Skip to content

Applications - Spring-Mass Systems

You will learn how to model a spring-mass system with damping and external forces, derive the governing differential equation, and interpret the different types of motion (underdamped, critically damped, overdamped).

Consider a mass mm attached to a spring with spring constant kk, with damping coefficient cc, and an optional external force F(t)F(t).

Newton’s second law gives:

md2xdt2=kxcdxdt+F(t)m \frac{d^2x}{dt^2} = -kx - c \frac{dx}{dt} + F(t)

Rearranged into standard form:

mx+cx+kx=F(t)mx'' + cx' + kx = F(t)

The behavior depends on the discriminant of the characteristic equation: c24mkc^2 - 4mk.

  • c24mk<0c^2 - 4mk \lt 0: Underdamped (oscillates while decaying)
  • c24mk=0c^2 - 4mk = 0: Critically damped (fastest non-oscillatory return)
  • c24mk>0c^2 - 4mk \gt 0: Overdamped (slow, non-oscillatory return)

The interactive 3D animation shows a spring-mass system with damped oscillation. Drag to rotate, scroll to zoom. The mass bounces on the spring and gradually settles to equilibrium.

Example 1: Underdamped Motion

A 2 kg mass is attached to a spring with k=50k = 50 N/m and damping c=4c = 4 Ns/m. Find the position function if released from rest at x=0.3x = 0.3 m.

Solution: Equation: 2x+4x+50x=02x'' + 4x' + 50x = 0, or x+2x+25x=0x'' + 2x' + 25x = 0

Characteristic equation: r2+2r+25=0    r=1±24ir^2 + 2r + 25 = 0 \implies r = -1 \pm \sqrt{24}i

Since c24mk=16400<0c^2 - 4mk = 16 - 400 \lt 0, this is underdamped.

x(t)=et(0.3cos(24t)+0.324sin(24t))x(t) = e^{-t}(0.3\cos(\sqrt{24}\,t) + \frac{0.3}{\sqrt{24}}\sin(\sqrt{24}\,t))

Example 2: Critically Damped

With m=1m = 1, k=9k = 9, find the damping cc for critical damping.

Solution: c2=4mk=36    c=6c^2 = 4mk = 36 \implies c = 6

The equation x+6x+9x=0x'' + 6x' + 9x = 0 has repeated root r=3r = -3.

Solution: x(t)=(C1+C2t)e3tx(t) = (C_1 + C_2 t)e^{-3t}

Example 3: With External Force (Resonance)

Solve x+4x=2cos(2t)x'' + 4x = 2\cos(2t) (forcing frequency matches natural frequency).

Solution: Since ω=2\omega = 2 matches the natural frequency, we get resonance. The particular solution grows as yp=12tsin(2t)y_p = \frac{1}{2}t\sin(2t), showing amplitude increasing linearly with time.

Spring-mass models are everywhere: car suspension systems, building sway during earthquakes, audio speakers, clocks, bungee jumping, and even biological systems like heartbeats or limb movement. Engineers use these equations to tune vehicle ride comfort, design earthquake-resistant buildings, and create stable control systems.

The standard form of the undamped spring-mass equation is:
When $c^2 - 4mk \lt 0$, the system is:
Critical damping occurs when:
Spring-mass systems are used to model:
Adding damping to a spring-mass system: