Two springs in parallel

Problem

A mass of $2{,}0\ \text{kg}$ is held horizontally in equilibrium by two springs attached in parallel to opposite walls, with constants $k_1 = 30\ \text{N/m}$ and $k_2 = 70\ \text{N/m}$, both initially at rest. An external force of $20\ \text{N}$ is applied towards one wall. By how much does the mass move?

Solution

The mass is connected to two springs attached to opposite walls. When the mass moves by an amount $x$ towards one wall, one spring stretches and the other compresses by the same length $x$: both react by opposing the displacement. The two elastic forces therefore point in the same direction (against the external force) and add together. This is the parallel-spring configuration.

The equivalent spring constant of the system is the sum of the two constants: $k_{\text{eq}} = k_1 + k_2 = 30 + 70 = 100\ \text{N/m}$

At equilibrium the external force is balanced by the overall elastic force: $F = k_{\text{eq}}\, x = (k_1 + k_2)\, x$

We obtain the displacement: $x = \frac{F}{k_1 + k_2} = \frac{20}{100}$

$\ev{x = 0{,}20\ \text{m} = 20\ \text{cm}}$

Note that the mass $m = 2{,}0\ \text{kg}$ does not enter the calculation of the equilibrium displacement: it only defines the body, but the final position depends solely on the ratio between the applied force and the overall stiffness of the springs.

Links

Topics: Dynamics Concepts: Elastic force and Hooke’s law · Newton’s second law Objects: Spring