Problem — Mass between two springs, equivalent K

Problem

A mass $m$ is connected between two springs of constants $K_1$ and $K_2$, fixed to two opposite walls. Show, starting from the force balance for a small displacement $x$, that the motion is harmonic with equivalent elastic constant $K_\text{eq} = K_1 + K_2$. Compare with the result for two springs in series.

Solution

We displace the mass by $x$ from the equilibrium position. Spring 1 is deformed and reacts with a force $-K_1 x$; spring 2 likewise reacts with $-K_2 x$. One spring pulls and the other pushes, but both point towards equilibrium, so the forces add up: $F = -K_1 x - K_2 x = -(K_1 + K_2)\, x$ Newton’s second law gives $m\,\ddx = -(K_1 + K_2)\, x$ which is the equation of harmonic motion $\ddx = -\omega^2 x$ with $\omega = \sqrt{\frac{K_1 + K_2}{m}}, \qquad T = 2\pi\sqrt{\frac{m}{K_1 + K_2}}$ The two springs arranged this way act in parallel, with equivalent constant $\ev{K_\text{eq} = K_1 + K_2}$ If instead the springs were in series (one after the other), the same force passes through both and the extensions add up: $1/K_\text{eq} = 1/K_1 + 1/K_2$, i.e. $K_\text{eq}^\text{serie} = \frac{K_1 K_2}{K_1 + K_2}$ which is smaller than the smaller of the two: the series system is softer, the parallel one stiffer.

Links

Topics: Oscillations and harmonic motion Concepts: Elastic force and Hooke’s law · Simple harmonic motion Skills: Symbolic set-up Objects: Spring