5.6 Energy, damping and resonance
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A2 content for Unit 5 (WPH15), specification points 148–153, including Core Practical 16. The defining equation of SHM, \(a=-\omega^2x\), and the period formulae are on the previous page, 5.5 Simple harmonic motion.
What this page covers
7 points
- The exchange of kinetic and potential energy in an undamped oscillator, and why total energy ∝ amplitude²
- Energy–displacement and energy–time graphs, and how to sketch them
- Damping: what it does to amplitude and energy; light, critical and heavy damping
- Free and forced oscillations, natural frequency and driving frequency
- Resonance, and how damping changes the resonance curve
- How damping and the plastic deformation of ductile materials protect buildings in earthquake zones and smooth a car's ride
- Core Practical 16: finding an unknown mass from the resonant frequencies of a mass–spring system
In one minute
6 points
- In an undamped oscillator, energy passes back and forth between kinetic and potential energy while the total stays constant, and that total is proportional to the amplitude squared.
- For a mass on a spring, \(E=\tfrac12kA^2\): halve the amplitude and only a quarter of the energy is left.
- Damping is a resistive force that transfers energy out of the oscillator, usually to thermal energy, so the amplitude decreases. Critical damping returns the system to equilibrium in the shortest time without oscillating.
- A free oscillation has no external driving force and happens at the natural frequency \(f_0\); a forced oscillation is driven by a periodic force and happens at the driving frequency.
- Resonance: when the driving frequency equals the natural frequency, energy is transferred most efficiently and the amplitude is a maximum.
- More damping makes the resonance peak lower, broader and at a slightly lower frequency.
Contents
12 sections
- How is energy shared in an undamped oscillation?
- What does damping do to an oscillation?
- What is the difference between free and forced oscillations?
- What is resonance, and how does damping change it?
- How do damping and plastic deformation protect buildings and vehicles?
- Core Practical 16: how do you find an unknown mass from resonant frequencies?
- Exam tips for 5.6
- Common misconceptions
- How this topic connects
- Key equations
- Check your understanding
- Specification checklist
How is energy shared in an undamped oscillation?
Before letting real oscillators lose energy, follow the energy in an ideal one. Every oscillator has two energy stores. One is the kinetic energy of the moving mass. The other is potential energy: elastic strain energy in a spring, or gravitational potential energy for a pendulum.
- At the extremes (\(x=\pm A\)) the mass is momentarily at rest, so its kinetic energy is zero and its potential energy is a maximum.
- At the equilibrium position (\(x=0\)) the speed is a maximum, so the kinetic energy is a maximum and the potential energy is zero (taking the equilibrium position as the zero of potential energy).
- With no damping, no energy leaves the system, so kinetic + potential energy is the same at every instant.
For a mass on a spring the potential energy is the elastic strain energy, the area under the force–extension graph. With \(F=kx\), that area is a triangle, \(\tfrac12Fx=\tfrac12kx^2\). At the extreme all the energy is stored this way, which gives the total energy:
\[E_p=\tfrac12kx^2\qquad E_{\text{total}}=\tfrac12kA^2\qquad E_k=\tfrac12k\left(A^2-x^2\right)\]Here \(k\) is the spring constant (N m⁻¹), \(x\) the displacement from equilibrium (m) and \(A\) the amplitude (m). These three are not in the formulae list: derive them from the elastic strain energy \(\Delta E_{el}=\tfrac12F\Delta x\) with \(F=kx\); the total is the strain energy at the extreme, \(x=A\).
Energy against time
Kinetic energy reaches a maximum twice in every cycle, once each way through the centre. Potential energy also peaks twice, once at each extreme. So graphs of \(E_k\) and \(E_p\) against time oscillate at twice the frequency of the displacement. They are never negative, they are out of step (one is a maximum when the other is zero), and their sum is a horizontal line: the total energy.
A trolley of mass 0.40 kg is attached to a spring of spring constant 25 N m⁻¹ and oscillates horizontally with negligible friction. The amplitude is 6.0 cm. Calculate (a) the total energy of the oscillation, (b) the maximum speed of the trolley and (c) its potential energy, kinetic energy and speed when its displacement is 3.0 cm.
(a) The total energy equals the strain energy at the extreme: \(E=\tfrac12kA^2=\tfrac12\times25\times0.060^2=\) 0.045 J
(b) At the centre all of it is kinetic: \(\tfrac12mv_{\max}^2=0.045\), so \(v_{\max}=\sqrt{2\times0.045\div0.40}=\) 0.47 m s⁻¹. Check with \(v_{\max}=\omega A\): \(\omega=\sqrt{k/m}=7.91\ \text{rad s}^{-1}\), and \(7.91\times0.060=0.47\ \text{m s}^{-1}\).
(c) \(E_p=\tfrac12\times25\times0.030^2=0.01125\ \text{J}\approx\) 0.011 J; \(E_k=0.045-0.01125=0.03375\ \text{J}\approx\) 0.034 J; \(v=\sqrt{2\times0.03375\div0.40}=\) 0.41 m s⁻¹.
Halfway out, only a quarter of the energy is potential, so three-quarters is still kinetic.
Potential energy depends on \(x^2\), so at \(x=A/2\) it is only ¼ of the total. Kinetic and potential energy are equal at \(x=A/\sqrt2\approx0.71A\).
What does damping do to an oscillation?
- Typical damping forces are air resistance, friction, drag in a liquid, and the internal friction of a material as it is repeatedly deformed.
- The energy is not destroyed. It is transferred, usually to thermal energy of the oscillator and its surroundings, sometimes to sound.
- Because \(E\propto A^2\), a fall in energy shows up as a fall in amplitude: new amplitude = √(energy fraction) × original amplitude.
- With light damping the period is almost unchanged; only the amplitude shrinks. For many systems the amplitude falls by the same fraction each cycle, which is an exponential decay.
Light, critical and heavy damping
| Degree of damping | What happens after the system is displaced and released |
|---|---|
| Light (under-damped) | oscillates with gradually decreasing amplitude, at almost the natural frequency |
| Critical | returns to equilibrium in the shortest possible time without oscillating |
| Heavy (over-damped) | does not oscillate, but creeps back to equilibrium more slowly than with critical damping |
A lightly damped oscillator has a period of 1.5 s. Its amplitude falls from 8.0 cm to 6.0 cm over 10 complete oscillations. When the amplitude was 8.0 cm the total energy was 0.064 J. Calculate (a) the energy when the amplitude is 6.0 cm, (b) the mean rate at which energy is transferred out of the oscillator and (c) the fraction of its energy lost in each cycle, assuming the amplitude falls by the same fraction every cycle.
(a) Energy ∝ amplitude²: \(E=0.064\times(6.0\div8.0)^2=0.064\times0.5625=\) 0.036 J
(b) \(\Delta E=0.064-0.036=0.028\ \text{J}\) in \(10\times1.5=15\ \text{s}\), so \(P=0.028\div15=\) 1.9 × 10⁻³ W (1.87 mW).
(c) The amplitude ratio per cycle is the tenth root of the overall ratio: \((6.0/8.0)^{1/10}=0.972\). The energy ratio per cycle is \(0.972^2=0.944\), so 5.6% of the energy is lost each cycle.
The amplitude has fallen by 25% but the energy by 44%. Always square the amplitude ratio before talking about energy.
What is the difference between free and forced oscillations?
- Natural frequency \(f_0\): the frequency at which a system oscillates when displaced and released. For a mass on a spring \(f_0=1/T\) with \(T=2\pi\sqrt{m/k}\); for a simple pendulum \(T=2\pi\sqrt{l/g}\).
- Driving frequency: the frequency of the external periodic force. Once the motion has settled (the steady state), a forced oscillator moves at the driving frequency, whatever its natural frequency.
- Examples of forced oscillation: a loudspeaker cone driven by the current in its coil, a vibration generator shaking a string, an engine shaking a car body, wind gusts on a tall chimney.
- Energy: a free oscillation that is damped dies away. A forced oscillation reaches a steady amplitude when the energy supplied by the driver each cycle equals the energy removed by damping each cycle.
The period formulae \(T=2\pi\sqrt{m/k}\), \(T=2\pi\sqrt{l/g}\), \(f=1/T\) and \(\omega=2\pi f\) are in the formulae list supplied in the exam (you met them on the simple harmonic motion page). Here \(m\) is the oscillating mass (kg), \(k\) the spring constant (N m⁻¹) and \(l\) the length of the pendulum (m).
A 0.50 kg mass hangs from a spring of spring constant 50 N m⁻¹. A vibration generator drives it at its natural frequency, and it oscillates with a steady amplitude of 4.0 cm. The generator is switched off, and the amplitude falls to 2.0 cm in 3.0 s. (a) Calculate the natural frequency. (b) Calculate the energy of the oscillation while it was driven. (c) Calculate the mean rate of energy loss in the 3.0 s after switching off, and explain what the generator was doing while the amplitude was steady.
(a) From the formulae list, \(f_0=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}=\dfrac{1}{2\pi}\sqrt{\dfrac{50}{0.50}}=\) 1.59 Hz
(b) \(E=\tfrac12\times50\times0.040^2=\) 0.040 J
(c) At 2.0 cm, \(E=\tfrac12\times50\times0.020^2=0.010\ \text{J}\), so \(P=(0.040-0.010)\div3.0=\) 0.010 W.
While the amplitude was steady, the generator was supplying energy at the same rate as damping removed it. Damping losses grow with amplitude, so at 4.0 cm that rate was greater than the 0.010 W average found here.
What is resonance, and how does damping change it?
Drive an oscillator at a range of frequencies, wait for the steady state at each one, and record the amplitude.
- At resonance the driving force is in step with the motion, so it does positive work on the oscillator throughout every cycle: energy is transferred from the driver to the oscillator most efficiently.
- The amplitude builds up until the energy lost to damping per cycle equals the energy supplied per cycle. With very little damping that takes a very large amplitude.
- Driven well below \(f_0\), the system just follows the driver with a small amplitude. Driven well above \(f_0\), it cannot keep up and the amplitude becomes very small.
How damping changes the resonance curve
- Lower peak: damping removes more energy per cycle, so the balance with the energy supplied is reached at a smaller amplitude.
- Broader peak: the system responds over a wider range of driving frequencies (it is less "selective").
- Peak at a slightly lower frequency than the natural frequency; with light damping the shift is negligible.
- Little change far from \(f_0\): all the curves start at the same low-frequency amplitude.
| Resonance that is useful | Resonance that is a problem |
|---|---|
| pushing a child on a swing once per swing; tuning a radio circuit to one station; musical instruments (air columns and strings); the quartz crystal in a clock; a microwave oven heating water molecules | a suspension bridge or footbridge driven by wind or footsteps; machine parts at certain engine speeds; a washing machine shaking during spin-up; buildings shaken by an earthquake at their natural frequency |
The body of a car has mass 1400 kg. Its suspension springs together act like a single spring of spring constant 2.0 × 10⁵ N m⁻¹. The car drives along a road with a series of identical humps 6.0 m apart. (a) Calculate the natural frequency of vertical oscillation of the car body. (b) Calculate the speed at which the humps would make the body resonate. (c) Explain why the bouncing is small at 5.0 m s⁻¹.
(a) \(f_0=\dfrac{1}{2\pi}\sqrt{\dfrac{2.0\times10^5}{1400}}=1.902\ \text{Hz}\approx\) 1.9 Hz
(b) The humps drive the body at \(f=\text{speed}\div\text{spacing}\); resonance happens when this equals \(f_0\): \(v=f_0\times\text{spacing}=1.902\times6.0=11.4\ \text{m s}^{-1}\approx\) 11 m s⁻¹ (about 41 km h⁻¹).
(c) At 5.0 m s⁻¹ the driving frequency is \(5.0\div6.0=0.83\ \text{Hz}\), well below 1.9 Hz, so the body is far from resonance and its amplitude is small.
Shock absorbers damp the suspension strongly (close to critical damping), so even at 11 m s⁻¹ the resonance peak is low and broad.
Write "the driving frequency equals the natural frequency", followed by "maximum amplitude" and "maximum (most efficient) transfer of energy". "The frequencies match" on its own is too vague.
How do damping and plastic deformation protect buildings and vehicles?
Engineers have two weapons against unwanted resonance: keep the natural frequency away from the driving frequency, and take energy out of the oscillation quickly.
Car suspension
- Springs let the wheels follow bumps without passing every jolt to the passengers. On their own, though, they would keep the car bouncing long after each bump, and resonate at certain speeds.
- A shock absorber (damper) beside each spring forces oil through small holes as the suspension moves. The viscous drag transfers the oscillation energy to thermal energy of the oil.
- The damping is set close to critical: after a bump the body returns to its normal height quickly with at most one small overshoot, and the resonance peak is low and broad.
- A worn shock absorber gives light damping: the car keeps bouncing after a bump and the tyres can lose grip.
Buildings in earthquake zones
- Earthquake ground motion contains a range of frequencies. A building whose natural frequency lies in that range is a forced oscillator near resonance, and its amplitude can grow until it is damaged.
- Avoid resonance: engineers design so that the building's natural frequencies are away from the strongest earthquake frequencies for that site. Flexible base isolators (layered rubber and steel bearings) under the building lower its natural frequency well below them.
- Add damping: fluid dampers in the frame, and large tuned masses near the top, absorb energy from the sway and transfer it to thermal energy, keeping the resonance peak low.
- Use plastic deformation of ductile materials: steel braces and dampers, and the lead core of some base isolators, are designed to yield. Each time they are stretched and compressed beyond the elastic limit, work is done deforming the metal permanently, and that energy is dissipated as thermal energy, not returned to the oscillation.
- The material must be ductile, able to deform plastically many times without fracturing. A brittle material would simply crack and absorb very little energy.
Why does a loop mean energy is lost? On loading, the work done on the damper is the area under the loading line. On unloading, the damper does work back, equal to the area under the unloading line. Because the unloading line sits lower, less work comes back than went in. The difference, the area enclosed by the loop, is the energy dissipated in each cycle.
During an earthquake the upper part of a building, of effective mass 4.0 × 10⁵ kg, sways with a natural frequency of 0.50 Hz and an amplitude of 0.10 m. When the ground motion stops, the building oscillates freely. In the first cycle of the free oscillation, each of two ductile steel dampers goes once round the loop in the figure above. Ignore all other energy losses. (a) Calculate the effective spring constant of the building. (b) Calculate the energy of the oscillation at 0.10 m amplitude. (c) Calculate the amplitude after the first cycle.
(a) Rearrange \(T=2\pi\sqrt{m/k}\) with \(T=1/f\): \(k=4\pi^2mf^2=4\pi^2\times4.0\times10^5\times0.50^2=3.95\times10^6\ \text{N m}^{-1}\approx\) 3.9 × 10⁶ N m⁻¹
(b) \(E=\tfrac12kA^2=\tfrac12\times3.948\times10^6\times0.10^2=19\,739\ \text{J}\approx\) 2.0 × 10⁴ J
(c) Energy dissipated = 2 × loop area = 2 × 1600 = 3200 J, so \(E=19\,739-3200=16\,539\ \text{J}\). Amplitude ∝ √energy: \(A=0.10\times\sqrt{16\,539\div19\,739}=0.0915\ \text{m}\approx\) 0.092 m
About 16% of the energy goes in one cycle, but the amplitude falls by less than 10%: the square-root effect of \(E\propto A^2\) again.
Saying that damping "reduces the natural frequency so there is no resonance" is wrong. Damping mainly lowers and broadens the resonance peak; the shift in frequency is small. Moving the natural frequency away from the driving frequency is a separate strategy (for example, base isolation).
Core Practical 16: how do you find an unknown mass from resonant frequencies?
This practical turns resonance into a weighing machine. For a mass on a spring, \(T=2\pi\sqrt{m/k}\), so squaring gives
\[T^2=\frac{4\pi^2}{k}\,m\]A graph of \(T^2\) against \(m\) for known masses is therefore a straight line of gradient \(4\pi^2/k\): a calibration line. Measure the resonant frequency with the unknown mass and read its mass off the line. The relation is not in the formulae list: derive it by squaring \(T=2\pi\sqrt{m/k}\).
- Apparatus: a spring hung from the vibrating rod of a vibration generator, driven by a signal generator; a mass hanger with slotted masses; a metre rule and pointer to judge the amplitude.
- Method: with a known mass on the spring, slowly increase the signal-generator frequency from well below resonance. Find the frequency at which the amplitude of the mass is greatest; approach it from above as well, and take the mean as the resonant frequency \(f\). Repeat for at least six masses, then for the unknown mass.
- Processing: \(T=1/f\); plot \(T^2\) against \(m\); draw the line of best fit.
- Check: displace each mass and time 20 free oscillations with a stopwatch. The natural frequency found this way should agree with the resonant frequency.
Practical issues
- The peak is hard to locate when it is broad. Keep damping low (no paddle or card on the mass), change the frequency in small steps near the peak and wait for the amplitude to settle at each setting.
- The spring has mass. Part of the spring moves with the load, so the line does not pass through the origin: it has a small positive intercept on the \(T^2\) axis. This does not affect the gradient or a value read off the line.
- Keep the amplitude small so the spring obeys Hooke's law and the coils never close up; stop the mass swinging sideways.
- Uncertainty: if \(f=1.45\ \text{Hz}\) is found to ±0.02 Hz, the percentage uncertainty in \(f\) is 1.4%, and in \(T^2\,(=1/f^2)\) it is 2 × 1.4% = 2.8%.
- Safety: clamp the stand firmly (counterweight on the base) and keep feet clear of slotted masses that could fall.
Use the specimen graph above to determine (a) the spring constant and (b) the unknown mass, whose resonant frequency was 1.45 Hz. (c) Explain the intercept on the \(T^2\) axis.
(a) Gradient from a large triangle on the line, from \(m=0.05\ \text{kg}\) to 0.45 kg: gradient \(=0.787\div0.40=1.97\ \text{s}^2\,\text{kg}^{-1}\). Gradient \(=4\pi^2/k\), so \(k=4\pi^2\div1.97=\) 20 N m⁻¹
(b) \(T^2=1\div1.45^2=0.476\ \text{s}^2\), so (reading across and down, or using the line's equation) \(m=(0.476-0.026)\div1.97=\) 0.23 kg
(c) The intercept, 0.026 s², is \(T^2\) for "zero load": the spring's own moving mass, equivalent to an extra \(0.026\div1.97\approx0.013\ \text{kg}\) on the end.
Reading the unknown mass from the calibration line, not from the formula with \(k\), automatically allows for the mass of the spring.
\(T^2=(4\pi^2/k)\,m\) has the form \(y=mx+c\), so the graph is a straight line whose gradient gives \(k\). A curve of \(T\) against \(m\) is much harder to read accurately.
Exam tips for 5.6
Energy ∝ amplitude². If the amplitude falls to 60%, the energy falls to 36%; if the energy halves, the amplitude falls to 71%. Write the ratio, then square it (or square-root it) as a separate step.
When you sketch kinetic or potential energy against time, draw two peaks in every period of the displacement, never go below zero, and draw a flat total-energy line. Against displacement, draw parabolas meeting the total at ±A.
For "how does increasing the damping affect the resonance curve?": the peak is lower, broader, and at a slightly lower frequency. Add that the amplitude far from resonance is barely changed.
"Damping removes energy" is incomplete. Say it is transferred from the oscillator (kinetic and potential energy) to thermal energy of the surroundings or the damper, or, for plastic deformation, dissipated as thermal energy in the deformed metal.
Critical damping: the system returns to equilibrium in the shortest time without oscillating. Both halves are needed; heavy damping also avoids oscillation but takes longer.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| Doubling the amplitude doubles the energy. | Potential energy at the extreme is \(\tfrac12kA^2\), which depends on \(A^2\). | Doubling the amplitude gives four times the energy. |
| At half the amplitude, kinetic and potential energy are equal. | \(E_p\propto x^2\), so at \(A/2\) it is only ¼ of the total. | They are equal at \(x=A/\sqrt2\approx0.71A\). |
| Damping slows the oscillation down, so the period gets longer as it dies away. | With light damping the period is almost exactly the natural period. | Light damping reduces the amplitude, not the frequency. |
| Heavy damping stops the system fastest. | Very large damping makes the return to equilibrium slow. | Critical damping gives the quickest return without oscillation. |
| A forced oscillator vibrates at its natural frequency. | In the steady state it follows the driver. | It oscillates at the driving frequency; the amplitude is largest when that equals the natural frequency. |
| Damping removes resonance. | The peak is still there, just lower and broader. | Damping limits the amplitude at resonance; it does not remove the peak. |
How this topic connects
5.5 Simple harmonic motion
The period formulae \(T=2\pi\sqrt{m/k}\) and \(T=2\pi\sqrt{l/g}\), and \(v_{\max}=\omega A\), all come from there; this page adds energy and what happens when the oscillator is damped or driven.
Hooke's law and elastic strain energy (Unit 1)
The spring's potential energy is the area under the force–extension graph, \(\tfrac12F\Delta x\); the energy dissipated by plastic deformation is the area enclosed by a loading–unloading loop.
Stationary waves (Unit 2) and practical skills (6.1)
A string or air column has many natural frequencies, its harmonics, and a vibration generator or a musician drives it into resonance at one of them. Linearising \(T=2\pi\sqrt{m/k}\) into a \(T^2\)–\(m\) graph, combining percentage uncertainties and explaining an intercept are typical of the A2 practical paper.
Key equations
| Equation | Use | In the formulae list? |
|---|---|---|
| \(T=2\pi\sqrt{m/k}\) | period of a mass–spring oscillator; \(m\) in kg, \(k\) in N m⁻¹ | yes |
| \(T=2\pi\sqrt{l/g}\) | period of a simple pendulum; \(l\) in m | yes |
| \(f=1/T,\ \ \omega=2\pi f\) | frequency (Hz) and angular frequency (rad s⁻¹) | yes |
| \(\Delta E_{el}=\tfrac12F\Delta x\) | elastic strain energy in a spring (J) | yes |
| \(E_{\text{total}}=\tfrac12kA^2\) | total energy of a mass–spring oscillator | no — derive it from \(\tfrac12F\Delta x\) with \(F=kA\) |
| \(E_k=\tfrac12k(A^2-x^2)\) | kinetic energy at displacement \(x\) | no — derive it: total − \(\tfrac12kx^2\) |
| \(E\propto A^2\) | energy and amplitude, any oscillator | no — learn it |
| \(T^2=(4\pi^2/k)\,m\); \(k=4\pi^2/\text{gradient}\) | Core Practical 16 graph | no — derive it by squaring \(T=2\pi\sqrt{m/k}\) |
| energy dissipated per cycle = area of force–extension loop | plastic deformation in a damper | no — learn it |
Check your understanding
1. The amplitude of an undamped oscillator is doubled. By what factor does its total energy change?
Energy ∝ amplitude², so it increases by a factor of \(2^2=\) 4.
2. At what displacement, as a fraction of the amplitude, are the kinetic and potential energies of a mass–spring oscillator equal?
\(\tfrac12kx^2=\tfrac12\times\tfrac12kA^2\), so \(x=A/\sqrt2\approx\) 0.71A.
3. State two ways in which increasing the damping changes the resonance curve of a forced oscillator.
Any two of: the peak amplitude is lower; the peak is broader; the peak occurs at a slightly lower frequency.
4. Explain, using the idea of resonance, why a small push once every swing can build up a large swing on a playground swing.
The pushes are a periodic driving force at the swing's natural frequency, so each push is in step with the motion and transfers energy to it. The amplitude grows until the energy lost to air resistance and friction each cycle equals the energy supplied by each push.
5. A 0.200 kg mass on a spring resonates at 1.60 Hz. Calculate the spring constant, ignoring the mass of the spring.
\(k=4\pi^2mf^2=4\pi^2\times0.200\times1.60^2=\) 20.2 N m⁻¹.
6. A lightly damped pendulum loses 19% of its energy in each cycle. By what fraction does its amplitude fall each cycle?
Energy kept = 0.81 of the previous value, so amplitude kept \(=\sqrt{0.81}=0.90\). The amplitude falls by 10% each cycle, much less than the 19% fall in energy.
Specification checklist
Numbers refer to the specification points.
- (148) I can explain what resonance is and give examples where it is useful and where it causes problems.
- (149) I can describe how to find an unknown mass from the resonant frequencies of a spring loaded with known masses, using a graph of \(T^2\) against \(m\), and analyse the results.
- (150) I can describe the interchange of kinetic and potential energy in an undamped oscillation, use \(E\propto A^2\), and sketch and interpret energy–displacement and energy–time graphs.
- (150) I can explain what happens to the energy of a damped oscillator and calculate energy changes from changes in amplitude.
- (151) I can distinguish between free and forced oscillations, and between natural frequency and driving frequency.
- (152) I can describe how the amplitude of a forced oscillation depends on the driving frequency, with a maximum when it equals the natural frequency, and how damping lowers, broadens and slightly shifts the resonance peak.
- (153) I can explain how damping and the plastic deformation of ductile materials reduce the amplitude of oscillations, including in buildings in earthquake zones and in car suspension; I can describe light, critical and heavy damping.
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