Mechanical Vibrations and Resonance

Washing machines that walk across the floor, bridges that sway under marching soldiers and engine mounts that hum at certain speeds are all examples of mechanical vibration. Engineers model them with surprisingly simple mathematics: a mass, a spring and a damper. Once you understand natural frequency and resonance, you can predict when a machine will shake itself apart and how to stop it.

The Spring-Mass-Damper Model

Picture a mass \( m \) (kg) on a spring of stiffness \( k \) (N/m) with a viscous damper of coefficient \( c \) (N·s/m). Newton's second law for the displacement \( x(t) \) gives

$$ m\ddot{x} + c\dot{x} + kx = F(t) $$

where \( F(t) \) is any external force. This one equation describes car suspensions, building floors, turbine blades and even the cone of a loudspeaker.

Natural frequency

With no damping and no external force, the mass oscillates at its undamped natural frequency:

$$ \omega_n = \sqrt{\frac{k}{m}}\ \ \text{(rad/s)}, \qquad f_n = \frac{\omega_n}{2\pi}\ \ \text{(Hz)} $$

A stiffer spring raises the frequency; a heavier mass lowers it.

Damping ratio

Damping is described by a dimensionless ratio

$$ \zeta = \frac{c}{2\sqrt{km}} $$

• \( \zeta < 1 \): underdamped, the system oscillates with decaying amplitude (most structures, with \( \zeta \) around 0.01 to 0.1).
• \( \zeta = 1 \): critically damped, returns to rest fastest without overshoot (good door closers).
• \( \zeta > 1 \): overdamped, slow return without oscillation.

Forced Vibration and Resonance

Now apply a harmonic force \( F(t) = F_0 \sin \omega t \), for instance from an unbalanced rotor. After transients die out, the mass vibrates at the forcing frequency with amplitude

$$ X = \frac{F_0/k}{\sqrt{(1 - r^2)^2 + (2\zeta r)^2}}, \qquad r = \frac{\omega}{\omega_n} $$

The quantity \( F_0/k \) is the static deflection, and the factor multiplying it is the magnification factor (MF). At \( r = 1 \) the term \( 1 - r^2 \) vanishes and

$$ \text{MF}_{r=1} = \frac{1}{2\zeta} $$

With \( \zeta = 0.02 \), the vibration is 25 times the static deflection. That is resonance: the forcing frequency matches the natural frequency, and each push adds energy in step with the motion.

Transmissibility

What engineers often care about is the force transmitted to the floor. The transmissibility ratio is

$$ TR = \frac{\sqrt{1 + (2\zeta r)^2}}{\sqrt{(1 - r^2)^2 + (2\zeta r)^2}} $$

A key result: \( TR < 1 \) only when \( r > \sqrt{2} \). Isolation mounts work by making the natural frequency well below the running frequency.

Solved Problem 1: Checking a Machine for Resonance

A 200 kg pump sits on four identical springs giving a combined stiffness of 800 kN/m. It runs at 600 rpm. Is it near resonance?

Step 1: Natural frequency.

$$ \omega_n = \sqrt{\frac{800\,000\ \text{N/m}}{200\ \text{kg}}} = \sqrt{4000\ \text{s}^{-2}} = 63.25\ \text{rad/s} $$

$$ f_n = \frac{63.25}{2\pi} = 10.07\ \text{Hz} $$

Step 2: Running frequency.

$$ f = \frac{600\ \text{rev/min}}{60\ \text{s/min}} = 10.0\ \text{Hz} $$

Step 3: Frequency ratio.

$$ r = \frac{10.0}{10.07} = 0.993 $$

Conclusion. The pump runs almost exactly at resonance. With light damping it would vibrate violently.

Sanity check. The static deflection is \( mg/k = (200)(9.81)/800\,000 = 2.45 \) mm. A handy rule is \( f_n \approx 15.76/\sqrt{\delta_{\text{st}}\ \text{(mm)}} \) Hz, giving \( 15.76/\sqrt{2.45} \approx 10.1 \) Hz, which matches.

Solved Problem 2: Redesigning the Speed

The pump's speed is raised to 1200 rpm. The mounts provide \( \zeta = 0.1 \), and the rotor's unbalance produces a force amplitude \( F_0 = 500 \) N. Find the damping coefficient, the vibration amplitude and the force transmitted to the floor.

Step 1: Damping coefficient.

$$ c = 2\zeta\sqrt{km} = 2(0.1)\sqrt{(800\,000)(200)} = 0.2 \times 12\,649 = 2530\ \text{N·s/m} $$

Step 2: Forcing frequency and ratio.

$$ \omega = \frac{1200 \times 2\pi}{60} = 125.66\ \text{rad/s}, \qquad r = \frac{125.66}{63.25} = 1.987 $$

Step 3: Magnification factor.

$$ 1 - r^2 = 1 - 3.948 = -2.948, \qquad 2\zeta r = 0.397 $$

$$ \text{MF} = \frac{1}{\sqrt{(-2.948)^2 + (0.397)^2}} = \frac{1}{\sqrt{8.691 + 0.158}} = \frac{1}{2.975} = 0.336 $$

Step 4: Amplitude.

$$ X = \frac{F_0}{k}\,\text{MF} = \frac{500\ \text{N}}{800\,000\ \text{N/m}}(0.336) = 2.10 \times 10^{-4}\ \text{m} = 0.21\ \text{mm} $$

Step 5: Transmitted force.

$$ TR = \sqrt{1 + 0.158}\ \times 0.336 = 1.076 \times 0.336 = 0.362 $$

$$ F_T = TR \times F_0 = 0.362 \times 500\ \text{N} \approx 181\ \text{N} $$

Sanity check. Because \( r \approx 2 > \sqrt{2} \), the mounts should isolate, and indeed only about 36% of the force reaches the floor. The amplitude of 0.21 mm is small, which is what we expect well above resonance where the mass's inertia resists the motion. Note that during start-up the pump must pass through 600 rpm; damping limits the peak there to \( \text{MF} = 1/(2 \times 0.1) = 5 \).

How Engineers Fight Resonance

1. Shift the natural frequency by changing stiffness or mass, so it is far from operating speeds.
2. Add damping with rubber mounts, viscous dampers or tuned mass dampers.
3. Remove the excitation by balancing rotors and aligning shafts.
4. Pass through quickly: run-up through critical speeds should be fast so amplitudes cannot build.

Did You Know?

• In 1831 soldiers marching across the Broughton Suspension Bridge near Manchester set it vibrating until it collapsed; armies now break step on bridges.
• The 1940 Tacoma Narrows Bridge failure is often blamed on simple resonance, but engineers now explain it as aerodynamic flutter, a self-excited instability.
• Taipei 101 has a tuned mass damper, a steel sphere of roughly 660 tonnes, hanging near its top to reduce wind-induced sway.

Key Takeaways

• Natural frequency \( \omega_n = \sqrt{k/m} \) is the heart of every vibration problem.
• Resonance occurs when forcing frequency equals natural frequency, and peak amplitude is limited only by damping, \( \text{MF} = 1/(2\zeta) \).
• Vibration isolation requires the frequency ratio \( r > \sqrt{2} \).
• Always convert rpm to rad/s or Hz before comparing frequencies.

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