New solution

Valter edited
View solution History
revision #20004 newer →
en/2.4.21.md +165 −0
@@ -0,0 +1,165 @@
+### 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
+
+$l_{min}=l_{0}$
+
+$l_{max}=l_{0}+\\frac{F}{k}$
+
+#### Answer
+
+[Insert a concise answer or boxed result]