Physics for ME · Module 10 of 16
Oscillations, Mechanical Waves, and Resonance
Anything with mass and stiffness has a favorite frequency. Drive it there and amplitudes grow: the most important warning in mechanical engineering.
Readiness check
From Modules 4 and 6, Math Modules 2 and 10. Tick only what you can do closed-notes.
- Use the spring model F = ks with units.
- Sketch A sin(ωt + φ) and name its anatomy (Math Module 2).
- Trade kinetic and potential energy in an audit.
- Recall ω = √(k/m) from the math course's oscillator.
- Convert between f, ω, and T.
The core idea
Restoring force plus inertia equals oscillation, at one natural frequency.
ωn = √(k/m)T = 2π√(m/k)Simple harmonic motion is the spring law inside Newton's second: x(t) = A sin(ωnt + φ). Damping drains the swing; periodic driving feeds it. When the drive frequency meets ωn, energy arrives in step every cycle: resonance.
The skills, taught in order
Anything with mass and stiffness oscillates at a natural frequency, and driving it there is the central hazard of machine design. These seven skills take you from one spring to the resonance audit and the modes of a continuous part.
10.1 Simple harmonic motion
When the restoring force is proportional to displacement, F = −kx, the motion is sinusoidal: x(t) = A cos(ωnt + φ) with ωn = √(k/m). Crucially the natural frequency does not depend on amplitude, which is why a pendulum clock keeps time as it winds down.
Micro-example. A mass that stretches its spring by a static amount s has ωn = √(g/s): the sag alone fixes the tune.
10.2 Energy in oscillation
An oscillator's energy is E = ½kA², trading between kinetic and potential twice per cycle. The speed peaks at the centre (all kinetic) and is zero at the turning points (all potential).
10.3 Pendulums
For small swings a pendulum is simple harmonic with T = 2π√(L/g), independent of mass and (to first order) of amplitude. Only its length and gravity set its period.
10.4 Damping
Real oscillations lose energy. Light (under-)damping lets the system ring down slowly; critical damping returns it fastest without overshoot; over-damping creeps back. For realistic damping the frequency shifts only slightly; damping mainly bleeds away amplitude.
10.5 Forced vibration and resonance
Drive an oscillator at a frequency near ωn and each push arrives in phase, so a small input builds a large response: resonance. Only damping caps the peak. Keeping excitation frequencies away from every ωn is the designer's job.
Micro-example. Small pushes timed to a child's swing build a large arc, while the same pushes at the wrong rate do nothing: resonance in a playground.
10.6 Mechanical waves
A wave carries a disturbance through a medium at speed v = fλ, while the medium itself only oscillates in place. On a stretched string or cable the speed is set by tension and mass per length, v = √(F/μ).
10.7 Standing waves and natural modes
Confine a wave between fixed ends and only certain wavelengths fit, giving standing waves at frequencies fn = n·v/(2L). A continuous part therefore has not one natural frequency but a whole series. Worked Example 2 finds them for a cable.
| System | Natural frequency or period |
|---|---|
| Spring-mass | ωn = √(k/m), T = 2π√(m/k) |
| Simple pendulum (small angle) | T = 2π√(L/g) |
| String, mode n (fixed ends) | fn = n·v/(2L), with v = √(F/μ) |
Engineering connection: direct preparation for Mechanical Vibrations (bridge reference Rao); machine mounts, rotor whirl, and acoustics. Pairs with Math Module 10 (the oscillator ODE) and Module 13 (spectra).
Worked example 1: the suspension's favorite speed
A quarter-car model carries m = 300 kg on a spring of k = 30 000 N/m. Find the natural frequency. The road has expansion joints every 12 m: at what speed does the car resonate?
- ProblemFind fn and the resonant road speed in Figure 1.
- Given / findm = 300 kg, k = 30 000 N/m, joint spacing λ = 12 m.
- AssumptionsQuarter-car single degree of freedom; light damping; one bump impulse per joint.
- Modelωn = √(k/m); the road forces the system at frequency v/λ; resonance when they match.
- Equationsωn = √(k/m) fn = ωn/2π v = fn·λ
- Solveωn = √100 = 10 rad/s, so fn = 10/2π = 1.59 Hz (T = 0.63 s). Resonant speed v = 1.59 × 12 = 19.1 m/s ≈ 69 km/h.
- CheckReal cars ride at 1 to 2 Hz: the 1.59 Hz result is in the right band. Faster or slower than 69 km/h moves the excitation off the peak: matching everyday experience of a "bad speed" on jointed roads.
- ConclusionThe shock absorber exists to flatten exactly this peak (the damping of Math Module 10). The same match-the-frequency audit governs machine mounts, pipeline supports, and rotor critical speeds.
Worked example 2: a wave on a tensioned cable
A cable of length L = 0.65 m and mass per length μ = 0.004 kg/m is stretched to a tension of 80 N and fixed at both ends. Find the speed of transverse waves along it and its fundamental (lowest) standing-wave frequency.
- ProblemFor the tensioned cable in Figure 2, find the transverse wave speed and the fundamental frequency.
- Given / findL = 0.65 m, μ = 0.004 kg/m, F = 80 N, fixed at both ends. Find v and f₁.
- AssumptionsUniform flexible cable, constant tension, small transverse amplitude, both ends held fixed (nodes).
- ModelTransverse waves travel at v = √(F/μ). A fixed-fixed span resonates when a whole number of half-wavelengths fits; the fundamental is one half-wave, λ₁ = 2L.
- Equationsv = √(F/μ) λ₁ = 2L f₁ = v/λ₁
- Solvev = √(80/0.004) = √20000 = 141 m/s. λ₁ = 2 × 0.65 = 1.30 m, so f₁ = 141.4/1.30 = 109 Hz. Higher modes follow as fn = n·f₁: 218 Hz, 327 Hz, and so on.
- CheckUnits: √(N ÷ (kg/m)) = √((kg·m/s²)(m/kg)) = √(m²/s²) = m/s. Tightening the cable raises both v and the pitch (v ∝ √F), exactly how a guitar string is tuned.
- ConclusionA continuous part has not one natural frequency but a whole series, its standing-wave modes. The same v = √(F/μ) and fn = n·v/(2L) govern belt and cable vibration, and the resonance warning of Example 1 now applies at every one of them.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Amplitude in the frequency formula | "Bigger swing, slower swing" claimed for SHM | "Does A appear in ω = √(k/m)?" | It does not: small-amplitude SHM keeps time regardless of amplitude. That is why pendulum clocks work. |
| Resonance read as a force amplifier | "The road hits harder at 69 km/h" | "What grows: the input, or the response?" | The input is unchanged; arriving in phase each cycle lets small inputs accumulate a large response. |
| Damping expected to shift ωn strongly | Retuning claimed from a fitted damper | "How far does ωd = ωn√(1−ζ²) move for small ζ?" | Barely, for realistic ζ. Damping mainly caps the peak; mass and stiffness set the tune. |
| Wave speed confused with particle speed | v = fλ applied to the bobbing material | "Is this the pattern's speed or the medium's?" | v = fλ is the pattern. The medium oscillates locally and goes nowhere. |
Practice ladder
A 4 kg mass hangs on a spring that stretches 50 mm under its weight. Find k, ωn, and fn.
Show answer
k = mg/s = 39.24/0.05 = 785 N/m. ωn = √(785/4) = 14.0 rad/s; fn = 2.23 Hz. The static deflection alone fixed the tune: ωn = √(g/s).
A pendulum clock runs slow by 2 minutes per day. Should its pendulum be lengthened or shortened, and by what fraction?
Show answer
Slow means T too long, so shorten. T ∝ √L: the period error is 2/1440 = 0.14%, so L must shrink by about 0.28% (twice the period fraction). Tiny adjustments, big timekeeping: the √L sensitivity from Math Module 4.
A 50 kg machine on its mounts has ωn = 25 rad/s. Its rotor runs at 1450 rpm. Is the operating point safely above resonance, and what happens during every start-up?
Show answer
Rotor frequency = 1450 × 2π/60 = 151.8 rad/s: about 6 times ωn, comfortably supercritical. But every start and stop sweeps the speed through 25 rad/s (239 rpm), so the machine crosses resonance twice per cycle: dampers and quick run-up exist for that moment.
Measure a real natural frequency: pluck a ruler clamped to a desk, time ten oscillations at two clamp lengths, and test the stiffness scaling (shorter overhang, higher frequency). Optionally verify with the phone spectrum app from Math Module 13.
What good work looks like
Two measured frequencies with timing method shown, the qualitative L-scaling confirmed, and one sentence connecting the experiment to turbine-blade or PCB vibration testing.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Write a one-page resonance audit of a real system (washing machine, fan on a shelf, footbridge video): the estimated ωn, the excitation source and frequency, the margin between them, and one fix (stiffen, soften, damp, or change speed).
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. Write ωn and T for a spring-mass system and a pendulum.
Spring-mass: ωn = √(k/m), T = 2π√(m/k). Pendulum: T = 2π√(L/g), small angles.
2. What does damping change, and what does it barely change?
It caps the resonance peak and decays free vibration; it barely shifts the frequency for small ζ.
3. State the resonance condition and why it is dangerous.
Drive frequency ≈ ωn: each cycle's energy arrives in phase, so amplitude builds until damping or failure stops it.
4. Relate wave speed, frequency, and wavelength.
v = fλ: the pattern's speed, set by the medium; f set by the source.
5. Why do continuous parts have many natural frequencies?
Each standing-wave pattern (mode) that fits the geometry has its own ωn: the mode shapes of Math Module 9, made physical.