations of a harmonic oscillator always occur symmetrically relative to the equilibrium position.

Therefore, the amplitude of these oscillations equals the distance from the initial position to the new equilibrium position:
$$A = \Delta x_{eq} = \frac{F}{2k}$$

From this, we immediately obtain the extreme lengths of the spring.
Minimum length (this is our initial extreme point from which the blocks started):
$$\boxed{l_{min} = l_0}$$

Maximum length (the opposite extreme point on the oscillation plot):
$$\boxed{l_{max} = l_0 + 2A = l_0 + 2\left(\frac{F}{2k}\right) = l_0 + \frac{F}{k}}$$