Vibrating Springs
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 7.3: “Applications of Second-Order Differential Equations” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/7-3-applications-of-second-order-differential-equations |
| Textbook used in class | Stewart, Calculus, Section 17.3: “Applications of Second-Order Differential Equations” |
Quick Reference
A spring-mass system satisfies: $$m x'' + c x' + k x = F(t),$$ where $m > 0$ is mass, $c \geq 0$ is the damping constant, $k > 0$ is the spring constant, and $F(t)$ is the external force. The natural frequency is $\omega_0 = \sqrt{k/m}$.
Discriminant $\Delta = c^2 - 4mk$ determines behavior for free vibration ($F=0$):
| $\Delta$ | Case | Behavior |
|---|---|---|
| $< 0$ | Underdamped | Oscillation with decaying amplitude |
| $= 0$ | Critically damped | Fastest return to equilibrium, no oscillation |
| $> 0$ | Overdamped | Exponential decay, no oscillation |
| $c = 0$ | Undamped | Pure oscillation at frequency $\omega_0$ |
Motivation
Attach a mass to a spring, pull it down, and release it. If there is no friction, the mass oscillates forever at a fixed frequency. Add friction (damping), and the oscillation gradually dies out -- the heavier the friction, the faster it dies. Drive the system by pushing rhythmically at just the right frequency, and the amplitude can grow without bound: resonance.
These qualitatively different behaviors all emerge from the same DE, with the discriminant $c^2 - 4mk$ determining which case applies. The DE is a direct translation of Newton’s second law: net force = mass $\times$ acceleration, where the forces are the spring restoring force $-kx$ and the damping friction $-cx'$.
Key Cases
Undamped free vibration ($c = 0$, $F = 0$): $$x'' + \omega_0^2 x = 0, \quad \omega_0 = \sqrt{k/m}.$$ $$x(t) = c_1\cos(\omega_0 t) + c_2\sin(\omega_0 t).$$ The system oscillates forever at frequency $\omega_0$ rad/s, period $2\pi/\omega_0$.
Underdamped ($c^2 < 4mk$): Characteristic roots $r = \alpha \pm \beta i$ with $\alpha = -c/(2m) < 0$ and $\beta = \sqrt{4mk-c^2}/(2m)$. $$x(t) = e^{\alpha t}(c_1\cos\beta t + c_2\sin\beta t).$$ Oscillation with amplitude decaying as $e^{\alpha t} \to 0$.
Critically damped ($c^2 = 4mk$): Repeated root $r = -c/(2m)$. $$x(t) = (c_1 + c_2 t)\,e^{-c/(2m)\cdot t}.$$ Returns to zero without oscillating, as quickly as possible.
Overdamped ($c^2 > 4mk$): Two distinct negative real roots $r_1 < r_2 < 0$. $$x(t) = c_1 e^{r_1 t} + c_2 e^{r_2 t} \to 0.$$ Returns to zero without oscillating, more slowly than critically damped.
Worked Example
A spring with $k = 4$ N/m holds a mass of $m = 1$ kg. No damping, no external force. Initial conditions: $x(0) = 2$ m, $x'(0) = 0$. Find $x(t)$.
$\omega_0 = \sqrt{4/1} = 2$. Equation: $x'' + 4x = 0$.
General solution: $x(t) = c_1\cos 2t + c_2\sin 2t$.
$x(0) = c_1 = 2$. $x'(t) = -2c_1\sin 2t + 2c_2\cos 2t$; $x'(0) = 2c_2 = 0 \Rightarrow c_2 = 0$.
$$x(t) = 2\cos 2t.$$
The mass oscillates between $x = 2$ and $x = -2$ with period $\pi$ seconds.
more damping always returns the system to equilibrium faster. Critically damped returns to zero faster than overdamped, even though overdamped has more friction. In the overdamped case, the extra friction also resists the motion toward equilibrium. The “fastest” behavior is achieved at the exact threshold $c^2 = 4mk$, which is why critical damping is the design target in shock absorbers and door closers.
Leveled Practice
Problem 1. A mass of $m = 2$ kg on a spring with $k = 8$ N/m and damping constant $c = 8$ N$\cdot$s/m. Classify the damping and write the general solution for $x(t)$.
Show answer
$\Delta = c^2 - 4mk = 64 - 4(2)(8) = 64 - 64 = 0$. Critically damped.
$r = -c/(2m) = -8/4 = -2$.
$x(t) = (c_1 + c_2 t)e^{-2t}$.