Dynamics · Chapter 10 of 10 · Advanced
Vibration and Time Response
A restoring force plus inertia gives oscillation. Add damping and forcing, and you can predict natural frequency, decay, and the resonance that destroys machines.
Readiness check
This capstone applies the equation of motion to oscillating systems. Tick only what you can do closed-notes.
- Write ΣF = ma for a mass on a spring.
- Recall the spring force F = kx.
- Solve a simple second-order differential equation.
- Work in rad/s and convert to Hz.
- Recognise an equilibrium position.
The core idea
A mass held by a spring oscillates at a natural frequency set by stiffness and mass; damping makes it decay, and forcing near that frequency makes it grow dangerously.
mẍ + cẋ + kx = F(t)ωn = √(k/m)ζ = c/(2√(km))Newton's law applied to a spring-mass system gives a second-order equation whose solution is an oscillation at ωn = √(k/m). Damping introduces the ratio ζ, which decides whether the motion oscillates and how fast it decays. A harmonic force drives a steady response that peaks sharply when its frequency matches ωn, the phenomenon of resonance.
The skills, taught in order
Vibration is the equation of motion solved for oscillation. Five skills run from the simplest free vibration to forcing, isolation, and energy shortcuts.
10.1 Free undamped vibration
With no damping or forcing, mẍ + kx = 0, whose solution is simple harmonic motion at the natural frequency ωn = √(k/m). The period is T = 2π/ωn and the frequency f = ωn/2π. Stiffer raises it; heavier lowers it. Gravity only shifts the equilibrium, not ωn.
10.2 Damped free vibration
Adding a damper gives mẍ + cẋ + kx = 0. The damping ratio ζ = c/(2√(km)) sets the character of the response.
| Damping ratio | Response | Example |
|---|---|---|
| ζ < 1 | underdamped: decaying oscillation | most structures |
| ζ = 1 | critically damped: fastest return, no overshoot | ideal door closer |
| ζ > 1 | overdamped: slow, no oscillation | heavy dashpot |
Underdamped systems oscillate at the damped frequency ωd = ωn√(1 − ζ²), slightly below ωn.
10.3 Forced vibration and resonance
A harmonic force produces a steady oscillation whose amplitude depends on the frequency ratio r = ω/ωn. The magnification factor M = 1/√[(1 − r²)² + (2ζr)²] peaks near r = 1: at resonance, even a small force builds a large response, limited only by damping.
10.4 Vibration isolation
The same curve guides isolation. Below r = √2 the mount amplifies the disturbance; above r = √2 it attenuates it. Good isolation runs the machine well above its mount's natural frequency, which is why engines sit on soft mounts.
| Frequency ratio r | Behaviour |
|---|---|
| r ≪ 1 | response follows the force (rigid) |
| r ≈ 1 | resonance: large amplitude |
| r > √2 | isolation: response attenuated |
10.5 The energy method and rigid-body vibration
For conservative systems the natural frequency follows from energy, by equating peak kinetic and potential energy, without writing the differential equation. The same idea handles rigid-body oscillations (a swinging pendulum, a rocking body) using I in place of m and a torsional or equivalent stiffness.
Engineering connection: machine mounts and isolators, earthquake and structural response, rotating-machinery balance, and vehicle suspension, feeding directly into system dynamics and control.
Worked example 1: a damped spring-mass system
An 8 kg mass hangs on a spring of stiffness 2000 N/m with a damper of coefficient 40 N·s/m. Find the natural frequency, the damping ratio, and the damped frequency, and classify the response.
- ProblemFind ωn, ζ, and ωd for the system in Figure 1, and classify it.
- Given / findm = 8 kg, k = 2000 N/m, c = 40 N·s/m. Find ωn, ζ, ωd, and the regime.
- AssumptionsLinear spring and viscous damper, one degree of freedom, small motion.
- ModelStandard damped oscillator: ωn from k and m, ζ from c relative to the critical value, ωd from both.
- Equationsωn = √(k/m) ζ = c/(2√(km)) ωd = ωn√(1 − ζ²)
- Solveωn = √(2000/8) = √250 = 15.81 rad/s (2.52 Hz). Critical damping cc = 2√(km) = 2√16000 = 253 N·s/m, so ζ = 40/253 = 0.158, underdamped. ωd = 15.81√(1 − 0.158²) = 15.61 rad/s.
- Checkζ well below 1 confirms a lightly damped, oscillating response, consistent with ωd only 1.3% below ωn. Light damping always leaves the damped frequency close to the natural one.
- ConclusionThe system rings at about 2.5 Hz and decays slowly. To kill oscillation quickly you would raise c toward the critical 253 N·s/m, the design target for a door closer or a suspension at rest.
Worked example 2: resonance and isolation
A 50 kg machine sits on mounts of total stiffness 200 kN/m with damping ratio 0.05. Find the speed at which it would resonate, and the magnification factor when it runs at 1800 rev/min.
- ProblemFind the resonant speed and the magnification factor at 1800 rev/min for the machine in Figure 2.
- Given / findm = 50 kg, k = 200 000 N/m, ζ = 0.05, operating speed 1800 rev/min. Find the resonant speed and M.
- AssumptionsSingle degree of freedom, harmonic forcing from rotating unbalance, linear mounts.
- ModelFind ωn; resonance occurs at r = 1; evaluate the magnification factor at the operating frequency ratio.
- Equationsωn = √(k/m) r = ω/ωn M = 1/√[(1 − r²)² + (2ζr)²]
- Solveωn = √(200 000/50) = √4000 = 63.2 rad/s, which is 604 rev/min, the resonant speed. At 1800 rev/min, ω = 188.5 rad/s, so r = 2.98. Then M = 1/√[(1 − 8.88)² + (0.30)²] = 1/7.89 = 0.13.
- CheckBecause r = 2.98 is well above √2, the machine is firmly in the isolation region, and M = 0.13 means only 13% of the static disturbance is transmitted. The danger is passing through 604 rev/min during start-up and shut-down.
- ConclusionSoft mounts give a low ωn, putting the running speed above resonance for good isolation. Engineers add damping mainly to survive the resonant crossing at start-up, not for the running condition.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Gravity changes ωn | Natural frequency made to depend on weight | "Does gravity shift equilibrium or stiffness?" | Gravity only shifts the equilibrium; ωn = √(k/m) is unchanged. |
| Confusing ωn and ωd | Damped and undamped frequencies swapped | "Is there damping?" | ωd = ωn√(1 − ζ²), slightly below ωn. |
| Running near resonance | Large vibration at r ≈ 1 | "How close is r to 1?" | Keep the operating frequency away from ωn, or add damping. |
| Stiffer mounts for isolation | Hard mounts raise ωn and worsen isolation | "Is r above or below √2?" | Isolation needs r > √2, so softer mounts (lower ωn) usually help. |
Practice ladder
A 4 kg mass on a 900 N/m spring vibrates freely. Find its natural frequency in Hz.
Show answer
ωn = √(900/4) = 15 rad/s; fn = 15/2π = 2.39 Hz. Stiffness up or mass down would raise it.
For the Worked Example 1 system, what damping coefficient c would make it critically damped?
Show answer
Critical damping is cc = 2√(km) = 2√(2000 × 8) = 253 N·s/m. The present 40 N·s/m is only 16% of this, which is why it oscillates rather than returning smoothly.
A simple pendulum of length 0.8 m swings with small amplitude. Find its natural frequency, using the energy or equation-of-motion result ωn = √(g/L).
Show answer
ωn = √(9.81/0.8) = 3.50 rad/s, so fn = 0.56 Hz and T = 1.79 s. The mass cancels, which is why pendulum period depends only on length, the basis of pendulum clocks.
Pick a real vibrating system (a washing machine, a footbridge, a phone on vibrate, an engine mount). Estimate its effective mass and stiffness, find ωn, and judge whether it operates near resonance or in isolation.
What good work looks like
An effective k and m identified, ωn computed, the operating frequency ratio found, and a clear statement of resonance risk or isolation with a damping comment.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Analyse one real vibrating system: find its natural frequency, its damping character, and whether its operating point is safe, ending with a design recommendation.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. Give the natural frequency of a spring-mass system.
ωn = √(k/m); period T = 2π/ωn.
2. Define the damping ratio and the three regimes.
ζ = c/(2√(km)); ζ < 1 underdamped, ζ = 1 critical, ζ > 1 overdamped.
3. Where does the forced response peak?
Near the frequency ratio r = 1, at resonance.
4. What frequency ratio is needed for isolation?
r > √2; beyond it the transmitted response is attenuated.
5. Does gravity change the natural frequency?
No; it only shifts the equilibrium position.
Textbook mapping
| Item | Mapping |
|---|---|
| Primary source | Meriam and Kraige, Engineering Mechanics: Dynamics (7th ed), Chapter 8 (Vibration and Time Response) |
| Cross-reference | Rao, Mechanical Vibrations · Hibbeler, Dynamics, Ch. 22 |
| Core topics | 10.1 Free undamped · 10.2 Damped free · 10.3 Forced and resonance · 10.4 Isolation · 10.5 Energy method |
| Engineering connection | Machine mounts, structural and seismic response, rotating-machinery balance, and suspension. |
| Read next | You have completed the Dynamics course. Return to the course hub or continue to System Dynamics and Control. |