Kvant Physics Problem 50

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Problem

F50. On a smooth horizontal table lie two identical cubes of mass $m$, connected by a spring of stiffness $k$ and length $l_0$ in its unstretched state. A constant force $F$, constant in both magnitude and direction, suddenly begins to act on the left cube (Fig. 1). Find the minimum and maximum distance

Fig. 1. Two identical cubes on a smooth horizontal table are connected by a spring; a horizontal force $F$ directed to the right is applied to the left cube.

Setup and Assumptions

The system consists of two identical cubes, each of mass $m$, placed on a smooth horizontal table. They are connected by an ideal spring of stiffness $k$ and natural length $l_0$. A constant horizontal force $F$ begins to act on the left cube at time $t=0$, directed to the right. The quantity of interest is the separation between the cubes as a function of time, and in particular its minimum and maximum values.

The table is assumed to be frictionless, the spring obeys Hooke’s law at all deformations, and the motion is purely horizontal so that vertical forces do not influence the dynamics. The reference frame attached to the table is inertial, and air resistance is neglected.

Physical Principles

Newton’s second law is applied separately to each cube. The left cube satisfies

$m x_1'' = F - k(x_1 - x_2 - l_0),$

while the right cube satisfies

$m x_2'' = k(x_1 - x_2 - l_0).$

The relative deformation of the spring is introduced as

$\xi = x_1 - x_2 - l_0.$

Subtracting the two equations gives

$m(x_1'' - x_2'') = F - 2k\xi,$

which leads to the equation of motion

$\xi'' + \frac{2k}{m}\xi = \frac{F}{m}.$

This is a forced harmonic oscillator with angular frequency

$\omega = \sqrt{\frac{2k}{m}}.$

The general solution has the form

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

The center-of-mass motion satisfies

$X_{\mathrm{cm}} = \frac{x_1 + x_2}{2}, \quad X_{\mathrm{cm}}'' = \frac{F}{2m},$

but this coordinate does not influence the relative extrema of the separation, which are fully determined by $\xi(t)$.

Derivation

The initial configuration corresponds to an unstretched spring with both cubes at rest. Therefore the initial conditions are

$\xi(0) = 0, \quad \xi'(0) = 0.$

Substituting the general solution into the condition at $t=0$ gives

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

which determines

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

Differentiating $\xi(t)$ yields

$\xi'(t) = -A\omega \sin(\omega t) + B\omega \cos(\omega t).$

Applying $\xi'(0)=0$ gives

$0 = B\omega,$

so

$B = 0.$

The relative deformation becomes

$\xi(t) = \frac{F}{2k} - \frac{F}{2k}\cos(\omega t) = \frac{F}{2k}\bigl(1 - \cos(\omega t)\bigr).$

Since $\cos(\omega t)$ varies between $-1$ and $1$, the factor $1 - \cos(\omega t)$ varies between $0$ and $2$. Therefore

$0 \le \xi(t) \le \frac{F}{k}.$

The separation between the cubes is

$L(t) = l_0 + \xi(t).$

Consequently, the minimum separation occurs when $\xi=0$, giving

$L_{\min} = l_0,$

and the maximum separation occurs when $\xi = \frac{F}{k}$, giving

$L_{\max} = l_0 + \frac{F}{k}.$