Edits to “Statement”, “Answer”

Valter edited Aug 13, 2026, 7:17 PM

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			### Statement

		−	$2.4.21.$ [Insert the problem statement]
		−
		−	### 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.

			![For problem $2.4.21$\|448x187, 35%](../../img/2.4.21/2.4.21.png)

			### 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):
			$$
			\\boxed{l_{max} = l_0 + 2A = l_0 + 2\\left(\\frac{F}{2k}\\right) = l_0 + \\frac{F}{k}}
			$$

			#### Answer

		@@ -159,7 +153,3 @@Answer
			$l_{min}=l_{0}$

			$l_{max}=l_{0}+\\frac{F}{k}$
		−
		−	#### Answer
		−
		−	[Insert a concise answer or boxed result]