Damping

So far we have studied idealised SHM — oscillations that continue forever with constant amplitude. In the real world, every oscillating system loses energy to its surroundings through friction, air resistance, or other dissipative forces. This causes the amplitude to decrease over time, a phenomenon called damping.

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Free and Forced Oscillations

Before studying damping, it is important to distinguish two types of oscillation:

Free oscillations occur when a system is displaced from equilibrium and then left to oscillate without any external driving force. The system oscillates at its natural frequency f₀, which depends only on the system's physical properties (mass, spring constant, length, etc.). A pendulum set swinging and left alone is undergoing free oscillation.

Forced oscillations occur when an external periodic force continuously drives the system. The system oscillates at the driving frequency f_d, which may or may not equal the natural frequency. A child being pushed on a swing is an example of forced oscillation.

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What Causes Damping?

Damping occurs when resistive forces act on the oscillating system, removing energy with each cycle. Common sources include:

• Air resistance — the moving object pushes through air molecules
• Friction at pivot points or contact surfaces
• Internal friction within the material (molecular forces resisting deformation)
• Viscous drag if the system moves through a liquid
• Electromagnetic damping (eddy currents in conducting materials moving through magnetic fields)

The energy removed from the oscillation is converted to thermal energy (heat) in the surroundings.

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Types of Damping

flowchart TD
    A["Damping Types"] --> B["Light Damping\n(Underdamping)"]
    A --> C["Critical Damping"]
    A --> D["Heavy Damping\n(Overdamping)"]
    B --> B1["System oscillates\nAmplitude decays exponentially\nFrequency ≈ f₀"]
    B --> B2["Examples:\n• Pendulum in air\n• Tuning fork\n• Guitar string"]
    C --> C1["Fastest return to\nequilibrium without\nany oscillation"]
    C --> C2["Examples:\n• Car suspension\n• Voltmeter needle\n• Door closer"]
    D --> D1["Slow return to\nequilibrium without\noscillation"]
    D --> D2["Examples:\n• Pendulum in honey\n• Over-tight door closer\n• Mass in thick oil"]

Light Damping (Underdamping)

In light damping, the resistive forces are small compared to the restoring force. The system continues to oscillate, but the amplitude decreases gradually with each cycle. The frequency of oscillation is very close to the natural frequency (slightly lower in practice, but the difference is usually negligible at A-Level).

The amplitude decreases exponentially: each successive peak is a fixed fraction of the previous one. On a displacement-time graph, the oscillation appears as a sinusoidal wave with an exponentially decaying envelope.

Examples:

• A pendulum swinging in air
• A tuning fork vibrating in air
• A guitar string after being plucked

Heavy Damping (Overdamping)

In heavy damping, the resistive forces are very large. The system returns to equilibrium slowly without oscillating. It simply creeps back towards the equilibrium position, taking a long time to get there.

Examples:

• A pendulum submerged in thick oil or treacle
• A door with an overly strong hydraulic closer
• A mass on a spring submerged in honey

Critical Damping

Critical damping is the special case between light and heavy damping where the system returns to equilibrium in the shortest possible time without oscillating. It is the fastest non-oscillatory return.

Critical damping is extremely important in engineering because many systems need to return to equilibrium quickly without overshooting.

Examples:

• Car suspension systems (designed to be close to critically damped)
• Bathroom door closers
• Needle in an analogue ammeter or voltmeter
• Gun recoil mechanisms

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Comparing the Three Types

Property	Light (Under)	Critical	Heavy (Over)
Oscillates?	Yes	No	No
Returns to equilibrium?	Eventually (amplitude → 0)	Yes, fastest without oscillation	Yes, but slowly
Overshoots equilibrium?	Yes, repeatedly	No (just reaches it)	No
Time to reach equilibrium	Long (many oscillations)	Shortest possible	Very long
Energy loss per cycle	Small	N/A (no cycle)	N/A (no cycle)

On a displacement-time graph:

• Light damping shows oscillations with exponentially decreasing peaks
• Critical damping shows a smooth curve that approaches zero quickly without crossing it
• Heavy damping shows a slow exponential approach to zero without crossing

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Damping and Energy

In a lightly damped system, the total energy decreases with each oscillation. Since E ∝ A², and the amplitude decreases exponentially, the energy also decreases exponentially.

After time t with light damping, the amplitude is approximately:

A(t) = A₀ × e^(−λt)

where λ is the damping constant and A₀ is the initial amplitude. This exponential decay means the system loses a larger amount of energy in the early oscillations and a smaller amount later.

Energy-Amplitude Relationship with Damping

After n half-lives of amplitude	Amplitude	Energy (∝ A²)
0	A₀	E₀
1	A₀/2	E₀/4
2	A₀/4	E₀/16
3	A₀/8	E₀/64

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Worked Example 1: Exponential Amplitude Decay

A lightly damped oscillator has an initial amplitude of 0.20 m. After 10 oscillations the amplitude has decreased to 0.10 m. Find the amplitude after 30 oscillations.

Solution:

In exponential decay, equal time intervals reduce the amplitude by the same factor.

After 10 oscillations: A = 0.20 × factor = 0.10, so factor = 0.5

After 20 oscillations: A = 0.10 × 0.5 = 0.05 m

After 30 oscillations: A = 0.05 × 0.5 = 0.025 m

The energy after 30 oscillations: E = E₀ × (0.025/0.20)² = E₀ × (1/8)² = E₀/64

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Worked Example 2: Energy Loss in Damped SHM

A 0.5 kg mass on a spring (k = 80 N m⁻¹) starts with amplitude 0.06 m. After 20 oscillations, the amplitude is 0.04 m. Calculate the energy dissipated.

Solution:

Initial energy: E₁ = ½kA₁² = ½ × 80 × 0.06² = 0.144 J

Final energy: E₂ = ½kA₂² = ½ × 80 × 0.04² = 0.064 J

Energy dissipated = E₁ − E₂ = 0.144 − 0.064 = 0.080 J

This energy has been converted to thermal energy in the surroundings.

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Practical Applications

Car Suspension

When a car goes over a bump, the springs compress and would normally oscillate up and down many times. This would be uncomfortable and dangerous. Shock absorbers (dampers) provide damping close to the critical value, allowing the car to return smoothly to its normal ride height after one or at most two oscillations.

If the damping is too light, the car bounces repeatedly. If too heavy, the suspension cannot respond quickly to bumps and the ride is harsh. Critical damping gives the optimal balance.

Door Closers

A fire door with a hydraulic closer is designed to shut automatically but not slam. The closer provides heavy or near-critical damping so the door returns to the closed position smoothly without oscillating back and forth.

Measuring Instruments

The needle on an analogue voltmeter or ammeter needs to reach the correct reading quickly without oscillating around it. Critical damping (often achieved using electromagnetic damping via eddy currents) allows the needle to settle rapidly.

Earthquake-Resistant Buildings

Modern buildings in earthquake zones incorporate damping mechanisms — massive pendulums (tuned mass dampers), hydraulic dampers, or friction dampers — to absorb oscillation energy during seismic events and reduce the amplitude of the building's vibration.

Musical Instruments

Musicians control damping constantly. Pressing a piano's sustain pedal lifts the dampers off the strings, allowing them to vibrate freely (light damping, long sustain). Releasing the pedal applies the dampers, stopping the vibrations quickly (heavy damping). A guitarist mutes strings by touching them lightly, applying damping to stop unwanted notes.

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The Q Factor (Quality Factor)

The Q factor is a dimensionless number that describes how underdamped an oscillator is — equivalently, how many oscillations it can sustain before the energy drops significantly:

Q = 2π × (energy stored)/(energy lost per cycle)

System	Typical Q Factor	Meaning
Pendulum in honey	~1	Heavily damped, barely oscillates
Car suspension	~0.5–1	Critically damped, no oscillation
Pendulum in air	~100–1000	Lightly damped, oscillates many times
Tuning fork	~1000	Very clean, sustained oscillation
Quartz crystal oscillator	~10⁵	Extremely lightly damped
Laser cavity	~10⁹	Almost no damping

A high Q factor means the system stores energy efficiently and oscillates many times before the amplitude decays significantly. This is desirable in clocks and oscillators but undesirable in suspension systems.

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Recognising Damping Types in Exam Questions

Exam questions may present a displacement-time graph and ask you to identify the type of damping:

• If you see oscillations with decreasing amplitude → light damping
• If you see a smooth return to zero without oscillation, relatively fast → critical damping
• If you see a very slow return to zero without oscillation → heavy damping

You may also be asked to sketch what the graph would look like under different damping conditions, or to explain which type of damping is desirable for a given application.

Common exam mistake: Stating that damping changes the frequency. For light damping, the frequency remains approximately equal to the natural frequency. Only the amplitude decreases. The frequency depends on the system parameters (m and k, or L and g), not on the damping level.

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A-Level Deep Dive: Damping

Spec mapping

Edexcel 9PH0 specification, Topic 13 — Oscillations introduces simple harmonic motion and then extends it to free, damped, and forced oscillations, including the qualitative description of light, critical and heavy damping and the effect of damping on resonance (refer to the official specification document for exact wording). Damping is the connective tissue that joins the idealised SHM of the early sub-strands to the real-world resonance phenomena examined later in the topic; it is also the bridge to Topic 4 (Mechanics — work and energy), because every damping question is ultimately a question about energy dissipation. Although Topic 13 sits in Paper 2 territory, damping concepts resurface in Topic 8 (Electric and magnetic fields) via electromagnetic damping (eddy currents) and in Topic 11 (Nuclear and particle physics) via the analogy with radioactive decay (both obey $N(t) = N_0 e^{-\lambda t}$N(t)=N0​e−λt). The Edexcel formula booklet provides $x = A\cos(\omega t)$x=Acos(ωt) and the SHM acceleration relation, but does not list the exponential amplitude decay $A(t) = A_0 e^{-\gamma t}$A(t)=A0​e−γt — the form must be recognised on inspection.

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Worked example with full mark scheme

Question (8 marks):

A pendulum of natural period $T_0 = 2.0\,\text{s}$T0​=2.0s is set swinging with an initial amplitude of $A_0 = 12.0\,\text{cm}$A0​=12.0cm. Air resistance causes the amplitude to follow $A(t) = A_0 e^{-\gamma t}$A(t)=A0​e−γt with damping constant $\gamma = 0.030\,\text{s}^{-1}$γ=0.030s−1.

(a) Calculate the amplitude after $t = 30\,\text{s}$t=30s. (2)

(b) Determine the fraction of the initial energy remaining at $t = 30\,\text{s}$t=30s. (3)

(c) The pendulum completes approximately one oscillation every 2.0 s. Find the number of complete oscillations required for the amplitude to fall to half its initial value. (3)

Solution with mark scheme:

(a) Step 1 — substitute into the exponential.

$A(30) = 0.120 \times e^{-0.030 \times 30} = 0.120 \times e^{-0.9}$A(30)=0.120×e−0.030×30=0.120×e−0.9

M1 — correct substitution into $A(t) = A_0 e^{-\gamma t}$A(t)=A0​e−γt with $\gamma t = 0.9$γt=0.9.

$e^{-0.9} \approx 0.4066$e−0.9≈0.4066, so $A(30) \approx 0.120 \times 0.4066 = 0.0488\,\text{m} \approx 4.9\,\text{cm}$A(30)≈0.120×0.4066=0.0488m≈4.9cm.

A1 — answer to 2 s.f. (or better) with units.

(b) Step 1 — relate energy to amplitude squared.

For SHM, $E = \tfrac{1}{2} k A^2$E=21​kA2, so $E \propto A^2$E∝A2.

M1 — stating or using $E \propto A^2$E∝A2.

Step 2 — form the energy ratio.

$\dfrac{E(30)}{E_0} = \left(\dfrac{A(30)}{A_0}\right)^2 = (e^{-0.9})^2 = e^{-1.8}$E0​E(30)​=(A0​A(30)​)2=(e−0.9)2=e−1.8.

M1 — squaring the amplitude ratio (equivalently doubling the exponent).

$e^{-1.8} \approx 0.165$e−1.8≈0.165, so about 16.5% of the initial energy remains.

A1 — final ratio quoted as a fraction or percentage.

(c) Step 1 — set the amplitude condition.

$\dfrac{A(t)}{A_0} = \dfrac{1}{2} \implies e^{-\gamma t} = 0.5$A0​A(t)​=21​⟹e−γt=0.5.

M1 — equation for the half-amplitude time.

Step 2 — solve for $t$t.

$-\gamma t = \ln(0.5) \implies t = \dfrac{\ln 2}{\gamma} = \dfrac{0.693}{0.030} \approx 23.1\,\text{s}$−γt=ln(0.5)⟹t=γln2​=0.0300.693​≈23.1s.

M1 — taking natural logs and isolating $t$t correctly.

Step 3 — convert to oscillations.

Number of oscillations $= t / T_0 = 23.1 / 2.0 \approx 11.5$=t/T0​=23.1/2.0≈11.5, so about 12 complete oscillations (the amplitude has just fallen below $A_0/2$A0​/2 at the 12th cycle).

A1 — sensible rounding to a whole number with brief justification.

Total: 8 marks.

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Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): A mass on a spring is set into vertical oscillation in air and then in a beaker of water. A motion sensor records the displacement-time graph in each case. The first graph shows decaying oscillations that persist for many cycles; the second shows the mass returning to equilibrium without oscillating, in roughly the same time as one period of the in-air motion.

(a) Identify the type of damping in each case and justify your answer using the displacement-time descriptions. (3)

(b) Explain, with reference to energy, why the amplitude of the in-air oscillation decreases between successive peaks. (3)

Mark scheme decomposition by AO:

(a)

• B1 (AO1.1) — in-air motion identified as light (under-)damping, because oscillations occur with gradually decreasing amplitude.
• B1 (AO1.1) — in-water motion identified as critical (or near-critical) damping, because the system returns to equilibrium without oscillating in approximately the shortest possible time.
• B1 (AO2.1) — explicit use of the data: "no overshoot, no oscillation, returns to equilibrium in a time comparable to one natural period" pinpoints critical rather than heavy damping.

(b)

• M1 (AO1.2) — work is done by the mass against drag/resistive forces.
• M1 (AO1.2) — kinetic and elastic potential energy are gradually converted to thermal energy in the surroundings.
• A1 (AO2.1) — since $E = \tfrac{1}{2}kA^2$E=21​kA2, a reduction in total mechanical energy must be accompanied by a reduction in amplitude.

Total: 6 marks split AO1 = 4, AO2 = 2. This pattern is typical: damping questions reward clean physical reasoning that links a graphical observation, an energy argument, and the SHM amplitude-energy relation in a single short paragraph.

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Synoptic links

Connects to:

1. Topic 13 — Simple harmonic motion (earlier sub-strand): damping is a perturbation of pure SHM. For light damping the equation of motion gains a velocity-dependent term, $m\ddot{x} + b\dot{x} + kx = 0$mx¨+bx˙+kx=0, and the solution becomes a sinusoid multiplied by an exponential envelope. The pure-SHM result $\omega_0 = \sqrt{k/m}$ω0​=k/m​ is recovered in the limit $b \to 0$b→0.

2. Topic 4 — Work, energy and power: the rate of energy dissipation by a damping force is $P_\text{diss} = bv^2$Pdiss​=bv2, exactly the "force $\times$× velocity" rule for power. Damping questions are work-energy questions in disguise.