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| ### Solution | | ### Solution |
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| Let us direct the X-axis along the table in the direction of the force $\\mathbf{F}$. | | 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: | | 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: |
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| $$ | | $$ |
| Ma = F | | Ma = F |
| $$ | | $$ |
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| where $M = 2m$ is the mass of the system. Hence, the acceleration of the center of mass is: | | where $M = 2m$ is the mass of the system. Hence, the acceleration of the center of mass is: |
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| $$ | | $$ |
| a = \\frac{F}{2m} | | a = \\frac{F}{2m} |
| $$ | | $$ |
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| 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: | | 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: |
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| $$ | | $$ |
| m\\ddot{x} = -k(x - l_0) + ma | | m\\ddot{x} = -k(x - l_0) + ma |
| $$ | | $$ |
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| Substituting the acceleration, we obtain: | | Substituting the acceleration, we obtain: |
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| $$ | | $$ |
| m\\ddot{x} = -k(x - l_0) + \\frac{F}{2} | | m\\ddot{x} = -k(x - l_0) + \\frac{F}{2} |
| $$ | | $$ |
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| Let us introduce the relative coordinate $\\xi = x - l_0$ and rewrite our equation of motion: | | Let us introduce the relative coordinate $\\xi = x - l_0$ and rewrite our equation of motion: |
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| $$ | | $$ |
| m\\ddot{\\xi} = -k\\xi + \\frac{F}{2} | | m\\ddot{\\xi} = -k\\xi + \\frac{F}{2} |
| $$ | | $$ |
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| 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: | | 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: |
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| $$ | | $$ |
| \\xi(t) = A \\cos(\\omega t + \\varphi), \\quad \\omega = \\sqrt{\\frac{k}{m}} | | \\xi(t) = A \\cos(\\omega t + \\varphi), \\quad \\omega = \\sqrt{\\frac{k}{m}} |
| $$ | | $$ |
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| We will look for a particular solution in the form $\\xi_{part}(t) = D$, where $D$ is a constant. Substituting this into the equation: | | We will look for a particular solution in the form $\\xi_{part}(t) = D$, where $D$ is a constant. Substituting this into the equation: |
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| $$ | | $$ |
| 0 = -kD + \\frac{F}{2} | | 0 = -kD + \\frac{F}{2} |
| $$ | | $$ |
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| From which | | From which |
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| $$ | | $$ |
| D = \\frac{F}{2k} | | D = \\frac{F}{2k} |
| $$ | | $$ |
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| Thus, the general solution of the differential equation is: | | Thus, the general solution of the differential equation is: |
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| $$ | | $$ |
| \\xi(t) = A \\cos(\\omega t + \\varphi) + \\frac{F}{2k} | | \\xi(t) = A \\cos(\\omega t + \\varphi) + \\frac{F}{2k} |
| $$ | | $$ |
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| 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 $\\dot{\\xi}(0) = 0$: | | 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 $\\dot{\\xi}(0) = 0$: |
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| $$ | | $$ |
| \\dot{\\xi}(t) = -\\omega A \\sin(\\omega t + \\varphi) | | \\dot{\\xi}(t) = -\\omega A \\sin(\\omega t + \\varphi) |
| $$ | | $$ |
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| Substituting $t=0$, we get: | | Substituting $t=0$, we get: |
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| $$ | | $$ |
| 0 = -\\omega A \\sin(\\varphi) | | 0 = -\\omega A \\sin(\\varphi) |
| $$ | | $$ |
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| Hence $\\varphi = 0$. Now let's find the amplitude of the oscillations: | | Hence $\\varphi = 0$. Now let's find the amplitude of the oscillations: |
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| $$ | | $$ |
| 0 = A + \\frac{F}{2k} | | 0 = A + \\frac{F}{2k} |
| $$ | | $$ |
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| Which yields | | Which yields |
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| $$ | | $$ |
| A = -\\frac{F}{2k} | | A = -\\frac{F}{2k} |
| $$ | | $$ |
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| So, the solution to the differential equation with the initial conditions is: | | So, the solution to the differential equation with the initial conditions is: |
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| $$ | | $$ |
| \\xi(t) = -\\frac{F}{2k} \\cos(\\omega t) + \\frac{F}{2k} | | \\xi(t) = -\\frac{F}{2k} \\cos(\\omega t) + \\frac{F}{2k} |
| $$ | | $$ |
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| Returning to the variable $x$: | | Returning to the variable $x$: |
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| $$ | | $$ |
| x(t) = l_0 - \\frac{F}{2k} \\cos(\\omega t) + \\frac{F}{2k} | | x(t) = l_0 - \\frac{F}{2k} \\cos(\\omega t) + \\frac{F}{2k} |
| $$ | | $$ |
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| The minimum distance between the blocks occurs when $\\cos(\\omega t) = 1$: | | The minimum distance between the blocks occurs when $\\cos(\\omega t) = 1$: |
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| $$ | | $$ |
| \\boxed{l_{min} = l_0 - \\frac{F}{2k} + \\frac{F}{2k} = l_0} | | \\boxed{l_{min} = l_0 - \\frac{F}{2k} + \\frac{F}{2k} = l_0} |
| $$ | | $$ |
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| The maximum distance between the blocks occurs when $\\cos(\\omega t) = -1$: | | The maximum distance between the blocks occurs when $\\cos(\\omega t) = -1$: |
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| $$ | | $$ |
| \\boxed{l_{max} = l_0 + \\frac{F}{2k} + \\frac{F}{2k} = l_0 + \\frac{F}{k}} | | \\boxed{l_{max} = l_0 + \\frac{F}{2k} + \\frac{F}{2k} = l_0 + \\frac{F}{k}} |
| $$ | | $$ |
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| ### Alternative Solution (via CM frame) | | ### Alternative Solution (via CM frame) |
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| Let us shift to a non-inertial frame of reference associated with the center of mass (CM) of the system. | | 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: | | The acceleration of the center of mass of the entire system is: |
| $$ | | $$ |
| a_{cm} = \\frac{F}{2m} | | a_{cm} = \\frac{F}{2m} |
| $$ | | $$ |
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| 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. | | 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. |
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| Then, in the CM frame, the following effective forces act on the blocks: | | 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 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). | | - On the right block: only the inertial force $\\frac{F}{2}$, directed in the opposite direction (away from the center of mass). |
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| We get a perfectly symmetric picture: in the CM frame, the blocks are pulled apart by equal effective forces $F_{eff} = \\frac{F}{2}$. | | We get a perfectly symmetric picture: in the CM frame, the blocks are pulled apart by equal effective forces $F_{eff} = \\frac{F}{2}$. |
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| Let's find the displacement of the equilibrium position in this frame. In the new equilibrium position, the elastic force balances this pulling force: | | 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} | | k \\Delta x_{eq} = \\frac{F}{2} \\implies \\Delta x_{eq} = \\frac{F}{2k} |
| $$ | | $$ |
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| 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. | | 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. |
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| Therefore, the amplitude of these oscillations equals the distance from the initial position to the new 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} | | A = \\Delta x_{eq} = \\frac{F}{2k} |
| $$ | | $$ |
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| From this, we immediately obtain the extreme lengths of the spring. | | From this, we immediately obtain the extreme lengths of the spring. |
| Minimum length (this is our initial extreme point from which the blocks started): | | Minimum length (this is our initial extreme point from which the blocks started): |
| $$ | | $$ |
| \\boxed{l_{min} = l_0} | | \\boxed{l_{min} = l_0} |
| $$ | | $$ |
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| Maximum length (the opposite extreme point on the oscillation plot): | | 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}} | | \\boxed{l_{max} = l_0 + 2A = l_0 + 2\\left(\\frac{F}{2k}\\right) = l_0 + \\frac{F}{k}} |
| $$ | | $$ |
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| #### Answer | | #### Answer |
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