Irodov 1.129: Series Springs Split the Stretch – Amit & Deepali Sinha Foundation

Problem Irodov 1.129

Two springs, of stiffness coefficients $k_1$ and $k_2$ , are joined end to end in series to form one composite spring. Find the minimum work needed to stretch this composite system by a total amount $\Delta l$ beyond its natural length.

The two springs in series, of stiffness k_1 and k_2, stretched by equal and opposite forces F at the free ends. Each spring's own extension is marked, Δ l_1 and Δ l_2. — The two springs in series, of stiffness $k_1$ and $k_2$ , stretched by equal and opposite forces $\vect F$ at the free ends. Each spring’s own extension is marked, $\Delta l_1$ and $\Delta l_2$ .

Given

$k_1$
$k_2$
$\Delta l$

Find

A_{\min}

, the minimum work to stretch the series system by

\Delta l

Solution

Step 1Why minimum work means a quasi-static stretch

Stretching the system any faster than quasi-statically leaves the springs with kinetic energy at the end, so the work supplied would exceed the elastic energy actually stored; the least work is done when the system is stretched slowly enough that it is always in equilibrium, so every joule supplied goes into the springs.

Step 2One force, two individual stretches

In series, the same force $F$ is transmitted through both springs at every instant, so each spring stretches according to its own Hooke’s law, and the two individual stretches add up to the system’s total stretch.

\Delta l = \Delta l_1+\Delta l_2 = \frac{F}{k_1}+\frac{F}{k_2}

Solving for $F$ in terms of the total stretch,

F = \frac{\Delta l}{\dfrac{1}{k_1}+\dfrac{1}{k_2}} = \frac{k_1k_2}{k_1+k_2}\,\Delta l

F = k\,\Delta l,\qquad k=\dfrac{k_1k_2}{k_1+k_2}

(1)

Step 3Work done stretching the composite spring

Because the process is quasi-static, at every intermediate stretch $x$ the applied force is the one in (1) evaluated at that $x$ , and the work supplied is the integral of that force over the displacement from $0$ to $\Delta l$ .

A_{\min} = \int_0^{\Delta l} k\,x\,dx = \left.\frac{kx^2}{2}\right|_0^{\Delta l}

A_{\min} = \dfrac12 k(\Delta l)^2,\qquad k=\dfrac{k_1k_2}{k_1+k_2}

(2)

Check

Dimensionally, $[k][\Delta l]^2 = (\mathrm{N/m})(\mathrm m^2) = \mathrm{N\!\cdot\! m} = \mathrm J$ , as required. If $k_1=k_2=k_0$ , (1) gives $k=k_0/2$ , so a pair of identical springs in series is half as stiff as either alone, exactly as expected since each spring then carries only half of $\Delta l$ for the same transmitted force.

Remark

The composite stiffness $k=k_1k_2/(k_1+k_2)$ is the same combination rule as two resistors in parallel, or two capacitors in series: whenever a single scalar (here, the same $F$ ) is shared between two elements that each respond additively (here, the individual stretches), the combined coefficient is the reciprocal sum. Springs in parallel instead share the same displacement and add their forces, giving $k=k_1+k_2$ : the series system here is always the more compliant of the two arrangements for the same pair of springs.