Forced Oscillations and Resonance

When an external periodic force continuously drives an oscillating system, the resulting motion is called a forced oscillation. The frequency at which the driving force oscillates is the driving frequency (f_d), and the system's inherent frequency of free oscillation is the natural frequency (f₀). The relationship between these two frequencies determines the amplitude of the response, leading to one of the most important phenomena in physics: resonance.

Forced Oscillations

Consider a mass on a spring that is being driven by an external periodic force — perhaps the top of the spring is attached to a mechanical oscillator that moves up and down at a controllable frequency f_d.

When the driving force is first applied, there is a brief transient phase where the system adjusts. After this, the system settles into a steady state where:

The system oscillates at the driving frequency f_d, not at its natural frequency
The amplitude of oscillation depends on how close f_d is to f₀
The phase relationship between the driving force and the displacement depends on f_d

Resonance

Resonance occurs when the driving frequency equals the natural frequency:

f_d = f₀

At resonance:

The amplitude of oscillation is at its maximum
Energy transfer from the driving force to the system is at its most efficient
The system absorbs energy from the driver at the maximum rate

The physical reason for maximum amplitude at resonance is that the driving force is always in the right direction at the right time. It pushes the system when it is already moving in that direction, adding energy each cycle. The amplitude builds up until the energy input per cycle from the driver equals the energy dissipated per cycle by damping.

Resonance Curves

If you plot the amplitude of the forced oscillation against the driving frequency, you get a resonance curve (also called a frequency-response curve).

Features of the Resonance Curve

At low driving frequencies (f_d ≪ f₀), the amplitude is approximately equal to the static displacement (the displacement that the driving force would produce if applied steadily)
As f_d approaches f₀, the amplitude increases dramatically
At f_d = f₀, the amplitude reaches its peak
As f_d increases beyond f₀, the amplitude decreases
At very high driving frequencies (f_d ≫ f₀), the amplitude tends towards zero — the system cannot keep up with the rapid driving force

flowchart TD
    A["Resonance Curve Features"] --> B["f_d ≪ f₀"]
    A --> C["f_d = f₀\n(Resonance)"]
    A --> D["f_d ≫ f₀"]
    B --> B1["Amplitude ≈ static displacement\nPhase difference ≈ 0°\nSystem follows driver"]
    C --> C1["Amplitude = MAXIMUM\nPhase difference = 90°\nMax energy transfer"]
    D --> D1["Amplitude → 0\nPhase difference → 180°\nSystem cannot follow driver"]

Effect of Damping on Resonance

Damping has a profound effect on the resonance curve:

Light Damping

The resonance peak is tall and narrow (high amplitude, sharply tuned)
The peak occurs very close to f₀
The system responds strongly only over a narrow range of driving frequencies

Moderate Damping

The resonance peak is lower and broader
The peak shifts very slightly below f₀
The system responds over a wider range of frequencies but with lower maximum amplitude

Heavy Damping

The resonance peak is very low and very broad
There may be no obvious peak at all — the curve is almost flat
The system barely responds to resonance

Key Principle

Increasing damping reduces the resonance peak amplitude and broadens the resonance curve.

This is crucial in engineering: adding damping to a system reduces the danger of destructive resonance but also reduces the system's responsiveness at its natural frequency.

Resonance Curve Comparison Table

Property	Light Damping	Moderate Damping	Heavy Damping
Peak amplitude	Very high	Moderate	Very low
Peak width	Narrow	Medium	Very broad
Peak frequency	≈ f₀	Slightly below f₀	Ill-defined
Response selectivity	Very selective	Moderate	Poor
Q factor	High	Medium	Low

Phase Relationship

The phase difference between the driving force and the displacement of the system varies with driving frequency:

Driving Frequency	Phase Difference	Physical Meaning
f_d ≪ f₀	Close to 0°	Displacement nearly in phase with force — system tracks the driver
f_d = f₀ (resonance)	90°	Displacement lags force by quarter cycle — velocity in phase with force
f_d ≫ f₀	Close to 180°	Displacement nearly in antiphase — system opposes the driver

At resonance, the displacement is 90° behind the driving force. This means the velocity (which leads displacement by 90°) is in phase with the driving force. When the force and velocity are in phase, the power transfer is maximised — this is why resonance produces maximum amplitude.

Key insight for exams: The reason resonance gives maximum amplitude is that the velocity is in phase with the driving force at f_d = f₀. This means the force always acts in the direction of motion, so work is always done positively on the system. At other frequencies, the force sometimes opposes the motion, reducing net energy transfer.

Worked Example 1: Resonance Peak Analysis

A mass-spring system has natural frequency 4.0 Hz. When driven at resonance with light damping, the peak amplitude is 15 cm. When the damping is increased, the peak amplitude drops to 6 cm.

(a) At what frequency does the peak occur in both cases?

Approximately 4.0 Hz in both cases. The natural frequency depends on m and k, not on damping.

(b) The lightly damped system is driven at 2.0 Hz. The amplitude is 2.5 cm. Estimate the amplitude at 2.0 Hz with increased damping.

With increased damping, amplitudes away from resonance are less affected than the peak. The off-resonance amplitude might be approximately 2.0 cm (slightly reduced but not dramatically).

Worked Example 2: Phase at Resonance

Explain why a child on a swing should be pushed at the highest point of the backward swing for maximum effect.

Solution:

At the highest point of the backward swing, the child is momentarily stationary and about to move forward. Pushing at this instant means the force acts in the direction the child is about to move — the force and velocity will be in phase for the forward swing. This is equivalent to driving at the natural frequency with the correct phase relationship (90° between force and displacement, 0° between force and velocity), which maximises energy transfer. Pushing at any other point in the cycle would mean the force sometimes opposes the motion.

Examples of Resonance

Tacoma Narrows Bridge (1940)

The most famous example of destructive resonance in engineering. Wind flowing across the bridge created periodic vortices that drove the bridge at its natural frequency of torsional oscillation. The amplitude grew until the bridge tore itself apart. Modern bridge design incorporates damping and aerodynamic features to prevent this.

Tuning Forks

When you strike a tuning fork, it vibrates at its natural frequency. If you hold it near another tuning fork of the same frequency, the sound waves from the first fork drive the second at its natural frequency — resonance occurs, and the second fork begins to vibrate audibly. A tuning fork of a different frequency will not respond.

Microwave Ovens

The microwave frequency (2.45 GHz) is close to a natural frequency of the water molecule's rotational mode. Water molecules in food absorb microwave energy resonantly, converting it to thermal energy and heating the food.

MRI Scanners

Magnetic Resonance Imaging works by driving hydrogen nuclei in the body at their natural precession frequency (the Larmor frequency) using radio frequency pulses. The nuclei resonate, absorbing energy. When the pulse stops, they re-emit this energy as radio signals that are detected and used to construct an image.

Barton's Pendulums

A classic demonstration of resonance uses Barton's pendulums: several pendulums of different lengths are hung from the same horizontal string, along with one heavy "driver" pendulum. When the driver is set swinging, the pendulum whose length (and therefore natural frequency) matches the driver responds with the largest amplitude. The others oscillate with smaller amplitudes. This shows that resonance occurs when f_d = f₀, and the response decreases as the mismatch between driving and natural frequency increases.

The responding pendulum also oscillates approximately 90° out of phase with the driver — exactly as the theory predicts for resonance.

Common Exam Mistakes on Resonance

Mistake	Correction
"At resonance, amplitude is infinite"	Amplitude is limited by damping — it reaches a finite maximum
"The system oscillates at f₀ during forced oscillation"	The system oscillates at f_d (the driving frequency), not f₀
"Damping changes the natural frequency"	Damping changes the resonance peak height, not f₀ (which depends on m and k)
"At resonance, force and displacement are in phase"	At resonance, force and velocity are in phase; force leads displacement by 90°
"Resonance only occurs at exactly f₀"	There is significant response near f₀; the bandwidth depends on damping

A-Level Deep Dive: Forced Oscillations and Resonance

Spec mapping

Edexcel 9PH0 specification Topic 13 — Oscillations covers free, damped and forced oscillations and the conditions under which resonance occurs (refer to the official specification document for exact wording). The forced-oscillation strand asks candidates to describe the variation of amplitude with driving frequency, to identify the resonance peak at $f_d = f_0$ , to describe how damping alters peak height and bandwidth, and to recognise resonance in physical contexts ranging from Barton's pendulums to microwave heating. Although Topic 13 sits in Paper 2 alongside thermodynamics and nuclear physics, the synoptic links into Topic 5 (electric circuits — LC tuning), Topic 11 (waves — standing waves on strings), and Topic 12 (gravitational fields — pendulum period) mean resonance reasoning surfaces across the whole 9PH0 assessment. The Edexcel formula booklet supplies $T = 2\pi\sqrt{m/k}$ and $T = 2\pi\sqrt{l/g}$ but does not give a closed expression for the Q factor; candidates must reason about peak sharpness qualitatively from the resonance curve.

Worked example with full mark scheme

Question (8 marks):

A mass-spring system has natural frequency $f_0 = 5.0\,\text{Hz}$ . It is driven by a sinusoidal force of constant peak magnitude across a range of driving frequencies. With light air damping the peak amplitude at resonance is $A_1 = 24\,\text{cm}$ and the resonance curve has a full-width-at-half-maximum bandwidth of $\Delta f_1 = 0.40\,\text{Hz}$ . The same system is then immersed in a viscous oil and re-driven; the new peak amplitude is $A_2 = 6.0\,\text{cm}$ and the bandwidth widens to $\Delta f_2 = 1.6\,\text{Hz}$ .

(a) Sketch the two resonance curves on the same axes, labelling the natural frequency and both peaks. (3)

(b) Estimate the Q factor ( $Q \approx f_0 / \Delta f$ ) in each case, and comment on what your answers tell you about the energy dissipated per cycle. (3)

(c) Explain, in terms of the phase relationship between the driving force and the velocity, why both peaks occur at approximately the same driving frequency. (2)

Solution with mark scheme:

(a) Step 1 — sketch axes and curves. Amplitude on the vertical axis, driving frequency on the horizontal axis. Both curves rise from a small static-displacement value at low $f_d$ , peak near $f_0 = 5.0\,\text{Hz}$ , and fall toward zero at high $f_d$ .

M1 — both curves drawn with peaks at (or very close to) $f_d = 5.0\,\text{Hz}$ .

M1 — the lightly damped curve drawn taller and narrower than the heavily damped curve.

A1 — peaks correctly labelled with $A_1 = 24\,\text{cm}$ and $A_2 = 6.0\,\text{cm}$ , and the natural frequency $f_0$ marked on the horizontal axis.

A common slip is drawing the heavily damped peak displaced noticeably below $f_0$ . The peak does shift slightly below $f_0$ as damping increases, but at A-Level it is acceptable to treat both peaks as occurring at $f_0$ to a good approximation.

(b) Step 1 — apply $Q \approx f_0 / \Delta f$ .

$Q_1 \approx \dfrac{5.0}{0.40} = 12.5, \qquad Q_2 \approx \dfrac{5.0}{1.6} \approx 3.1$

M1 — correct substitution into the bandwidth definition for at least one case.

A1 — both Q values computed correctly.

Step 2 — interpret. A higher Q factor means less energy dissipated per cycle relative to the energy stored. The lightly damped system loses only a small fraction of its energy per oscillation, allowing amplitude to build to a sharp, tall peak. The viscous-oil case dissipates a much larger fraction per cycle, capping the peak amplitude.

A1 — explicit link between high Q and low fractional energy loss per cycle.

(c) Step 1 — phase argument. Resonance occurs when the velocity of the oscillator is in phase with the driving force, so the work done by the driver per cycle is maximised. This phase condition (force leading displacement by 90°) is set by the system's inertia and stiffness, which depend on $m$ and $k$ — not on damping. Damping primarily controls how much energy is dissipated per cycle, hence the peak height, while leaving the resonant frequency essentially unchanged.

M1 — identifies the in-phase force-and-velocity condition as the resonance criterion.

A1 — concludes that since $f_0 = (1/2\pi)\sqrt{k/m}$ depends only on $m$ and $k$ , the peak frequency is approximately unchanged by damping.

Total: 8 marks (M4 A4, split as shown).

Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): A vertical mass-spring system of mass $m = 0.20\,\text{kg}$ and spring constant $k = 80\,\text{N\,m}^{-1}$ is suspended in air and driven by a small motor whose driving frequency can be varied.

(a) Calculate the natural frequency of the system. (2)

(b) The motor is set to a driving frequency of $3.2\,\text{Hz}$ , then slowly increased to $3.2\,\text{Hz}$ above resonance, then decreased back through resonance. Describe and explain how the steady-state amplitude varies through this sweep, with reference to the phase relationship between driving force and displacement. (4)

Mark scheme decomposition by AO:

(a)

M1 (AO2.2) — using $f_0 = (1/2\pi)\sqrt{k/m}$ with the correct substitution $f_0 = (1/2\pi)\sqrt{80/0.20}$ .
A1 (AO1.2) — $f_0 \approx 3.18\,\text{Hz}$ (accept 3.2 Hz to 2 s.f.).

(b)

M1 (AO1.1) — identifies that the amplitude is small and the displacement is approximately in phase with the force at low $f_d$ .
M1 (AO1.2) — identifies that the amplitude rises sharply as $f_d \to f_0$ , peaking at resonance with displacement 90° behind force (so velocity in phase with force).
A1 (AO2.1) — identifies that beyond resonance the amplitude falls, and the displacement tends toward 180° out of phase (antiphase) with the force as $f_d$ becomes large.
A1 (AO3.2) — explicitly states that the amplitude curve traced through the sweep is symmetric about resonance (to a good approximation for light damping), and that the maximum energy transfer per cycle is what produces the peak.

Total: 6 marks split AO1 = 3, AO2 = 2, AO3 = 1. Edexcel Paper 2 questions on Topic 13 routinely combine a numerical opener (a) with an extended descriptive part (b) requiring the candidate to weave together amplitude variation, phase, and energy reasoning.

Synoptic links

Connects to:

Topic 13 — SHM and damping (within the same topic): the resonance curve is the amplitude response of a damped, driven harmonic oscillator. The lightly damped limit is the closest to the idealised undamped SHM described earlier in the topic.
Topic 5 — Electric circuits (LC and RLC resonance): an LC circuit oscillates at $f_0 = 1/(2\pi\sqrt{LC})$ , and an RLC circuit driven by an AC source has a resonance curve identical in shape to a mechanical oscillator's. The resistor R plays the role of damping; high-Q LC tank circuits are the heart of radio tuning.
Topic 11 — Standing waves: a string fixed at both ends has a discrete set of natural frequencies (the harmonics). When driven at one of these, large-amplitude standing waves appear — the same physics as resonance, distributed across a continuous medium.