13  Spring Force

Author

Theresa Honein

Published

January 1, 2026

13.1 Introduction

TipThink!

Question: What does it mean for a spring to be linear?

Consider a mass \(m\) suspended by a spring causing an extension \(\delta\). Suspending a mass \(2m\) causes double the extension, \(2\delta\).

Hooke’s Law: The force generated by a spring is assumed to be linearly proportional to its extension/compression.

Consider a block of mass \(m\) attached to a linear spring of stiffness \(k\) and unstretched length \(\ell_0\). Denote the current length by \(\ell\) and the stretch by \(\varepsilon = \ell-\ell_0\).

TipThink!

Question: What is the stretch in each case? What is the spring force?

  1. Unstretched: \(\varepsilon = 0\), no spring force.

  2. Extended: \(\varepsilon = \ell - \ell_0 > 0\). Spring force \({\bf F}_s = -k\varepsilon{\bf E}_x\) (pointing left, toward equilibrium).

  3. Compressed: \(\varepsilon = \ell - \ell_0 < 0\). Spring force \({\bf F}_s = -k\varepsilon{\bf E}_x\) (pointing right, toward equilibrium).

ImportantNote! Prescribing the spring force

The spring force always seeks to return the particle to the unstretched state. The same expression applies whether the spring is extended or compressed.

13.2 Formalism

In general:

General spring configuration
TipThink!

Question: What is the prescription of the spring force in this general case?

…

The stretch is: \[\begin{align} \varepsilon = \lnorm{\bf r}-{\bf r}_A\rnorm-\ell_0, \end{align}\] where \(\ell_0\) is the unstretched length. The spring force is: \[\begin{align} {\bf F}_s = -K\lp\lnorm{\bf r}-{\bf r}_A\rnorm-\ell_0\rp\frac{{\bf r}-{\bf r}_A}{\lnorm{\bf r}-{\bf r}_A\rnorm}. \end{align}\] This direction is correct in both tension and compression.

13.2.1 Example: Horizontal Simple Harmonic Oscillator

Derive the equations of motion with the origin at the fixed end of the spring.

TipThink!

Question: How does the equation of motion change if the origin is taken at the free end when the spring is unstretched?

The equation of motion is \[\begin{align} \ddot{x}+\frac{k}{m}x = 0, \end{align}\] where \(x\) is measured from the undeformed position.

13.2.2 Example: Vertical Simple Harmonic Oscillator

By analogy, the equation of motion of a vertical harmonic oscillator is: \[\begin{align} \ddot{x}+\frac{k}{m}x = 0, \end{align}\] where \(x\) is measured from the static equilibrium position.

See this video for intuition on static deflection.

13.3 Summary

The spring force on a mass \(m\) at position \(\mathbf{r}\), attached to a spring (stiffness \(K\), unstretched length \(\ell_0\)) with base at \(\mathbf{r}_A\): \[\begin{align} \mathbf{F}_s = -K(\|\mathbf{r}-\mathbf{r}_A\|-\ell_0)\,\frac{\mathbf{r}-\mathbf{r}_A}{\|\mathbf{r}-\mathbf{r}_A\|}. \end{align}\] Taking the origin at the static equilibrium position yields \(\ddot x + (k/m)x = 0\) with \(\omega_n=\sqrt{k/m}\).

13.4 Lecture Videos

13.5 Exercises

The following problems are from Set 09 – Spring Force (Chapter 8 of MKB). In each problem, find the equation of motion of the block.

1. [08-002] (ans. \(\omega_n = 12\) rad/s, \(f_n = 1.910\) Hz)

2. [08-004] First determine the static deflection of the spring; take the origin at the statically deflected position. (ans. \(\delta_{st} = 0.200\) m, \(\tau = 0.898\) s, \(v_{\max} = 0.7\) m/s)

3. [08-019] (ans. \(\ddot y + \frac{3k}{mL^2}y = 0\))