Problem — Mass on two springs in parallel

Problem

A mass $m = 1\;\text{kg}$ is held between two identical springs ($K = 100\;\text{N/m}$ each) attached to two opposite walls, at rest in the initial configuration. Determine the angular frequency of the oscillations along the axis of the springs.

Solution

We displace the mass by an amount $x$ along the axis. One spring stretches by $x$ and pulls back with force $-Kx$; the other compresses by $x$ and pushes back with force $-Kx$. The two forces point in the same direction (restoring), so they add up:

$F = -Kx - Kx = -2Kx$

The system is equivalent to a single spring with effective constant $K_\text{eq} = 2K = 200\;\text{N/m}$. The angular frequency is:

$\omega = \sqrt{\frac{K_\text{eq}}{m}} = \sqrt{\frac{2K}{m}} = \sqrt{\frac{200}{1}}\;\text{rad/s}$

$\ev{\omega = \sqrt{\frac{2K}{m}} \approx 14{,}14\;\text{rad/s}}$

Links

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