Irodov 1.152: A Sudden Force Doubles the Stretch – Amit & Deepali Sinha Foundation

Problem Irodov 1.152

Two bars rest on a smooth horizontal plane, connected in line by a weightless spring of stiffness $\varkappa$ whose natural (unstretched) length is $l_0$ . Starting from rest, with the spring at its natural length, a constant horizontal force $F$ is applied to one bar, directed away from the other along the line joining them. Find the maximum and minimum separation the bars reach during the ensuing motion:

when the two bars have equal mass;
when the bars have masses $m_1$ and $m_2$ , with the force $F$ applied to the bar of mass $m_2$ .

The two bars joined by the spring, at rest at the natural length l_0 when the constant force F begins to act on the bar of mass m_2; no wall is present, both bars are free. — The two bars joined by the spring, at rest at the natural length $l_0$ when the constant force $\vect F$ begins to act on the bar of mass $m_2$ ; no wall is present, both bars are free.

Given

$m_1$
$m_2$
$\varkappa$
$l_0$
$F$

Find

(a)

l_{\max}

l_{\min}

for

m_1=m_2

; (b)

l_{\max}

l_{\min}

for general

m_1,m_2

Solution

Step 1Coordinates and the forces on each bar

Let $x_1$ and $x_2$ be the positions of the two bars along the line of the spring, measured in the direction of $\vect F$ , with bar 2 (the one $F$ acts on) ahead of bar 1. The separation is $x_2-x_1$ , and its excess over the natural length is the spring’s extension.

e = x_2-x_1-l_0

(1)

A stretched spring ( $e>0$ ) pulls each bar toward the other with a force of magnitude $\varkappa e$ ; a compressed one ( $e<0$ ) pushes them apart, which the same expression covers by its sign. Bar 1 feels only the spring; bar 2 feels the spring and the applied force $F$ . The plane is smooth, so nothing else acts along the line of motion.

\begin{align*}m_1\ddot x_1 &= \varkappa e \\ m_2\ddot x_2 &= F-\varkappa e\end{align*}

\ddot x_1 = \dfrac{\varkappa e}{m_1},\qquad \ddot x_2 = \dfrac{F}{m_2}-\dfrac{\varkappa e}{m_2}

(2)

Step 2The relative coordinate and the reduced mass

Differentiating (1) twice ( $l_0$ is constant) and substituting the two accelerations from (2),

\begin{align*}\ddot e &= \ddot x_2-\ddot x_1 \\ &= \frac{F}{m_2}-\frac{\varkappa e}{m_2}-\frac{\varkappa e}{m_1} \\ &= \frac{F}{m_2}-\varkappa e\left(\frac{1}{m_1}+\frac{1}{m_2}\right)\end{align*}

The bracket is the reciprocal of the reduced mass $\mu$ of the pair, the single mass that makes a two-body relative motion look like a one-body problem.

\mu = \dfrac{m_1m_2}{m_1+m_2},\qquad \dfrac{1}{\mu} = \dfrac{1}{m_1}+\dfrac{1}{m_2}

(3)

Multiplying the equation for $\ddot e$ through by $\mu$ and using (3),

\begin{align*}\mu\ddot e &= \frac{\mu F}{m_2}-\varkappa e \\ &= \frac{m_1F}{m_1+m_2}-\varkappa e\end{align*}

The constant term is the share of $F$ that drives the relative motion; naming it $F_1$ gives the equation of one body of mass $\mu$ , coordinate $e$ , under a constant force and a spring.

\mu\ddot e = F_1-\varkappa e,\qquad F_1=\dfrac{m_1F}{m_1+m_2}

(4)

Step 3Energy of the relative motion

Multiplying (4) by $\dot e$ turns each side into a time derivative: the left side of $\frac12\mu\dot e^2$ , the right side of $F_1e-\frac12\varkappa e^2$ .

\begin{align*}\mu\dot e\,\ddot e &= F_1\dot e-\varkappa e\,\dot e \\ \frac{d}{dt}\left(\frac12\mu\dot e^2\right) &= \frac{d}{dt}\left(F_1e-\frac12\varkappa e^2\right)\end{align*}

Integrating from the start, where the bars are at rest at the natural length ( $e=0$ , $\dot e=0$ ), to any later instant, both sides start from zero.

\begin{align*}\frac12\mu\dot e^2-0 &= \left(F_1e-\frac12\varkappa e^2\right)-0\end{align*}

\frac12\mu\dot e^2 = F_1e-\frac12\varkappa e^2

(5)

Step 4Turning points of the separation

The left side of (5) cannot be negative, so the motion is confined to the values of $e$ where the right side is non-negative, and the extremes of the separation are where $\dot e=0$ .

\begin{align*}F_1e-\frac12\varkappa e^2 &= 0 \\ e\left(F_1-\frac12\varkappa e\right) &= 0\end{align*}

The two roots are the two turning points: $e=0$ , the starting separation, and the far one.

e_{\max} = \dfrac{2F_1}{\varkappa} = \dfrac{2m_1F}{\varkappa(m_1+m_2)}

(6)

Between them the right side of (5) is positive, so $e$ oscillates between $0$ and $e_{\max}$ : the separation never drops below $l_0$ and never exceeds $l_0+e_{\max}$ .

l_{\max} = l_0+\dfrac{2m_1F}{\varkappa(m_1+m_2)},\qquad l_{\min}=l_0

(7)

Step 5Equal masses

For $m_1=m_2=m$ , the driving share $F_1$ of (4) and the far turning point (6) specialize.

\begin{align*}F_1 &= \frac{mF}{m+m} = \frac{F}{2} \\ e_{\max} &= \frac{2F_1}{\varkappa} = \frac{F}{\varkappa}\end{align*}

l_{\max} = l_0+\dfrac{F}{\varkappa},\qquad l_{\min}=l_0

(8)

Check

As $m_1\to0$ (the bar without the applied force carries no inertia), $F_1\to0$ and $l_{\max}\to l_0$ : with nothing to resist, bar 1 is dragged along instantly and the spring never stretches. As $m_2\to0$ (the driven bar carries no inertia), $F_1\to F$ and $l_{\max}\to l_0+2F/\varkappa$ : bar 1 behaves as an immovable wall and bar 2 is a single mass on a spring suddenly loaded by $F$ , the classic overshoot-to-twice-the-static-stretch result.

Remark

The bars never approach closer than $l_0$ : the applied force only ever stretches the spring, so the separation’s lower turning point coincides with the initial, undeformed state and the motion is a one-sided oscillation between $l_0$ and $l_{\max}$ . This is the same doubling phenomenon as a mass suddenly hung on a vertical spring, which sags to twice its eventual equilibrium extension before the first return; here the “equilibrium” extension $F_1/\varkappa$ is set by $F$ acting through the reduced-mass dynamics rather than by weight.