Senior 5 practice book · Unit 2 of 11

Damped and Forced Oscillations

50 questions that complete the Senior 5 quiz for this unit: 20 core and 30 challenge. Easy and hard questions are mixed. Try each question, then tap “Show answer”.

Common misconceptions

Damping slows an oscillator down, so its period gets longer and longer until it stops.Light damping reduces the AMPLITUDE; the period stays almost the same all the time (ω′ is only slightly less than ω₀).
At resonance the amplitude becomes infinite.Every real system has some damping. The amplitude grows until the energy lost per cycle equals the energy supplied per cycle, so it stays finite.
A forced oscillator always vibrates at its own natural frequency.Once the starting transient has died away, a forced oscillator vibrates at the frequency of the DRIVER; it only has its largest amplitude when the two frequencies are equal.
A damped oscillator loses the same number of centimetres of amplitude on every swing.With exponential damping it loses the same FRACTION of its amplitude in equal times, so the drops get smaller as the amplitude gets smaller.
The more damping there is, the faster a displaced system comes back to rest.This is only true up to critical damping. Beyond it (heavy damping) the system creeps back more and more slowly.

What this unit covers

Topics marked new are not tested much in the quiz, so this book gives them extra questions.

Free and damped oscillations: resistive forces and where the energy goes
Light (under-), critical and heavy (over-) damping
Exponential decay of amplitude A = A₀e^−λt and the decay constant
Energy of a damped oscillator: E ∝ A², fraction of energy lost per cycle
Equation of motion of a damped oscillator: m d²x/dt² + b dx/dt + kx = 0
Damped angular frequency ω′ = √(ω₀² − (b/2m)²) and the critical damping coefficient b = 2√(km)
Forced oscillations, natural frequency and resonance
Effect of damping on the resonance curve (height, width and position of the peak)
Useful and harmful resonance and how unwanted vibrations are reduced
Phase difference between the driving force and the driven oscillator
Barton's pendulums and laboratory demonstrations of resonance
Measuring damping from data: ratio of successive amplitudes and ln A against t

Go to the questions

Questions (1–50)

Easier and harder questions are mixed, just like in a real exam. The 30 harder ones are marked ★ Challenge.

1True or false
The natural frequency of a playground swing depends on how hard it is pushed.
Show answer
Answer: False
The natural frequency depends on the swing's length (and g), not on the push; a harder push only increases the amplitude.
Common mistakeLearners often mix up amplitude and frequency; pushing harder makes the swing go higher, not faster.
2True or false · ★ Challenge
A tuning fork struck inside a vacuum chamber would keep vibrating for ever, because there is no air resistance.
Show answer
Answer: False
Without air the damping is much smaller, but internal friction in the metal still turns some of the vibration energy into thermal energy, so it slowly stops.
Common mistakeLearners often treat air resistance as the only resistive force; internal friction in the material also damps vibrations.
3True or false · ★ Challenge
When a system is driven very slowly (far below its natural frequency), it moves almost in phase with the driver.
Show answer
Answer: True
At low frequency the spring simply follows the driver: displacement and driving force reach their maxima together, with almost no lag.
Common mistakeLearners often think the system always lags by a quarter cycle; the lag depends on the driving frequency.
4True or false
In Barton's pendulums, the light pendulum with the same length as the driver lags behind the driver by about a quarter of a cycle.
Show answer
Answer: True
At resonance the driven pendulum's displacement lags the driver by π/2, so it passes the middle when the driver is at an end.
Common mistakeLearners often expect the resonating pendulum to swing exactly in step with the driver; that only happens for the much shorter pendulums, whose natural frequency is well above the driving frequency.
5True or false · ★ Challenge
Increasing the damping reduces the amplitude at resonance much more than it reduces the amplitude at driving frequencies far from resonance.
Show answer
Answer: True
Far from resonance the amplitude is set mainly by the spring (low f) or the mass (high f); near resonance it is limited only by damping, so damping matters most there.
Common mistakeLearners often think damping lowers the whole curve by the same amount; its effect is concentrated at the peak.
6True or false · ★ Challenge
If the amplitudes of successive swings of a damped oscillator all fall by the same RATIO, the damping is exponential.
Show answer
Answer: True
A = A₀e^−λt gives A(t + T)/A(t) = e^−λT, the same ratio for every cycle.
Common mistakeLearners often think a constant ratio means a constant loss in centimetres; a constant ratio means the drops get smaller each time.
7True or false
Light damping makes the period of an oscillator slightly longer than its undamped period.
Show answer
Answer: True
ω′ = √(ω₀² − (b/2m)²) is a little smaller than ω₀, so T′ = 2π/ω′ is a little longer than T₀.
Common mistakeLearners often say damping does not affect the period at all; it has a small effect, which grows as the damping increases.
8Multiple choice · ★ Challenge
At resonance, the driving force on a lightly damped oscillator is in phase with which quantity?
1. AThe oscillator's displacement
2. BThe oscillator's acceleration
3. CThe oscillator's velocity
4. DThe spring's restoring force
Show answer
Answer: C. The oscillator's velocity
The displacement lags by π/2, so the velocity (a quarter cycle ahead of the displacement) is in phase with the force; force × velocity is always positive and energy transfer is greatest.
Common mistakeChoosing displacement is tempting because 'in phase' sounds like the best match, but a force in phase with displacement would do as much negative work as positive work.
9Fill in the blank
A bathroom scale whose pointer swings back and forth several times before settling on the reading is ______ damped.
Show answer
Answer: lightly (under-)
Oscillation with a decreasing amplitude before settling is the sign of light (under-) damping.
Common mistakeWriting 'critically' is wrong because a critically damped pointer would not swing past the reading at all.
10Multiple choice · ★ Challenge
Once Barton's pendulums have settled, what can be said about the frequencies of the light pendulums?
1. AEach swings at its own natural frequency
2. BOnly the resonating pendulum moves; the others stay still
3. CThey all swing at the frequency of the driver
4. DThe shorter pendulums swing at higher frequencies than the driver
Show answer
Answer: C. They all swing at the frequency of the driver
All the light pendulums are forced oscillators, so in the steady state they vibrate at the driving frequency; only their amplitudes and phases differ.
Common mistakeChoosing 'own natural frequency' confuses free oscillations with forced ones; the driver sets the frequency.
11True or false · ★ Challenge
If a damped oscillator loses 10% of its energy every cycle, it also loses 10% of its amplitude every cycle.
Show answer
Answer: False
E ∝ A², so A ∝ √E: keeping 90% of the energy keeps √0.90 = 95% of the amplitude, a loss of only about 5%.
Common mistakeLearners often apply the same percentage to amplitude and energy; the square relation makes the energy fall faster.
12Fill in the blank
In Barton's experiment, the heavy pendulum that sets all the light pendulums swinging is called the ______ pendulum.
Show answer
Answer: driver (driving)
The heavy bob stores much energy and acts as the source of the periodic force passed on through the string.
Common mistakeLearners sometimes call it the 'resonant' pendulum; resonance describes the light pendulum that responds most strongly, not the driver.
13Multiple choice · ★ Challenge
The needle of an analogue voltmeter is designed to be critically damped. Why?
1. ASo that it swings past the reading a few times, showing it is free to move
2. BSo that it moves very slowly and cannot ever overshoot
3. CSo that it has no damping and follows rapid changes
4. DSo that it settles on the reading quickly without swinging about it
Show answer
Answer: D. So that it settles on the reading quickly without swinging about it
Critical damping returns the needle to its final position in the shortest time without overshooting, so the reading can be taken at once.
Common mistakeChoosing 'very slowly and cannot overshoot' describes heavy damping; it avoids overshoot but makes the reader wait.
14Fill in the blank
A hospital MRI scanner uses resonance of the nuclei of ______ atoms in the water of the body, in a strong magnetic field.
Show answer
Answer: hydrogen
Radio waves at the right frequency make hydrogen nuclei (protons) in the body resonate; the signals they give out are used to build the image.
Common mistakeWriting 'oxygen' is tempting because water contains oxygen, but it is the hydrogen nuclei (single protons) that resonate in MRI.
15Multiple choice · ★ Challenge
A new footbridge has a natural frequency of 2.0 Hz, close to walking pace. Which change would move its natural frequency well ABOVE walking pace?
1. AAdding a heavy concrete layer to the deck
2. BMaking the span longer and more flexible, so that it bends more easily
3. CAdding steel beams that make the deck much stiffer
4. DRemoving the dampers fixed under the deck
Show answer
Answer: C. Adding steel beams that make the deck much stiffer
f₀ = (1/2π)√(k/m): increasing the stiffness k raises f₀. Adding mass or making the deck more flexible lowers it; dampers do not change f₀ much.
Common mistakeChoosing 'heavy concrete' may seem to make the bridge stronger, but more mass LOWERS the natural frequency.
16Multiple choice
In A = A₀e^−λt, what is the SI unit of the decay constant λ?
1. As⁻¹
2. Bs
3. Cm/s
4. Dit has no unit
Show answer
Answer: A. s⁻¹
The power of e must have no unit, so λt has no unit and λ must be measured in s⁻¹.
Common mistakeChoosing 'no unit' forgets that t has a unit (s), so λ must cancel it.
17Multiple choice · ★ Challenge
What is the SI unit of the damping coefficient b in F = −bv?
1. AN/m
2. Bkg/s
3. Ckg s
4. DN/s
Show answer
Answer: B. kg/s
b = F/v, so its unit is N ÷ (m/s) = N s/m = (kg m/s²)(s/m) = kg/s.
Common mistakeChoosing N/m confuses b with the spring constant k, which is force per unit DISPLACEMENT, not per unit velocity.
18Multiple choice · ★ Challenge
After some time, the amplitude of a damped oscillator has fallen to 30% of its starting value. What percentage of its starting energy is left?
1. A30%
2. B55%
3. C70%
4. D9%
Show answer
Answer: D. 9%
E ∝ A², so E/E₀ = 0.30² = 0.09 = 9%.
Common mistakeChoosing 30% assumes energy is proportional to amplitude; it is proportional to the square of the amplitude.
19Multiple choice
A microwave oven heats food quickly. Which explanation uses the idea of resonance?
1. AMicrowaves heat the metal walls of the oven, and the hot walls then heat the food
2. BWater molecules absorb energy strongly at the microwave frequency and move more
3. CMicrowaves reflect from the food and travel round the oven many times
4. DThe food's large mass makes it vibrate slowly at a low frequency
Show answer
Answer: B. Water molecules absorb energy strongly at the microwave frequency and move more
The microwaves (about 2.45 GHz) make the water molecules twist to and fro with the field; they absorb this energy strongly, so their random motion (internal energy) increases.
Common mistakeChoosing 'the walls get hot' is wrong: the metal walls reflect microwaves and stay fairly cool; the energy goes into the water in the food.
20Short answer · ★ Challenge
Write the equation of motion of a damped mass–spring system that is also driven by a periodic force F₀ cos(ωt), and state what each term represents.
Show answer
Model answer: m d²x/dt² + b dx/dt + kx = F₀ cos(ωt). m d²x/dt² is mass × acceleration; b dx/dt is the damping force (proportional to velocity); kx is the spring's restoring force; F₀ cos(ωt) is the external driving force of amplitude F₀ and angular frequency ω.
Common mistakeLearners often put the driving force with the same sign as the restoring force; the driving force is external and appears on the other side of the equation.
21Multiple choice · ★ Challenge
A damped oscillator has amplitude A = A₀e^−0.050t. What percentage of its starting energy remains after 10 s?
1. A61%
2. B50%
3. C37%
4. D14%
Show answer
Answer: C. 37%
E ∝ A², so E = E₀e^−2λt = E₀e^{−2 × 0.050 × 10} = E₀e⁻¹ = 0.37E₀.
Common mistakeChoosing 61% uses e^−λt, which is the AMPLITUDE fraction; the energy decays twice as fast, with 2λ.
22Multiple choice
A car's shock absorber pushes a piston through oil to damp the bouncing of the car. Into which form is most of the oscillation energy transferred?
1. AGravitational potential energy of the car
2. BThermal energy in the oil
3. CElastic energy stored permanently in the springs
4. DChemical energy in the oil
Show answer
Answer: B. Thermal energy in the oil
The viscous oil resists the piston; work done against this friction becomes internal (thermal) energy, so the oil warms up.
Common mistakeChoosing 'elastic energy in the springs' forgets that the springs give their stored energy back each cycle; only friction removes energy for good.
23Short answer · ★ Challenge
Loudspeaker cones are made heavily damped. Explain why a sharp resonance would be a problem in a loudspeaker.
Show answer
Model answer: A loudspeaker must reproduce all frequencies of music and speech equally. A sharp resonance would make notes near its natural frequency much louder than others, and the cone would keep 'ringing' after a sound stopped. Heavy damping flattens the response and stops the cone quickly.
Common mistakeLearners often think resonance always improves sound; for a loudspeaker an even (flat) response matters more than loudness at one frequency.
24Multiple choice
In the damped oscillator equation m d²x/dt² + b dx/dt + kx = 0, what does the term b dx/dt represent?
1. AThe resistive force, proportional to the velocity
2. BThe restoring force exerted on the mass by the spring
3. CThe mass times acceleration
4. DThe external driving force
Show answer
Answer: A. The resistive force, proportional to the velocity
dx/dt is the velocity, so b dx/dt is a drag (damping) force proportional to the speed, as for a piston moving through oil.
Common mistakeChoosing 'restoring force' confuses it with kx, which depends on displacement, not velocity.
25Short answer · ★ Challenge
A learner writes: 'The amplitude of a damped swing halves in 8 s, so after 16 s it will be zero.' Correct the statement.
Show answer
Model answer: In exponential decay the amplitude halves in EVERY 8 s. After 16 s it is ½ × ½ = ¼ of the starting value, after 24 s one eighth, and so on. In theory it never becomes exactly zero.
Common mistakeThe learner treats the decay as linear (the same loss in each 8 s); exponential decay loses the same fraction, not the same amount.
26Multiple choice · ★ Challenge
A loose panel on a minibus rattles loudly when the engine turns at 1800 revolutions per minute, but is quiet at 1200 and 2400 revolutions per minute. What is the natural frequency of the panel closest to?
1. A1800 Hz
2. B3.0 Hz
3. C188 Hz
4. D30.0 Hz
Show answer
Answer: D. 30.0 Hz
The engine drives the panel once per revolution: 1800 ÷ 60 = 30 revolutions per second = 30 Hz. The panel resonates, so its natural frequency is about 30 Hz.
Common mistakeChoosing 188 Hz gives the angular speed in rad/s (2π × 30); a frequency in hertz is revolutions per second.
27Multiple choice
In Barton's pendulums, a heavy driver pendulum and several light paper-cone pendulums of different lengths hang from the same tight string. Which light pendulum swings with the largest amplitude?
1. AThe shortest one, because it swings fastest
2. BThe one whose length is equal to that of the driver
3. CThe longest one, because it swings through the largest arc
4. DThe one hanging closest to the driver
Show answer
Answer: B. The one whose length is equal to that of the driver
The pendulum of the same length has the same natural frequency as the driver, so it resonates and absorbs the most energy.
Common mistakeChoosing 'the closest one' assumes distance matters; the energy travels along the string to all of them, and it is the matching frequency that matters.
28Fill in the blank · ★ Challenge
A pendulum of period 2.0 s has successive amplitudes of 20.0 cm, 16.0 cm and 12.8 cm. Its decay constant λ is about ______ s⁻¹.
Show answer
Answer: 0.11
Each period the amplitude is multiplied by 0.80, so e^−2.0λ = 0.80 and λ = ln(1.25) ÷ 2.0 = 0.223 ÷ 2.0 = 0.11 s⁻¹.
Common mistakeWriting 0.10 s⁻¹ uses the 20% drop directly instead of ln(1/0.80); and forgetting to divide by the period gives 0.22 s⁻¹.
29Short answer
Give two examples where damping is useful and one example where damping is a nuisance.
Show answer
Model answer: Useful: car or moto taxi shock absorbers; door closers; the needle of a measuring meter; dampers in tall buildings to reduce swaying. Nuisance: a guitar or inanga string stops ringing too soon; a clock pendulum or a playground swing loses amplitude and needs extra pushes.
Common mistakeLearners often list only harmful examples; engineers often ADD damping on purpose to stop unwanted oscillation.
30Short answer · ★ Challenge
A classroom door has a closer that is underdamped. Describe what happens when someone lets go of the open door, and explain how the closer should be adjusted.
Show answer
Model answer: The door swings shut, passes the closed position (it bangs or bounces against the frame) and oscillates a few times before settling. The damping should be increased to about critical, so that the door closes in the shortest time without swinging back; too much damping would make it close very slowly.
Common mistakeLearners often say 'add as much damping as possible'; over-damping makes the door creep shut, which is also a fault.
31Fill in the blank
When b/2m becomes equal to ω₀, the damped angular frequency ω′ = √(ω₀² − (b/2m)²) becomes ______, so the system no longer oscillates.
Show answer
Answer: zero
ω′ = √(ω₀² − ω₀²) = 0: this is critical damping, where the system returns to equilibrium without oscillating.
Common mistakeWriting ω₀ forgets that the damping term is subtracted; at critical damping it cancels ω₀ completely.
32Short answer · ★ Challenge
Some tall buildings contain a large mass on springs and dampers near the top (a tuned mass damper). Explain how this reduces swaying in strong wind or an earthquake.
Show answer
Model answer: The damper's natural frequency is tuned to that of the building. When the building starts to sway, the mass oscillates out of phase with it and pushes against the motion. The dampers turn the energy of this motion into thermal energy, so energy is removed from the building and its amplitude stays small.
Common mistakeLearners often think the heavy mass simply makes the building harder to move; it works because it is tuned to the building's frequency and moves against it.
33Multiple choice · ★ Challenge
A spring of constant 450 N/m is fixed to a vibrator running at 5.0 Hz. What mass should be hung on the spring so that it resonates?
1. A0.46 kg
2. B18 kg
3. C14 kg
4. D0.073 kg
Show answer
Answer: A. 0.46 kg
Resonance when f₀ = 5.0 Hz: ω = 2π × 5.0 = 31.4 rad/s, and m = k/ω² = 450 ÷ 987 = 0.46 kg.
Common mistakeChoosing 18 kg uses k/f² instead of k/(2πf)²; the formula ω² = k/m needs the angular frequency.
34Short answer
A singer can make a wine glass shatter by singing one note loudly. State two conditions needed for this to happen.
Show answer
Model answer: 1) The frequency of the note must equal the natural frequency of the glass, so it resonates. 2) The sound must be loud enough (and the glass lightly damped) for the amplitude of vibration to build up beyond what the glass can bear.
Common mistakeLearners often say any loud note will break the glass; without matching the natural frequency the energy transfer stays small.
35Multiple choice · ★ Challenge
The peaks of a damped oscillation on a data logger are 10.0 cm, 8.0 cm, 6.4 cm and 5.12 cm. What should the next peak be?
1. A4.10 cm
2. B3.84 cm
3. C3.12 cm
4. D4.61 cm
Show answer
Answer: A. 4.10 cm
Each peak is 0.80 of the one before (8.0/10.0 = 6.4/8.0 = 0.80), so the next is 5.12 × 0.80 = 4.10 cm.
Common mistakeChoosing 3.84 cm subtracts the last drop (1.28 cm) again; exponential damping keeps the RATIO constant, not the difference.
36Multiple choice · ★ Challenge
A 1.5 kg mass hangs on a spring of constant 600 N/m inside an oil-filled damper. What damping coefficient b makes the system exactly critically damped?
1. A30 kg/s
2. B60 kg/s
3. C20 kg/s
4. D1800 kg/s
Show answer
Answer: B. 60 kg/s
Critical damping when b/2m = ω₀, i.e. b = 2√(km) = 2√(600 × 1.5) = 2 × 30 = 60 kg/s.
Common mistakeChoosing 30 kg/s forgets the factor 2 in b = 2√(km), which comes from b/2m = √(k/m).
37Fill in the blank
For an oscillator with λ = 0.10 s⁻¹, the amplitude falls to 1/e (about 37%) of its starting value after ______ s.
Show answer
Answer: 10
A = A₀e^−λt = A₀/e when λt = 1, so t = 1/λ = 1 ÷ 0.10 = 10 s.
Common mistakeWriting 6.9 s gives the half-time ln 2/λ; the 1/e time is simply 1/λ.
38Multiple choice · ★ Challenge
Pendulums P and Q start with the same amplitude. P has λ = 0.020 s⁻¹ and Q has λ = 0.080 s⁻¹. After 25 s, what is the ratio of their amplitudes, A_P : A_Q?
1. A4.0
2. B1.5
3. C4.5
4. D0.22
Show answer
Answer: C. 4.5
A_P/A_Q = e^{−0.020 × 25} ÷ e^{−0.080 × 25} = e^{0.060 × 25} = e^1.5 = 4.5.
Common mistakeChoosing 4.0 takes the ratio of the decay constants; the amplitudes depend exponentially on λt, not in proportion to λ.
39Multiple choice
A lightly damped oscillator is driven exactly at its natural frequency. By how much does its displacement lag behind the driving force?
1. AA quarter of a cycle (π/2 rad)
2. BNothing: they are exactly in phase
3. CHalf a cycle (π rad)
4. DA whole cycle (2π rad)
Show answer
Answer: A. A quarter of a cycle (π/2 rad)
At resonance the displacement lags the driving force by π/2, which puts the VELOCITY in phase with the force, so the force does positive work all the time.
Common mistakeChoosing 'in phase' describes very slow driving, far below resonance; at resonance the lag is a quarter of a cycle.
40Short answer · ★ Challenge
For a lightly damped system, describe how the phase difference between the driver and the driven oscillator changes as the driving frequency is increased from far below the natural frequency f₀ to far above it.
Show answer
Model answer: Far below f₀ the oscillator moves in phase with the driver (lag ≈ 0). As the frequency rises the lag increases; at f₀ the oscillator lags by π/2 (a quarter cycle). Far above f₀ the lag approaches π, so the oscillator moves opposite to the driver. With light damping the change from 0 to π happens sharply near f₀.
Common mistakeLearners often think the phase jumps suddenly from 0 to π with no in-between value; it passes through π/2 at resonance.
41Multiple choice · ★ Challenge
A mass on a spring is driven at a frequency much HIGHER than its natural frequency. How does its motion compare with the motion of the driver?
1. ALarge amplitude, moving almost exactly in step with the driver at all times
2. BSmall amplitude, moving almost in step with the driver, with hardly any lag
3. CLarge amplitude, lagging the driver by a quarter of a cycle
4. DSmall amplitude, moving almost opposite to the driver (π out of phase)
Show answer
Answer: D. Small amplitude, moving almost opposite to the driver (π out of phase)
Far above resonance the mass cannot follow the rapid driving; its amplitude is small and its displacement lags by nearly π, so it moves opposite to the driver.
Common mistakeChoosing 'in step' describes driving far BELOW resonance; the phase lag grows from 0 through π/2 at resonance to almost π at high frequency.
42Short answer
A wall clock's pendulum keeps the same amplitude for weeks, although friction acts on it. Explain where the energy comes from and why the amplitude does not keep growing.
Show answer
Model answer: A spring or battery gives the pendulum a small push once every swing (through the escapement). At the steady amplitude, the energy supplied in each cycle exactly equals the energy lost to friction in each cycle, so the amplitude stays constant. If the amplitude grew, the losses would grow too until they balanced the supply again.
Common mistakeLearners often think the pendulum needs no energy because it is 'undamped'; every real pendulum is damped and must be topped up.
43Multiple choice · ★ Challenge
A 0.80 kg mass on a spring of constant 320 N/m is damped with b = 19.2 kg/s. Using ω′ = √(k/m − (b/2m)²), what is the angular frequency of its damped oscillations?
1. A16.0 rad/s
2. B20.0 rad/s
3. C12.0 rad/s
4. D23.3 rad/s
Show answer
Answer: A. 16.0 rad/s
ω₀² = k/m = 320 ÷ 0.80 = 400 s⁻² and b/2m = 19.2 ÷ 1.6 = 12 s⁻¹, so ω′ = √(400 − 144) = √256 = 16.0 rad/s.
Common mistakeChoosing 20.0 rad/s gives the undamped ω₀; the damping term (b/2m)² must be subtracted before taking the square root.
44Multiple choice
Two resonance curves are drawn for the same mass–spring system. Curve X has a tall, narrow peak at 4.9 Hz; curve Y has a low, broad peak at 4.6 Hz. Which statement is correct?
1. AX has more damping, because its peak is taller
2. BY has a heavier mass, because a heavier mass always gives a lower and broader peak
3. CY has more damping, and more damping moves the peak to a slightly lower frequency
4. DX has more damping, because its peak is at a higher frequency
Show answer
Answer: C. Y has more damping, and more damping moves the peak to a slightly lower frequency
Damping lowers and widens the peak, and shifts it slightly below the natural frequency; Y shows all three effects.
Common mistakeChoosing 'X, because its peak is taller' reverses the effect: damping removes energy, so it LOWERS the peak.
45Short answer · ★ Challenge
In a Barton's pendulums demonstration, the paper cones are replaced by heavier brass rings of the same size, so the light pendulums are much less damped. Predict two changes in what is seen.
Show answer
Model answer: 1) The pendulum whose length matches the driver swings with a much larger amplitude: the resonance peak is taller and sharper. 2) The pendulums with other lengths swing with relatively small amplitudes, and the change of phase, from nearly in step (short pendulums) to nearly opposite (long pendulums), happens more suddenly near the matching length.
Common mistakeLearners often predict that heavier bobs change the frequency at resonance; a pendulum's natural frequency depends on its length, not its mass.
46Multiple choice · ★ Challenge
A learner records the amplitude of a damped pendulum: 12.0 cm at t = 0 and 9.0 cm at t = 5.0 s. Assuming exponential decay, what does she predict at t = 10.0 s?
1. A6.00 cm
2. B6.75 cm
3. C4.50 cm
4. D7.50 cm
Show answer
Answer: B. 6.75 cm
In each 5.0 s the amplitude is multiplied by 9.0/12.0 = 0.75, so at 10 s A = 9.0 × 0.75 = 6.75 cm.
Common mistakeChoosing 6.0 cm subtracts 3.0 cm again (a linear fall); exponential decay multiplies by the same factor in equal times.
47Multiple choice
The amplitude of a damped oscillator obeys A = A₀e^−λt. Which graph gives a straight line?
1. AA against t
2. BA against ln t
3. C1/A against t
4. Dln A against t
Show answer
Answer: D. ln A against t
Taking logs: ln A = ln A₀ − λt, which has the form y = c + mx, a straight line of gradient −λ.
Common mistakeChoosing A against t forgets that exponential decay gives a curve whose slope keeps getting smaller.
48Short answer · ★ Challenge
One wheel of a car carries 300 kg and has a suspension spring of constant 3.0 × 10⁴ N/m. (a) What damping coefficient would make the wheel's motion critically damped? (b) Describe the motion if the shock absorber only gave half this value.
Show answer
Model answer: (a) b = 2√(km) = 2√(3.0 × 10⁴ × 300) = 2√(9.0 × 10⁶) = 2 × 3000 = 6.0 × 10³ kg/s. (b) With b smaller than the critical value the system is underdamped: after a bump the car bounces up and down a few times with decreasing amplitude before settling.
Common mistakeLearners often use b = √(km) and get 3000 kg/s; the critical condition b/2m = √(k/m) gives b = 2√(km).
49Short answer
A 1.0 kg mass on a spring of constant 100 N/m has a small damping coefficient b = 0.20 kg/s. Compare its damped angular frequency with its undamped one. What does this show about light damping?
Show answer
Model answer: ω₀ = √(100/1.0) = 10 rad/s; b/2m = 0.10 s⁻¹, so ω′ = √(100 − 0.01) = 9.9995 rad/s. The two are practically the same: light damping reduces the amplitude but hardly changes the frequency or period.
Common mistakeLearners often expect any damping to change the frequency noticeably; the change depends on (b/2m)², which is tiny for light damping.
50Short answer · ★ Challenge
A graph of ln(A/cm) against t for a damped pendulum is a straight line from ln A = 2.30 at t = 0 to ln A = 1.10 at t = 30 s. Find the decay constant λ, the starting amplitude and the time for the amplitude to halve.
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Model answer: ln A = ln A₀ − λt, so λ = −gradient = (2.30 − 1.10) ÷ 30 = 0.040 s⁻¹. A₀ = e^2.30 ≈ 10 cm. Half-time t½ = ln 2/λ = 0.693 ÷ 0.040 ≈ 17 s.
Common mistakeLearners often read the starting amplitude as 2.30 cm; the intercept is ln A₀, so A₀ = e^2.30.