Standing Waves

A standing wave is a wave pattern that appears to remain fixed in position rather than traveling through a medium. Standing waves are produced when two waves with the same frequency, wavelength, and approximately equal amplitude travel in opposite directions and interfere.

Standing waves are important in many vibrating systems, including:

• guitar, violin, and piano strings
• organ pipes and wind instruments
• air columns
• microwave cavities
• antennas
• bridges and buildings
• laboratory wave experiments

Standing waves also provide a clear connection between interference, resonance, wavelength, and frequency.

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How Standing Waves Form

Imagine a wave traveling along a stretched string.

When the wave reaches a fixed end, it reflects and travels back along the string. The incoming wave and reflected wave now travel in opposite directions.

The two waves overlap and interfere.

At certain frequencies, this interference creates a stable pattern in which some points remain stationary while other points oscillate strongly.

This pattern is a standing wave.

The individual waves are still traveling. It is the resulting interference pattern that appears stationary.

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Conditions for a Stable Standing Wave

A clear standing-wave pattern is produced when the two interfering waves have:

• the same frequency
• the same wavelength
• similar or equal amplitudes
• opposite directions of travel

The waves continuously undergo both constructive and destructive interference.

At particular locations, the interference is always destructive.

At other locations, the interference alternates to produce maximum oscillation.

These locations are called nodes and antinodes.

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Nodes

A node is a point on a standing wave where the displacement is always zero.

At a node, the two traveling waves interfere destructively.

If one wave attempts to displace the medium upward, the other produces an equal downward displacement.

The resulting displacement is:

zero

Nodes therefore appear stationary.

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Antinodes

An antinode is a point where the oscillation has maximum amplitude.

At an antinode, the interfering waves produce maximum constructive interference during the oscillation.

The particles of the medium move through the largest range.

Therefore:

• nodes → minimum displacement
• antinodes → maximum displacement

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Nodes and Antinodes Alternate

Along a standing wave, nodes and antinodes alternate.

A typical pattern is:

Node — Antinode — Node — Antinode — Node

The distance relationships are important.

Distance between adjacent nodes:

λ/2

Distance between adjacent antinodes:

λ/2

Distance from a node to the nearest antinode:

λ/4

where λ represents wavelength.

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Determining Wavelength from a Standing Wave

Suppose adjacent nodes are separated by:

0.40 m

Since adjacent nodes are separated by λ/2:

λ/2 = 0.40 m

Therefore:

λ = 0.80 m

If the wave speed is 120 m/s:

v = fλ

120 = f(0.80)

f = 150 Hz

Therefore:

f = 150 Hz

Standing-wave patterns can therefore be used to determine wavelength and frequency experimentally.

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Waves on a String

Consider a string fixed at both ends.

The ends cannot move, so both ends must be:

nodes

Only certain wavelengths can fit between these fixed endpoints.

This means the string cannot form a stable standing wave at every possible frequency.

Instead, it has a set of allowed frequencies called its natural frequencies.

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Fundamental Frequency

The simplest standing-wave pattern on a string is called the fundamental mode or first harmonic.

For a string fixed at both ends, the fundamental pattern contains:

• a node at each end
• one antinode in the middle
• half of one wavelength along the string

Therefore:

L = λ₁/2

so:

λ₁ = 2L

where:

• L = length of the string
• λ₁ = wavelength of the fundamental

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Fundamental Frequency of a String

Using:

v = fλ

and:

λ₁ = 2L

we obtain:

f₁ = v/(2L)

Therefore:

f₁ = v/(2L)

The fundamental frequency is the lowest natural frequency at which the string can form a standing wave.

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Example: Fundamental Frequency

A string is:

0.80 m

long.

Wave speed along the string is:

240 m/s

Fundamental wavelength:

λ₁ = 2L

λ₁ = 2(0.80)

λ₁ = 1.60 m

Now:

f₁ = v/λ₁

f₁ = 240/1.60

f₁ = 150 Hz

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Harmonics

Higher-frequency standing-wave patterns are called harmonics.

For a string fixed at both ends:

first harmonic = fundamental

second harmonic = 2 × fundamental frequency

third harmonic = 3 × fundamental frequency

and so on.

In general:

fₙ = nf₁

where n is the harmonic number.

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Harmonics on a String

For a string fixed at both ends:

The general wavelength relationship is:

λₙ = 2L/n

Therefore:

fₙ = nv/(2L)

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Understanding the Harmonic Number

For a string fixed at both ends, the harmonic number tells us how many half-wavelengths fit along the string.

First harmonic:

L = λ/2

Second harmonic:

L = 2(λ/2)

Third harmonic:

L = 3(λ/2)

In general:

L = nλ/2

This leads directly to:

λₙ = 2L/n

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Nodes and Antinodes in String Harmonics

For a string fixed at both ends:

First harmonic:

• 2 nodes
• 1 antinode

Second harmonic:

• 3 nodes
• 2 antinodes

Third harmonic:

• 4 nodes
• 3 antinodes

In general, for the nth harmonic:

• number of antinodes = n
• number of nodes = n + 1

including the fixed endpoints.

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Worked Example: Third Harmonic

A string has length:

L = 1.20 m

Find the wavelength of its third harmonic.

Use:

λₙ = 2L/n

λ₃ = 2(1.20)/3

λ₃ = 2.40/3

λ₃ = 0.80 m

If the wave speed is:

v = 200 m/s

then:

f₃ = v/λ₃

f₃ = 200/0.80

f₃ = 250 Hz

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Finding the Fundamental from a Higher Harmonic

Suppose the fourth harmonic of a string is:

480 Hz

Since:

f₄ = 4f₁

480 = 4f₁

Therefore:

f₁ = 120 Hz

The other harmonics would include:

f₂ = 240 Hz

f₃ = 360 Hz

f₄ = 480 Hz

f₅ = 600 Hz

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Standing Waves in Air Columns

Standing waves can also form in columns of air.

This occurs in musical instruments such as:

• flutes
• clarinets
• recorders
• trumpets
• trombones
• organ pipes

The boundary conditions depend on whether the end of the air column is:

• open
• closed

This changes the allowed standing-wave patterns.

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Open Ends and Closed Ends

For the common displacement representation of air-column standing waves:

At an open end, air particles can move freely.

Therefore, an open end is a:

displacement antinode

At a closed end, air particles cannot move freely.

Therefore, a closed end is a:

displacement node

This distinction determines which wavelengths fit inside the air column.

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Pressure Nodes and Antinodes

Sound waves can also be described using variations in pressure.

For pressure standing waves:

• an open end is approximately a pressure node
• a closed end is a pressure antinode

Therefore, it is important to know whether a diagram represents:

particle displacement

or:

pressure variation

The node and antinode locations are reversed between these two descriptions.

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Open-Open Air Columns

An air column open at both ends has displacement antinodes at both ends.

For the fundamental mode:

L = λ₁/2

Therefore:

λ₁ = 2L

and:

f₁ = v/(2L)

This is the same mathematical relationship as a string fixed at both ends, even though the boundary conditions are physically different.

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Harmonics in an Open-Open Pipe

An ideal open-open pipe supports:

• first harmonic
• second harmonic
• third harmonic
• fourth harmonic
• and so on

The relationships are:

λₙ = 2L/n

and:

fₙ = nv/(2L)

Therefore:

fₙ = nf₁

All integer harmonics are possible in the ideal model.

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Example: Open Pipe

An open pipe has length:

L = 0.85 m

Take the speed of sound as:

v = 340 m/s

Fundamental wavelength:

λ₁ = 2L

λ₁ = 1.70 m

Fundamental frequency:

f₁ = 340/1.70

f₁ = 200 Hz

Second harmonic:

f₂ = 400 Hz

Third harmonic:

f₃ = 600 Hz

Fourth harmonic:

f₄ = 800 Hz

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Closed-Open Air Columns

Now consider an air column that is:

• closed at one end
• open at the other

For displacement:

• closed end = node
• open end = antinode

The shortest standing-wave pattern that satisfies these conditions contains:

one quarter of a wavelength

Therefore:

L = λ₁/4

so:

λ₁ = 4L

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Fundamental Frequency of a Closed-Open Pipe

Using:

v = fλ

and:

λ₁ = 4L

we obtain:

f₁ = v/(4L)

This is half the fundamental frequency of an open-open pipe of the same length, assuming the same wave speed.

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Harmonics in a Closed-Open Pipe

A closed-open pipe supports only the odd harmonics in the ideal model.

Its frequencies are:

f₁, 3f₁, 5f₁, 7f₁, ...

The allowed frequencies can be written:

fₙ = nv/(4L)

where:

n = 1, 3, 5, 7, ...

The allowed wavelengths are:

λₙ = 4L/n

for odd n.

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Why Even Harmonics Are Missing

A closed-open pipe must always have:

• a displacement node at the closed end
• a displacement antinode at the open end

Patterns corresponding to the even harmonics of the fundamental cannot satisfy both boundary conditions simultaneously.

Therefore, the allowed modes correspond to:

1st harmonic

3rd harmonic

5th harmonic

7th harmonic

and so on.

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Important Harmonic Numbering

The second allowed mode of a closed-open pipe is actually the:

third harmonic

The third allowed mode is the:

fifth harmonic

This distinction is important.

For example:

First mode → f₁

Second mode → 3f₁

Third mode → 5f₁

Do not automatically assume that the second mode means the second harmonic.

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Example: Closed Pipe

A closed-open pipe has length:

L = 0.50 m

Take:

v = 340 m/s

Fundamental frequency:

f₁ = v/(4L)

f₁ = 340/[4(0.50)]

f₁ = 340/2

f₁ = 170 Hz

The next allowed frequency is the third harmonic:

f₃ = 3(170)

f₃ = 510 Hz

The next is:

f₅ = 5(170)

f₅ = 850 Hz

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Comparing Standing-Wave Systems

This table is one of the most useful summaries for solving standing-wave problems.

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Resonance

Every vibrating system has one or more natural frequencies.

If an external periodic force drives the system at or near one of these natural frequencies, the system can oscillate with a much larger amplitude.

This phenomenon is called:

resonance

Examples include:

• pushing a playground swing at the correct rhythm
• vibrating a guitar string
• blowing across an air column
• tuning a radio circuit
• vibration of mechanical structures

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Natural Frequency

A natural frequency is a frequency at which a system tends to oscillate when disturbed.

For example, if a stretched string is plucked, it naturally vibrates using combinations of its allowed frequencies.

These frequencies are determined by physical properties such as:

• length
• tension
• mass
• shape
• material
• boundary conditions

Different systems therefore have different natural frequencies.

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Driving Frequency

A driving frequency is the frequency of an external force acting repeatedly on a system.

Suppose a swing has a natural frequency.

If someone pushes it randomly, the motion may remain relatively small.

If the pushes occur at an appropriate frequency and phase, energy is transferred efficiently to the swing.

The amplitude increases.

This is resonance.

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Maximum Amplitude at Resonance

When the driving frequency is close to the natural frequency:

f_drive ≈ f_natural

the amplitude becomes large.

In an idealized system:

maximum response occurs at resonance

Real systems experience damping, so the exact frequency of maximum amplitude can differ slightly from the undamped natural frequency.

For introductory problems, it is normally treated as occurring at the natural frequency.

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Energy Transfer During Resonance

Resonance occurs because energy is transferred efficiently from the driving source to the oscillating system.

If the driving force repeatedly acts in a way that reinforces the motion, the system gains energy.

The amplitude increases until energy supplied by the driving force is balanced by energy losses.

These losses may result from:

• friction
• air resistance
• internal deformation
• sound production
• thermal energy transfer

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Damping

Damping is the loss of mechanical energy from an oscillating system.

Greater damping generally:

• reduces the maximum resonant amplitude
• makes the resonance less sharp
• causes oscillations to die away more quickly

A lightly damped system can produce a much larger resonant response than a heavily damped system.

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Resonance Curve

If amplitude is plotted against driving frequency, the graph typically shows a peak near the natural frequency.

Far from resonance:

amplitude is relatively small

Near resonance:

amplitude increases

At resonance:

amplitude reaches its maximum

The width and height of the peak depend strongly on damping.

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Resonance and Musical Instruments

Musical instruments depend heavily on resonance.

When a guitar string is plucked, the string vibrates at its natural frequencies.

The body of the guitar also vibrates and transfers energy to the surrounding air.

In wind instruments, the air column resonates at particular frequencies determined partly by:

• tube length
• whether the ends are open or closed
• air temperature
• instrument geometry

This produces the instrument's characteristic notes.

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Harmonics and Musical Tone

Two instruments can play the same fundamental frequency but sound different.

This occurs partly because they produce different combinations and strengths of harmonics.

The fundamental largely determines the perceived pitch.

The mixture of harmonics contributes to the instrument's:

timbre

or tone quality.

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Changing the Length of a String

For a string:

f₁ = v/(2L)

If wave speed remains constant, decreasing L increases the fundamental frequency.

Therefore:

shorter string → higher frequency

longer string → lower frequency

This is why pressing a guitar string against a fret changes its pitch.

The vibrating portion of the string becomes shorter.

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Changing Wave Speed on a String

Wave speed on a stretched string depends on tension and linear mass density.

A commonly used relationship is:

v = √(T/μ)

where:

• T = tension
• μ = mass per unit length

Increasing tension increases wave speed.

Since:

f₁ = v/(2L)

increasing the tension generally increases the resonant frequencies.

This is why tightening a musical string raises its pitch.

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Changing the Length of an Air Column

For an open-open pipe:

f₁ = v/(2L)

For a closed-open pipe:

f₁ = v/(4L)

In both cases:

longer air column → lower fundamental frequency

shorter air column → higher fundamental frequency

This principle is used in wind instruments.

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Speed of Sound and Temperature

The speed of sound in air depends on temperature.

At typical room conditions, it is approximately:

340–343 m/s

As air temperature increases, the speed of sound increases.

Since:

f = v/λ

changes in sound speed can slightly change the resonant frequencies of air columns.

This is one reason temperature can affect the tuning of some instruments.

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Worked Problem: String Harmonics

A string is:

L = 1.50 m

long and has wave speed:

v = 180 m/s

Find the fundamental frequency.

f₁ = v/(2L)

f₁ = 180/[2(1.50)]

f₁ = 180/3

f₁ = 60 Hz

Third harmonic:

f₃ = 3f₁

f₃ = 3(60)

f₃ = 180 Hz

Third-harmonic wavelength:

λ₃ = v/f₃

λ₃ = 180/180

λ₃ = 1.0 m

Check:

λ₃ = 2L/3

λ₃ = 3/3

λ₃ = 1.0 m

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Worked Problem: Open Air Column

An open-open pipe is:

L = 0.75 m

long.

Take:

v = 330 m/s

Fundamental frequency:

f₁ = 330/[2(0.75)]

f₁ = 330/1.50

f₁ = 220 Hz

Third harmonic:

f₃ = 3(220)

f₃ = 660 Hz

Third-harmonic wavelength:

λ₃ = 2L/3

λ₃ = 1.50/3

λ₃ = 0.50 m

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Worked Problem: Closed-Open Air Column

A pipe closed at one end has:

L = 0.40 m

and:

v = 320 m/s

Fundamental frequency:

f₁ = v/(4L)

f₁ = 320/[4(0.40)]

f₁ = 320/1.60

f₁ = 200 Hz

The next allowed harmonic is the third:

f₃ = 3(200)

f₃ = 600 Hz

The next is the fifth:

f₅ = 5(200)

f₅ = 1000 Hz

There is no second or fourth harmonic in the ideal closed-open model.

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Worked Problem: Identifying a Harmonic from Nodes

A string fixed at both ends shows:

5 antinodes

Since the nth harmonic has n antinodes:

n = 5

Therefore, the string is vibrating in its:

fifth harmonic

It will also contain:

6 nodes

including the two endpoints.

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Worked Problem: Finding Wavelength from Nodes

Adjacent nodes in a standing wave are:

0.25 m apart

Since:

node-to-node distance = λ/2

then:

λ/2 = 0.25

λ = 0.50 m

If:

v = 150 m/s

then:

f = v/λ

f = 150/0.50

f = 300 Hz

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Standing Waves Do Not Transport Energy Like Traveling Waves

A traveling wave carries energy from one region to another.

In an ideal standing wave, there is no net energy transfer along the medium averaged over time.

Energy is instead exchanged locally between different forms as parts of the medium oscillate.

This is another important distinction between traveling and standing waves.

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Standing Waves in Structures

Standing-wave and resonance effects are important in engineering.

Structures such as:

• bridges
• towers
• aircraft
• machines
• buildings

have natural frequencies.

If repeated forces occur near one of those frequencies, large oscillations may develop.

Engineers therefore analyze natural frequencies and resonance when designing structures.

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Preventing Unwanted Resonance

Engineers can reduce dangerous resonance by:

• increasing damping
• changing the mass of the system
• changing stiffness
• changing dimensions
• altering natural frequencies
• isolating vibrations

The goal is often to prevent external driving frequencies from matching important natural frequencies.

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Useful Problem-Solving Relationships

For all waves:

v = fλ

For a string fixed at both ends:

λₙ = 2L/n

fₙ = nv/(2L)

For an open-open pipe:

λₙ = 2L/n

fₙ = nv/(2L)

For a closed-open pipe:

λₙ = 4L/n

fₙ = nv/(4L)

where n = 1, 3, 5, 7...

Distance between adjacent nodes:

λ/2

Distance between adjacent antinodes:

λ/2

Distance from node to nearest antinode:

λ/4

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A Reliable Strategy for Standing-Wave Problems

First identify the system:

• string fixed at both ends
• open-open pipe
• closed-open pipe

Then determine the boundary conditions.

Ask:

• Where must nodes occur?
• Where must antinodes occur?
• Which harmonics are allowed?

Next determine how much of a wavelength fits into the length L.

Then:

1. calculate λ
2. use v = fλ
3. calculate f
4. check that the harmonic number is allowed
5. include correct units

Drawing the standing-wave pattern can make the problem much easier.

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Common Mistakes

Thinking a standing wave is not made from traveling waves

Standing waves result from interference between waves traveling in opposite directions.

Confusing nodes and antinodes

Nodes have zero displacement. Antinodes have maximum displacement.

Thinking adjacent nodes are one wavelength apart

Adjacent nodes are separated by:

λ/2

Thinking a node and neighboring antinode are λ/2 apart

They are separated by:

λ/4

Assuming L = λ for the fundamental of a fixed string

For the first harmonic:

L = λ/2

so:

λ = 2L

Using open-open formulas for a closed-open pipe

Closed-open fundamental:

λ = 4L

not 2L.

Including even harmonics in an ideal closed-open pipe

Only odd harmonics occur:

1, 3, 5, 7...

Calling the second allowed mode of a closed-open pipe the second harmonic

It is the:

third harmonic

Confusing displacement and pressure nodes in air columns

At an open end there is approximately a displacement antinode but a pressure node.

Assuming resonance means frequency increases

Resonance describes a large response when the driving frequency matches or approaches a natural frequency.

Ignoring units

Length should normally be converted to metres when using wave speed in m/s.

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Key Terms

Standing wave — stationary interference pattern produced by waves of the same frequency traveling in opposite directions.

Interference — combination of overlapping waves.

Constructive interference — interference that produces increased displacement.

Destructive interference — interference that reduces or cancels displacement.

Node — point in a standing wave where displacement remains zero.

Antinode — point where the oscillation has maximum amplitude.

Fundamental frequency — lowest natural frequency of a vibrating system.

First harmonic — fundamental mode of vibration.

Harmonic — allowed frequency related to the fundamental frequency.

Natural frequency — frequency at which a system naturally tends to oscillate.

Resonance — large-amplitude response produced when a system is driven near one of its natural frequencies.

Driving frequency — frequency of an external periodic force.

Damping — reduction of oscillation caused by energy loss.

Air column — column of air capable of supporting sound-wave standing patterns.

Open-open pipe — air column open at both ends.

Closed-open pipe — air column closed at one end and open at the other.

Boundary condition — physical restriction determining whether a node or antinode must occur at a boundary.

Displacement node — location where particle displacement is zero.

Pressure node — location where sound-pressure variation is minimal.

Mode — particular allowed standing-wave pattern of a system.

Timbre — characteristic quality of a sound influenced by its harmonic content.

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Key Takeaways

• Standing waves form when waves of the same frequency and wavelength travel in opposite directions and interfere.
• Nodes are locations of zero displacement.
• Antinodes are locations of maximum displacement.
• Adjacent nodes are separated by λ/2.
• Adjacent antinodes are separated by λ/2.
• A node and its nearest antinode are separated by λ/4.
• Strings fixed at both ends have nodes at both ends.
• The fundamental wavelength of a fixed string is λ₁ = 2L.
• The fundamental frequency of a fixed string is f₁ = v/(2L).
• Fixed strings support all integer harmonics.
• For strings, fₙ = nf₁ and λₙ = 2L/n.
• An ideal open-open air column also supports all integer harmonics.
• An open end is a displacement antinode and approximately a pressure node.
• A closed end is a displacement node and approximately a pressure antinode.
• A closed-open pipe has fundamental wavelength λ₁ = 4L.
• A closed-open pipe has fundamental frequency f₁ = v/(4L).
• Ideal closed-open pipes support only odd harmonics.
• Resonance occurs when a system is driven near one of its natural frequencies.
• Resonance produces a large amplitude because energy is transferred efficiently to the oscillating system.
• Damping limits resonant amplitude.
• Standing waves explain important aspects of musical instruments, acoustics, engineering, and mechanical vibration.

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Check Your Understanding

1. Explain how two traveling waves can combine to produce a standing wave.
2. Describe the difference between a node and an antinode.
3. The distance between two adjacent nodes is 0.60 m. Determine the wavelength.
4. A node and its nearest antinode are 0.20 m apart. Determine the wavelength.
5. A string fixed at both ends is 1.2 m long. Determine the wavelength of its fundamental mode.
6. A 0.80 m string has a wave speed of 240 m/s. Calculate its fundamental frequency.
7. The fundamental frequency of a string is 120 Hz. Determine the frequencies of its second, third, and fifth harmonics.
8. A string fixed at both ends displays four antinodes. Which harmonic is present? How many nodes are there?
9. A string is 1.5 m long and supports waves traveling at 300 m/s. Calculate the wavelength and frequency of its third harmonic.
10. An open-open pipe is 0.50 m long. Using a sound speed of 340 m/s, calculate its fundamental frequency and second harmonic frequency.
11. A closed-open pipe is 0.50 m long. Using a sound speed of 340 m/s, calculate its fundamental frequency and the next two allowed harmonic frequencies.
12. Explain why an ideal closed-open pipe does not support the second harmonic.
13. Explain the difference between displacement nodes and pressure nodes at the open and closed ends of an air column.
14. Describe resonance in terms of natural frequency, driving frequency, energy transfer, and amplitude.
15. A vibrating system has a natural frequency of 250 Hz. Predict what happens to its amplitude as a driving frequency is gradually increased from 100 Hz through 250 Hz and then to 400 Hz. Explain how damping would affect the maximum response.

Waves on Strings

A stretched string can carry transverse waves. When the string is fixed at both ends, waves traveling along the string reflect from the ends and interfere with one another. At certain frequencies, this interference produces stable standing-wave patterns.

This behaviour explains how many stringed musical instruments produce sound, including:

• guitars
• violins
• cellos
• harps
• pianos

The frequencies produced by a string depend mainly on its length, tension, and mass per unit length.

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Waves Traveling Along a String

If one end of a stretched string is disturbed, a transverse wave travels along it.

The particles of the string move perpendicular to the direction in which the wave travels.

For example, the wave may travel horizontally while sections of the string move vertically.

The speed of the wave depends on the physical properties of the string rather than simply on how quickly the source vibrates.

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Reflection at a Fixed End

When a traveling pulse reaches a fixed end, it reflects.

A reflected pulse at a fixed end is inverted.

If the original pulse is upward, the reflected pulse is downward.

With continuous waves, the incident and reflected waves travel in opposite directions along the same string.

They then overlap and interfere.

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Formation of a Standing Wave

A standing wave forms when two waves with the same frequency and wavelength travel in opposite directions and interfere.

On a string fixed at both ends:

• the original wave travels toward an end
• the wave reflects
• the reflected wave travels back
• the two waves overlap
• interference produces nodes and antinodes

At particular frequencies, the reflected wave reinforces a stable standing-wave pattern.

These frequencies are the natural frequencies of the string.

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Nodes

A node is a point on a standing wave that remains at zero displacement.

The two ends of a string fixed at both ends must always be nodes because the string cannot move there.

At a node, the two traveling waves continuously interfere destructively.

Nodes can also occur between the ends of the string in higher harmonics.

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Antinodes

An antinode is a point where the string oscillates with maximum amplitude.

Antinodes occur halfway between neighboring nodes.

Therefore:

node → antinode → node → antinode → node

The string does not physically remain in the curved shape shown in a standing-wave diagram. Instead, it oscillates between opposite extreme positions.

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Spacing of Nodes and Antinodes

For a standing wave:

Adjacent nodes are separated by:

λ/2

Adjacent antinodes are separated by:

λ/2

A node and its nearest antinode are separated by:

λ/4

These relationships allow wavelength to be determined from a standing-wave pattern.

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Example: Finding Wavelength from Nodes

Two neighboring nodes are:

0.35 m apart

Since:

node spacing = λ/2

then:

λ/2 = 0.35

Therefore:

λ = 0.70 m

If the wave speed is 140 m/s:

v = fλ

140 = f(0.70)

f = 200 Hz

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Why Only Certain Standing Waves Are Possible

Both ends of the string must be nodes.

Therefore, the string length must contain an exact whole number of half-wavelengths.

The condition is:

L = nλ/2

where:

• L = string length
• λ = wavelength
• n = 1, 2, 3, 4, ...

This boundary condition means that only certain wavelengths and frequencies can form stable standing waves.

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The Fundamental Mode

The simplest possible standing wave has:

• a node at each end
• one antinode
• one half-wavelength along the string

This is called the:

fundamental mode

or:

first harmonic

For the fundamental:

L = λ₁/2

Therefore:

λ₁ = 2L

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Fundamental Frequency

All waves obey:

v = fλ

For the fundamental:

λ₁ = 2L

Therefore:

f₁ = v/(2L)

So:

f₁ = v/(2L)

The fundamental frequency is the lowest natural frequency of the string.

It usually provides the main perceived pitch of a vibrating musical string.

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Example: Fundamental Frequency

A string has:

L = 0.75 m

and wave speed:

v = 240 m/s

Fundamental wavelength:

λ₁ = 2L

λ₁ = 2(0.75)

λ₁ = 1.50 m

Frequency:

f₁ = v/λ₁

f₁ = 240/1.50

f₁ = 160 Hz

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Higher Harmonics

A string can also vibrate with more than one half-wavelength fitting along its length.

These patterns are called higher harmonics.

Second harmonic:

two half-wavelengths fit along the string

Third harmonic:

three half-wavelengths fit along the string

Fourth harmonic:

four half-wavelengths fit along the string

In general:

L = nλₙ/2

---

Wavelength of the nth Harmonic

Rearranging:

L = nλₙ/2

gives:

λₙ = 2L/n

As harmonic number increases, wavelength decreases.

For example:

---

Frequency of the nth Harmonic

Since:

v = fλ

and:

λₙ = 2L/n

then:

fₙ = nv/(2L)

Because:

f₁ = v/(2L)

we can also write:

fₙ = nf₁

Therefore:

f₂ = 2f₁

f₃ = 3f₁

f₄ = 4f₁

and so on.

---

Harmonic Patterns

For a string fixed at both ends:

The two fixed ends are included when counting nodes.

---

Identifying a Harmonic from a Diagram

Suppose a standing-wave pattern contains:

4 antinodes

For a string fixed at both ends:

number of antinodes = harmonic number

Therefore:

n = 4

The pattern is the:

fourth harmonic

It contains:

5 nodes

including the endpoints.

---

Worked Harmonic Example

A string is:

L = 1.20 m

and waves travel along it at:

v = 180 m/s

Find the fundamental frequency.

f₁ = v/(2L)

f₁ = 180/[2(1.20)]

f₁ = 180/2.40

f₁ = 75 Hz

Third harmonic:

f₃ = 3f₁

f₃ = 3(75)

f₃ = 225 Hz

Third-harmonic wavelength:

λ₃ = 2L/3

λ₃ = 2(1.20)/3

λ₃ = 0.80 m

Check:

v = fλ

v = 225(0.80)

v = 180 m/s

---

Wave Speed on a String

The wave speed on a stretched string depends mainly on:

• tension
• mass per unit length

The relationship is:

v = √(T/μ)

where:

• v = wave speed in m/s
• T = tension in N
• μ = mass per unit length in kg/m

The symbol μ is the Greek letter mu.

---

Mass per Unit Length

Mass per unit length is also called linear mass density.

It describes how much mass is contained in each metre of string.

The relationship is:

μ = m/L

where:

• m = mass of the string
• L = length of the string

Units:

kg/m

A thicker or denser string generally has a larger μ than a thin, light string.

---

Example: Linear Mass Density

A 2.0 m string has mass:

0.010 kg

Then:

μ = m/L

μ = 0.010/2.0

μ = 0.0050 kg/m

This value can then be used to calculate wave speed.

---

Effect of Tension on Wave Speed

Wave speed is:

v = √(T/μ)

If μ remains constant:

increasing tension increases wave speed

A tightly stretched string transmits disturbances faster than the same string under lower tension.

However, because T is inside a square root, doubling tension does not double wave speed.

---

Example: Changing Tension

Suppose the tension increases by a factor of 4.

Since:

v ∝ √T

then:

v_new/v_old = √4

Therefore:

v_new = 2v_old

The wave speed doubles.

---

Effect of Linear Mass Density on Wave Speed

From:

v = √(T/μ)

we see:

v ∝ 1/√μ

Therefore:

larger μ → lower wave speed

smaller μ → higher wave speed

A heavier string per unit length carries waves more slowly when tension is unchanged.

---

Example: Changing Linear Mass Density

Suppose μ becomes four times larger while tension remains unchanged.

Then:

v_new/v_old = 1/√4

Therefore:

v_new = 1/2 v_old

The wave speed is halved.

---

Combining Wave Speed and Fundamental Frequency

We know:

f₁ = v/(2L)

and:

v = √(T/μ)

Substituting gives:

f₁ = (1/2L)√(T/μ)

This equation shows the three major factors controlling the fundamental frequency of a string:

• length L
• tension T
• linear mass density μ

---

Effect of String Length on Frequency

From:

f₁ = (1/2L)√(T/μ)

if tension and linear mass density remain constant:

f₁ ∝ 1/L

Therefore:

longer string → lower frequency

shorter string → higher frequency

If the vibrating length is halved:

frequency doubles

---

Example: Changing Length

A string has fundamental frequency:

220 Hz

Its vibrating length is reduced to half while tension and μ remain constant.

Since:

f ∝ 1/L

the new frequency is:

2(220)

= 440 Hz

Halving the vibrating length raises the frequency by one octave.

---

Effect of Tension on Frequency

If L and μ remain constant:

f ∝ √T

Therefore:

higher tension → higher frequency

lower tension → lower frequency

If tension becomes four times larger:

frequency changes by:

√4 = 2

Therefore:

frequency doubles

---

Effect of Mass per Unit Length on Frequency

If L and T remain constant:

f ∝ 1/√μ

Therefore:

heavier string → lower frequency

lighter string → higher frequency

If μ becomes four times larger:

frequency becomes:

1/√4 = 1/2

Therefore:

frequency is halved

---

Comparing the Three Factors

The complete relationship is:

f₁ = (1/2L)√(T/μ)

---

Proportional Reasoning

Suppose the string length doubles.

Since:

f ∝ 1/L

the frequency becomes:

1/2 as large

Suppose tension increases by a factor of 9.

Since:

f ∝ √T

the frequency becomes:

3 times larger

Suppose μ increases by a factor of 9.

Since:

f ∝ 1/√μ

the frequency becomes:

1/3 as large

Recognizing these proportional relationships can often avoid lengthy calculations.

---

Worked Example: Wave Speed from Tension

A string has:

T = 100 N

μ = 0.0040 kg/m

Calculate wave speed.

v = √(T/μ)

v = √(100/0.0040)

v = √25,000

v ≈ 158 m/s

---

Worked Example: Fundamental Frequency from String Properties

A string has:

L = 0.80 m

T = 144 N

μ = 0.0040 kg/m

First calculate wave speed:

v = √(144/0.0040)

v = √36,000

v ≈ 190 m/s

Now:

f₁ = v/(2L)

f₁ = 190/[2(0.80)]

f₁ = 190/1.60

f₁ ≈ 119 Hz

---

Worked Example: Higher Harmonic

Using the same string with:

f₁ ≈ 119 Hz

find the fourth harmonic.

f₄ = 4f₁

f₄ ≈ 4(119)

f₄ ≈ 476 Hz

Alternatively:

f₄ = 4v/(2L)

which gives the same result.

---

Solving for Tension

The wave-speed equation can be rearranged.

Starting with:

v = √(T/μ)

Square both sides:

v² = T/μ

Therefore:

T = μv²

This allows the required tension to be calculated when wave speed and linear mass density are known.

---

Example: Finding Tension

A string carries waves at:

v = 200 m/s

and has:

μ = 0.0030 kg/m

Then:

T = μv²

T = 0.0030(200²)

T = 0.0030(40,000)

T = 120 N

---

Finding Tension from Frequency

Suppose:

L = 0.60 m

μ = 0.0050 kg/m

and the desired fundamental frequency is:

f₁ = 200 Hz

First use:

v = 2Lf₁

v = 2(0.60)(200)

v = 240 m/s

Then:

T = μv²

T = 0.0050(240²)

T = 0.0050(57,600)

T = 288 N

---

String Length and Musical Pitch

Stringed instruments use the relationship between length and frequency.

On a guitar or violin, pressing a string against the fingerboard or fretboard reduces the length of string that is free to vibrate.

Since:

f ∝ 1/L

a shorter vibrating length produces a higher frequency.

This allows one string to produce many different notes.

---

Guitar Strings

A guitar string is fixed approximately at:

• the bridge
• the nut or selected fret

These locations act as boundaries for the vibrating portion of the string.

When the player presses the string against a fret:

• vibrating length decreases
• fundamental wavelength decreases
• fundamental frequency increases
• pitch rises

This is one of the main ways guitarists select notes.

---

Guitar Tuning

A guitar's tuning pegs change the tension in the strings.

Increasing tension:

• increases wave speed
• increases fundamental frequency
• raises pitch

Decreasing tension:

• decreases wave speed
• decreases frequency
• lowers pitch

The player adjusts tension until the string reaches the required frequency.

---

Different Guitar Strings

Guitar strings are not all identical.

Lower-pitched strings generally have greater mass per unit length.

Since:

f ∝ 1/√μ

a larger μ helps produce a lower fundamental frequency without requiring an extremely long string.

This is why bass strings are usually thicker than treble strings.

---

Violin Strings

A violin operates using the same physical principles.

The musician changes pitch by pressing a string against the fingerboard.

This changes the effective vibrating length.

Unlike a guitar, a violin typically has no frets, so the player must control the vibrating length very precisely using finger position.

Small changes in length produce changes in frequency and therefore pitch.

---

Piano Strings

A piano also uses stretched strings, but the strings are struck by hammers rather than plucked or bowed.

Different notes use strings with different combinations of:

• length
• tension
• linear mass density

Low-frequency piano strings are often:

• longer
• heavier per unit length
• wound with additional material

High-frequency strings are generally:

• shorter
• thinner

These design choices allow a piano to produce a very large range of frequencies.

---

Fundamental and Harmonics in Musical Sound

When a string is plucked, struck, or bowed, it generally does not vibrate in only one mode.

Several standing-wave modes can exist at the same time.

The resulting sound can contain:

• fundamental frequency
• second harmonic
• third harmonic
• fourth harmonic
• higher harmonics

The fundamental strongly influences the perceived pitch.

The relative strengths of the harmonics help determine the instrument's timbre.

---

Harmonics and Timbre

Suppose two instruments both play:

440 Hz

Both may have the same fundamental frequency.

However, one might have strong second and fourth harmonics while another has stronger third and fifth harmonics.

The waveforms and sounds will therefore be different.

This is one reason a violin and guitar can play the same note but remain easy to distinguish.

---

Harmonic Nodes on Musical Strings

Musicians can deliberately touch a string lightly at particular positions to force a node to occur there.

This suppresses some vibration modes while allowing particular harmonics to become prominent.

For example, lightly touching the midpoint of a string creates a node there.

The fundamental cannot exist because its midpoint is an antinode.

The second harmonic, however, naturally has a node at the midpoint.

This allows the second harmonic to be emphasized.

---

Resonance and the Instrument Body

A vibrating string alone moves relatively little air and therefore produces limited sound.

Musical instruments use other structures to improve sound production.

For example:

• a guitar uses its body and soundboard
• a violin uses its body
• a piano uses a large soundboard

The vibrating string transfers energy to these structures.

Their vibrations move more air and produce a much stronger sound.

Resonance plays an important role in this energy transfer.

---

Why Instruments Need a Soundboard or Body

A thin string has a small surface area.

Although it vibrates rapidly, it does not push a large volume of air.

When its vibration is transferred to a larger surface:

• more air is moved
• stronger pressure variations are produced
• the sound becomes louder

The instrument body also affects which frequencies are amplified most strongly, contributing to timbre.

---

Real Strings vs Ideal Strings

The simple equations assume an ideal string that is:

• perfectly flexible
• uniform
• under constant tension
• fixed exactly at both ends

Real strings are not perfectly ideal.

Factors such as:

• stiffness
• thickness
• imperfect supports
• material properties

can cause small deviations from the simple harmonic model.

Nevertheless, the ideal model provides an excellent approximation for many situations.

---

A Complete String Problem

A string has:

L = 1.00 m

T = 180 N

μ = 0.0050 kg/m

First calculate wave speed:

v = √(T/μ)

v = √(180/0.0050)

v = √36,000

v ≈ 190 m/s

Fundamental frequency:

f₁ = v/(2L)

f₁ = 190/2

f₁ ≈ 95 Hz

Third harmonic:

f₃ = 3f₁

f₃ ≈ 285 Hz

Third-harmonic wavelength:

λ₃ = 2L/3

λ₃ ≈ 0.667 m

---

A Complete Comparison Problem

Two strings have the same length and linear mass density.

String B has four times the tension of String A.

Since:

f ∝ √T

then:

f_B/f_A = √(T_B/T_A)

f_B/f_A = √4

f_B/f_A = 2

Therefore:

String B has twice the fundamental frequency of String A.

---

Changing More Than One Variable

Sometimes several string properties change simultaneously.

Suppose:

• string length is doubled
• tension becomes four times larger
• μ remains unchanged

Since:

f ∝ √T/L

the tension change multiplies frequency by:

√4 = 2

The length change multiplies frequency by:

1/2

Overall:

2 × 1/2 = 1

Therefore:

the fundamental frequency remains unchanged

This type of reasoning is useful when comparing strings.

---

Comparing Two Strings

For two strings:

f₁ = (1/2L)√(T/μ)

Therefore, frequency depends on all three variables.

A shorter string does not necessarily have a higher frequency if its tension or linear mass density is also different.

When comparing real strings, consider:

• L
• T
• μ

rather than looking at only one variable.

---

A Reliable Strategy for String Problems

First identify what is known:

• L
• T
• μ
• v
• λ
• f
• harmonic number n

Then determine which relationship is most useful.

For wave speed:

v = √(T/μ)

For all waves:

v = fλ

For standing waves on a string:

λₙ = 2L/n

For harmonic frequency:

fₙ = nf₁

For fundamental frequency:

f₁ = v/(2L)

or:

f₁ = (1/2L)√(T/μ)

Then:

1. convert units if necessary
2. identify the harmonic
3. calculate the required quantity
4. include units
5. check whether the result matches the expected trend

---

Checking Answers Using Physical Reasoning

After calculating, ask whether the result makes sense.

If tension increased, frequency should generally:

increase

If string length increased, frequency should:

decrease

If μ increased, frequency should:

decrease

If harmonic number increased, frequency should:

increase

and wavelength should:

decrease

These qualitative checks can catch calculation errors.

---

Common Mistakes

Assuming the fundamental wavelength equals the string length

For a string fixed at both ends:

λ₁ = 2L

Forgetting that both ends are nodes

Fixed endpoints cannot oscillate.

Counting nodes instead of antinodes to identify the harmonic

For the nth harmonic:

• antinodes = n
• nodes = n + 1

Thinking adjacent nodes are one wavelength apart

They are separated by:

λ/2

Using the wrong harmonic wavelength

Use:

λₙ = 2L/n

Assuming wave speed depends on frequency

For a given ideal string under fixed conditions, wave speed is determined mainly by T and μ.

Thinking doubling tension doubles frequency

Because:

f ∝ √T

doubling tension multiplies frequency by √2, not 2.

Thinking doubling μ halves frequency

Doubling μ changes frequency by:

1/√2

not 1/2.

Confusing total string mass with linear mass density

Use:

μ = m/L

Using grams instead of kilograms

Convert mass to kilograms before calculating μ in kg/m.

Ignoring the vibrating length

Pressing a string against a fret changes the effective vibrating length.

Assuming a musical string vibrates only at its fundamental

Real strings normally contain several harmonics simultaneously.

---

Key Terms

Vibrating string — stretched string capable of supporting transverse waves and standing-wave patterns.

Standing wave — stationary interference pattern formed by waves traveling in opposite directions.

Incident wave — wave traveling toward a boundary.

Reflected wave — wave that travels away from a boundary after reflection.

Node — point of zero displacement in a standing wave.

Antinode — point of maximum displacement in a standing wave.

Fundamental frequency — lowest natural frequency of a vibrating system.

First harmonic — fundamental mode of a standing wave.

Harmonic — allowed standing-wave frequency that is an integer multiple of the fundamental for an ideal string fixed at both ends.

Natural frequency — frequency at which a system naturally tends to oscillate.

Wave speed — speed at which a disturbance travels through the string.

Tension — pulling force acting along a stretched string.

Linear mass density — mass per unit length of a string, represented by μ.

Vibrating length — portion of a string that is free to oscillate between its effective fixed boundaries.

Pitch — perception associated mainly with the fundamental frequency of a sound.

Timbre — sound quality influenced by the mixture and relative strength of harmonics.

Resonance — strong response when a system is driven near one of its natural frequencies.

Soundboard — larger vibrating surface that helps transfer energy from strings to the surrounding air.

---

Key Takeaways

• Waves on stretched strings are transverse waves.
• Reflection and interference can create standing waves on strings.
• A string fixed at both ends must have a node at each end.
• Standing waves occur only at particular allowed wavelengths and frequencies.
• The fundamental mode contains one antinode and two nodes.
• For the fundamental, λ₁ = 2L.
• For the nth harmonic, λₙ = 2L/n.
• The nth harmonic has n antinodes and n + 1 nodes.
• Harmonic frequencies follow fₙ = nf₁ for an ideal string.
• Wave speed and frequency are related by v = fλ.
• Wave speed on a stretched string is v = √(T/μ).
• Linear mass density is μ = m/L.
• Increasing tension increases wave speed and frequency.
• Increasing linear mass density decreases wave speed and frequency.
• Increasing vibrating length decreases the fundamental frequency.
• Fundamental frequency can be written as f₁ = (1/2L)√(T/μ).
• String instruments change pitch by controlling length, tension, and linear mass density.
• Pressing a guitar or violin string shortens its vibrating length and raises its frequency.
• Tuning pegs change tension and therefore frequency.
• Thick or heavy strings have greater linear mass density and generally produce lower frequencies.
• Musical strings usually vibrate in several harmonic modes simultaneously.
• Harmonic content contributes to the timbre of an instrument.
• Instrument bodies and soundboards improve the transfer of vibrational energy to the surrounding air.

---

Check Your Understanding

1. Explain how reflection and interference produce a standing wave on a string fixed at both ends.
2. Why must both ends of a string fixed at both ends be nodes?
3. A standing wave has adjacent nodes separated by 0.45 m. Determine its wavelength.
4. A string is 0.90 m long. Determine the wavelength of its fundamental mode.
5. A string is 1.20 m long. Determine the wavelengths of its second, third, and fourth harmonics.
6. A string has a fundamental frequency of 150 Hz. Determine the frequencies of its second, third, fifth, and eighth harmonics.
7. A standing-wave pattern contains 6 antinodes. Identify the harmonic and determine the total number of nodes.
8. A 0.80 m string carries waves at 320 m/s. Calculate its fundamental frequency.
9. For the string in Question 8, calculate the frequency and wavelength of the fourth harmonic.
10. A string has tension 125 N and linear mass density 0.0050 kg/m. Calculate the wave speed.
11. A string is 1.5 m long and has mass 0.012 kg. Calculate its linear mass density.
12. The tension in a string is increased from T to 9T while its length and linear mass density remain unchanged. By what factor do its wave speed and fundamental frequency change?
13. Two strings have the same length and tension. String B has four times the linear mass density of String A. Compare their wave speeds and fundamental frequencies.
14. Explain how a guitarist can raise the pitch of a string by changing either its vibrating length or its tension.
15. A string has L = 0.75 m, T = 192 N, and μ = 0.0030 kg/m. Calculate its wave speed, fundamental frequency, third-harmonic frequency, and third-harmonic wavelength.

Beats and Interference

When two or more waves meet, they combine according to the principle of superposition. The resulting displacement at any point is the sum of the individual wave displacements at that point.

This interaction is called interference.

Interference can make a wave temporarily larger, smaller, or even cancel it completely. In sound waves, interference affects the pressure variations reaching our ears, changing the amplitude and therefore the perceived loudness or sound intensity.

A particularly interesting form of sound interference occurs when two sounds have slightly different frequencies. Their interference produces regular changes in loudness called beats.

---

The Principle of Superposition

The principle of superposition states:

When two or more waves overlap, the resultant displacement is the algebraic sum of the individual displacements.

For two waves:

y_resultant = y₁ + y₂

This means that waves do not normally destroy one another permanently when they overlap.

Instead:

1. the waves meet
2. their displacements combine
3. interference occurs
4. the waves continue travelling afterward

The effect depends strongly on the relative phase of the waves.

---

Constructive Interference

Constructive interference occurs when waves overlap so that their displacements reinforce each other.

For example, if two crests arrive at the same place at the same time:

positive displacement + positive displacement = larger positive displacement

Similarly:

trough + trough = deeper trough

For two identical waves of amplitude A that are perfectly in phase:

resultant amplitude = 2A

---

Waves in Phase

Two waves are in phase when corresponding parts of the waves occur together.

For example:

• crest meets crest
• trough meets trough
• compression meets compression
• rarefaction meets rarefaction

For two identical sinusoidal waves, a phase difference of:

0°

produces maximum constructive interference.

A phase difference of:

360°

also represents the same relative phase after one complete cycle.

---

Destructive Interference

Destructive interference occurs when overlapping waves have displacements in opposite directions.

For example:

positive displacement + negative displacement = smaller resultant displacement

If two waves have equal amplitudes and are exactly opposite in phase:

resultant amplitude = 0

This is called complete destructive interference.

---

Waves Out of Phase

Two identical sinusoidal waves are completely out of phase when their phase difference is:

180°

Then:

• crest meets trough
• trough meets crest

If their amplitudes are equal, complete cancellation occurs at that location.

If the amplitudes are different, cancellation is only partial.

For example:

Wave 1 amplitude = 5 units

Wave 2 amplitude = 3 units in the opposite direction

Resultant amplitude:

5 − 3 = 2 units

---

Constructive and Destructive Interference Compared

Interference can also occur anywhere between these two extremes.

---

Interference of Sound Waves

Sound is a longitudinal wave involving variations in pressure.

When two sound waves reach the same location, their pressure variations combine.

Constructive interference can produce larger pressure variations.

Destructive interference can produce smaller pressure variations.

As a result, interference can create locations where sound is:

• louder
• quieter
• nearly cancelled

---

Amplitude and Sound Intensity

The intensity of a wave is related to the square of its amplitude:

I ∝ A²

where:

• I = intensity
• A = amplitude

This means changes in amplitude can produce much larger changes in intensity.

If the amplitude doubles:

I_new/I_old = (2A)²/A²

I_new = 4I_old

So doubling wave amplitude corresponds to four times the intensity, assuming the waves are otherwise comparable.

---

Worked Example: Constructive Interference

Two identical coherent sound waves each have amplitude:

A = 2.0 units

They arrive perfectly in phase.

Resultant amplitude:

A_resultant = 2.0 + 2.0

A_resultant = 4.0 units

Compared with one wave, the amplitude has doubled.

Since:

I ∝ A²

the intensity at perfect constructive interference is:

4 times the intensity of either individual wave

---

Worked Example: Destructive Interference

Two waves have amplitudes:

A₁ = 6 units

A₂ = 4 units

They arrive exactly opposite in phase.

Resultant amplitude magnitude:

A_resultant = |6 − 4|

A_resultant = 2 units

The waves partially cancel.

Complete cancellation would require equal amplitudes as well as the correct phase relationship.

---

Path Difference and Interference

Interference also depends on how far each wave travels before reaching a particular point.

The difference between the two travel distances is called the path difference.

For two coherent waves, constructive interference occurs when the path difference is:

Δd = nλ

where:

n = 0, 1, 2, 3, ...

Examples include:

0, λ, 2λ, 3λ ...

---

Destructive Interference and Path Difference

For two coherent waves initially in phase, destructive interference occurs when:

Δd = (n + ½)λ

Examples include:

λ/2, 3λ/2, 5λ/2 ...

The extra half-wavelength causes the waves to arrive out of phase.

This explains why sound can be louder at some locations and quieter at others.

---

Interference Patterns in Rooms

Sound waves can reach a listener by different paths.

For example, sound may travel:

• directly from a speaker
• after reflecting from a wall
• after reflecting from the floor
• after reflecting from the ceiling

The waves can then interfere.

At some locations:

constructive interference → greater amplitude

At other locations:

destructive interference → smaller amplitude

This can create uneven sound levels around a room.

---

What Are Beats?

Beats are regular variations in sound amplitude produced when two sound waves with slightly different frequencies interfere.

Suppose one source produces:

440 Hz

and another produces:

442 Hz

The waves repeatedly move:

in phase → out of phase → in phase → out of phase

The listener hears:

loud → quiet → loud → quiet

This repeating variation in loudness is the beat pattern.

---

Why Beats Form

Consider two waves that begin in phase.

At first:

crest + crest → constructive interference

The sound is relatively loud.

Because the frequencies are slightly different, one wave gradually moves ahead of the other.

Eventually:

crest + trough → destructive interference

The sound becomes relatively quiet.

The faster wave continues gaining phase.

Eventually the waves become aligned again:

crest + crest → constructive interference

The sound becomes loud again.

This cycle repeats.

---

Beat Frequency

The number of loudness cycles heard each second is called the beat frequency.

For two frequencies f₁ and f₂:

f_beat = |f₁ − f₂|

The absolute value is used because beat frequency is positive.

---

Worked Example: Beat Frequency

Two tuning forks have frequencies:

f₁ = 440 Hz

f₂ = 444 Hz

Beat frequency:

f_beat = |444 − 440|

f_beat = 4 Hz

The listener hears approximately:

4 beats per second

---

Worked Example: Close Frequencies

Two speakers produce:

f₁ = 512 Hz

f₂ = 515 Hz

Then:

f_beat = |515 − 512|

f_beat = 3 Hz

The loudness rises and falls three times each second.

---

Finding an Unknown Frequency

Beat frequency can also be used to determine an unknown frequency.

Suppose a known tuning fork produces:

440 Hz

An unknown source produces:

5 beats per second

Then:

|f_unknown − 440| = 5

Therefore:

f_unknown = 435 Hz

or:

f_unknown = 445 Hz

Beat frequency alone does not tell us whether the unknown frequency is above or below the known frequency.

Additional information is needed.

---

Resolving the Two Possible Frequencies

Suppose the unknown frequency produces 5 beats per second with a 440 Hz reference.

Possible frequencies are:

435 Hz or 445 Hz

Now suppose the unknown source is adjusted slightly upward in frequency.

If the beat frequency becomes smaller, the unknown frequency was probably below 440 Hz and is moving toward it.

If the beat frequency becomes larger, it may have been above 440 Hz and is moving farther away.

This principle is extremely useful when tuning musical instruments.

---

Beats and Musical Tuning

Musicians can use beats to compare the frequency of an instrument with a known reference frequency.

Suppose a tuning reference produces:

440 Hz

A string being tuned produces:

443 Hz

Beat frequency:

f_beat = |443 − 440|

f_beat = 3 Hz

The musician hears three beats each second.

---

Tuning by Reducing Beat Frequency

The musician adjusts the instrument.

Suppose the frequencies change:

443 Hz → 442 Hz → 441 Hz → 440 Hz

The beat frequencies become:

3 Hz → 2 Hz → 1 Hz → 0 Hz

As the instrument approaches the reference frequency, the beats become slower.

When the frequencies match:

f_beat = |440 − 440| = 0 Hz

The beats disappear.

This tells the musician that the frequencies match closely.

---

Why Beats Are Useful for Tuning

The human ear may have difficulty deciding whether one sustained tone is slightly higher or lower than another.

However, slow changes in loudness can be easy to detect.

Beats therefore transform a small frequency difference into a noticeable rhythm in amplitude.

The closer the two frequencies become:

smaller frequency difference → slower beats

When they match:

frequency difference = 0 → no beats

---

Tuning a Guitar

Suppose a reference tone is:

330 Hz

A guitar string produces:

326 Hz

Beat frequency:

f_beat = |330 − 326|

f_beat = 4 Hz

The guitarist tightens the string, raising its frequency.

At 328 Hz:

f_beat = 2 Hz

At 329.5 Hz:

f_beat = 0.5 Hz

At 330 Hz:

f_beat = 0 Hz

The disappearing beats indicate that the frequencies have matched.

---

Tuning Pianos

Beat patterns are also important in piano tuning.

A piano tuner can compare:

• strings with reference frequencies
• different strings producing related pitches
• intervals between notes

The rate of beating provides information about small differences in frequency.

In practical musical tuning, some intervals are intentionally adjusted according to the tuning system being used, so not every useful tuning comparison necessarily aims for zero beats.

---

The Beat Envelope

When two similar frequencies interfere, the resulting waveform has a rapidly oscillating wave whose amplitude slowly rises and falls.

The outer shape showing the amplitude variation is called the envelope.

The envelope corresponds to the pattern of:

loud → quiet → loud → quiet

The beat frequency describes how often the loudness maxima repeat.

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Mathematical Description of Beats

Suppose two equal-amplitude waves are:

y₁ = A cos(2πf₁t)

and:

y₂ = A cos(2πf₂t)

Using superposition:

y = y₁ + y₂

The combination can be expressed as:

y = 2A cos[π(f₁ − f₂)t] cos[π(f₁ + f₂)t]

This shows two important features.

The rapidly changing part corresponds approximately to the average frequency:

f_average = (f₁ + f₂)/2

The slowly changing amplitude produces the beats.

The audible beat frequency is:

f_beat = |f₁ − f₂|

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Average Frequency Example

Suppose:

f₁ = 500 Hz

f₂ = 506 Hz

Average frequency:

f_average = (500 + 506)/2

f_average = 503 Hz

Beat frequency:

f_beat = |506 − 500|

f_beat = 6 Hz

The listener therefore hears a tone near 503 Hz whose loudness varies about six times each second.

---

Beat Period

The time between successive loudness maxima is the beat period.

Since:

T = 1/f

the beat period is:

T_beat = 1/f_beat

For a beat frequency of 4 Hz:

T_beat = 1/4

T_beat = 0.25 s

A loudness maximum occurs approximately every 0.25 seconds.

---

Worked Example: Beat Period

Two sources produce:

f₁ = 300 Hz

f₂ = 302.5 Hz

Beat frequency:

f_beat = |302.5 − 300|

f_beat = 2.5 Hz

Beat period:

T_beat = 1/2.5

T_beat = 0.40 s

The sound reaches a loudness maximum approximately every 0.40 seconds.

---

Beats Require Similar Frequencies

The classic beat effect is most easily heard when the two frequencies are relatively close together.

If the frequency difference becomes large, the ear may begin to distinguish two separate tones rather than hearing a clear slow beat pattern.

Therefore, beats are particularly useful for detecting small frequency differences.

---

Coherence and Stable Interference

For a stable interference pattern, waves should have a predictable phase relationship.

Such waves are described as coherent.

Two independent sound sources may not maintain perfect coherence for long periods.

However, waves derived from related or stable sources can produce clear interference effects.

Beat formation does not require the frequencies to be identical; in fact, the small frequency difference is what produces the changing phase relationship.

---

Sound Interference in Speakers

Suppose two speakers produce the same frequency.

At one location, the sound waves may arrive in phase.

The listener experiences constructive interference.

At another location, the path difference may cause the waves to arrive out of phase.

The listener experiences destructive interference.

As a person moves around the room, the sound intensity may therefore change.

---

Noise-Cancelling Headphones

Active noise cancellation uses destructive interference.

A microphone detects unwanted sound.

Electronics produce another sound wave designed to oppose it.

Ideally:

noise wave + opposing wave → reduced resultant amplitude

The unwanted sound becomes quieter.

This is sometimes described as producing anti-noise.

---

Why Noise Cancellation Is Not Perfect

Perfect cancellation would require the cancelling wave to have exactly the correct:

• frequency
• amplitude
• phase
• timing

at the listener's ear.

Real environments contain complicated sounds arriving from many directions.

Therefore, noise cancellation usually reduces unwanted sound rather than eliminating every sound completely.

It tends to be especially effective for predictable, lower-frequency background noise.

---

Interference in Concert Halls

Sound interference is an important consideration in the design of:

• concert halls
• theatres
• classrooms
• recording studios
• stadiums

Reflected sound waves can interfere with direct sound.

Poor design can produce areas where certain frequencies become:

• unusually loud
• unusually quiet
• unclear

Acoustic design attempts to control these effects.

---

Multiple Speakers

Large venues often use many speakers.

If their sound arrives at different times or phases, interference can reduce sound quality.

Engineers therefore consider:

• speaker position
• distance
• timing
• phase
• frequency response
• reflections

Digital systems can introduce carefully controlled delays to coordinate sound arriving from different speakers.

---

Interference and Standing Waves

Reflected waves can interfere with incoming waves.

Under suitable conditions, this produces standing waves.

Standing waves contain:

• nodes
• antinodes

At a node, the wave amplitude is very small.

At an antinode, the amplitude is large.

Standing waves in rooms can therefore make certain frequencies noticeably louder or quieter at different locations.

---

Interference in Musical Instruments

Musical instruments depend on wave interference.

Reflections inside:

• strings
• pipes
• air columns

can produce standing waves.

Constructive interference reinforces frequencies that satisfy the instrument's resonance conditions.

This helps determine the frequencies that the instrument can produce strongly.

---

Sound Intensity and Distance

Interference should not be confused with the normal decrease in intensity as sound spreads.

A sound can become quieter because:

• energy spreads over a larger area
• energy is absorbed
• destructive interference occurs

These are different physical processes.

When analyzing a sound situation, it is important to identify which mechanism is responsible.

---

Real-World Example: Aircraft Cabins

Aircraft cabins contain relatively steady low-frequency noise from:

• engines
• airflow
• mechanical systems

Active noise-cancelling headphones can analyze this noise and generate an opposing waveform.

Destructive interference reduces the pressure variations reaching the listener.

The result is a quieter perceived environment.

---

Real-World Example: Vehicle Noise Control

Active noise-control systems can also be used in vehicles.

Microphones detect unwanted sound.

Speakers generate controlled sound waves.

The waves interfere destructively with selected noise frequencies.

The system must continuously adjust because vehicle conditions change with:

• speed
• engine operation
• road surface
• passenger position

---

Real-World Example: Speaker Placement

Imagine two speakers producing the same 600 Hz tone.

A listener at one position may receive both waves in phase.

The sound is relatively loud.

A second listener may be positioned where one wave travels an additional half-wavelength.

The waves can arrive approximately 180° out of phase.

The second listener may hear a much quieter sound.

This demonstrates that interference depends on location.

---

Worked Example: Wavelength and Destructive Interference

Suppose sound has:

f = 680 Hz

and the speed of sound is:

v = 340 m/s

Using:

v = fλ

Wavelength:

λ = v/f

λ = 340/680

λ = 0.50 m

For the first destructive interference condition:

Δd = λ/2

Δd = 0.50/2

Δd = 0.25 m

A path difference of 0.25 m can therefore produce destructive interference if the waves began in phase.

---

Worked Example: Constructive Interference

Using the same wavelength:

λ = 0.50 m

Constructive interference can occur for:

Δd = nλ

Possible path differences include:

0 m, 0.50 m, 1.00 m, 1.50 m ...

At these positions, the waves can arrive in phase.

---

Moving Through an Interference Pattern

Imagine walking through a room while two coherent speakers produce the same tone.

As your position changes:

• path difference changes
• phase difference changes
• interference changes

You might hear:

loud → quiet → loud → quiet

This is an interference pattern caused by position.

It should not be confused with beats, which result from waves having slightly different frequencies.

---

Beats vs Spatial Interference

These two effects can sound superficially similar but have different causes.

This distinction is important.

---

Interference Does Not Destroy Energy

When destructive interference produces a low-amplitude region, the energy has not simply disappeared.

Interference changes how wave energy is distributed through space.

In a larger wave system, regions of destructive interference are associated with energy being redistributed elsewhere.

This is another example of conservation of energy.

---

Interference in Other Waves

Interference is not limited to sound.

The principle of superposition applies to many types of waves, including:

• water waves
• electromagnetic waves
• light
• radio waves
• microwaves

The details differ, but constructive and destructive interference remain fundamental wave behaviours.

---

A Complete Beat Problem

Two sound sources produce:

f₁ = 438 Hz

f₂ = 442 Hz

Beat frequency:

f_beat = |442 − 438|

f_beat = 4 Hz

Average frequency:

f_average = (442 + 438)/2

f_average = 880/2

f_average = 440 Hz

Beat period:

T_beat = 1/4

T_beat = 0.25 s

Therefore, the listener hears a tone near 440 Hz whose amplitude rises and falls four times each second.

---

A Tuning Problem

A piano string is compared with a:

440 Hz

reference tone.

Initially, the musician hears:

6 beats per second

Possible string frequencies are:

434 Hz or 446 Hz

The musician slightly increases the string frequency.

The beat frequency falls to:

4 Hz

This indicates that the string is moving toward 440 Hz.

Therefore, the original string frequency was:

434 Hz

After adjustment:

436 Hz

would produce 4 beats per second.

The musician continues increasing the frequency until the beats disappear at approximately:

440 Hz

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Common Mistakes

Thinking interference permanently changes the original waves

The waves superpose while overlapping and then continue propagating.

Thinking constructive interference means two waves have the same frequency

Constructive interference describes how their displacements combine at a particular place and time.

Thinking destructive interference always produces zero amplitude

Complete cancellation requires appropriate phase and equal amplitudes.

Confusing beats with ordinary interference patterns

Beats are produced by slightly different frequencies and create a time-varying amplitude.

Adding frequencies to find beat frequency

Beat frequency is the absolute difference:

f_beat = |f₁ − f₂|

Using the average frequency as the beat frequency

The average frequency describes the approximate pitch of the rapidly oscillating sound, not the beat rate.

Thinking zero beats means there is no sound

Zero beats means the frequencies match. The sound itself can still be strong.

Assuming a beat frequency identifies an unknown frequency uniquely

An unknown frequency may be either above or below the reference frequency.

Thinking destructive interference violates conservation of energy

Interference redistributes wave energy rather than destroying it.

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Key Terms

Superposition — principle stating that overlapping wave displacements add algebraically.

Interference — combination of overlapping waves according to superposition.

Constructive interference — interference in which wave displacements reinforce one another.

Destructive interference — interference in which wave displacements oppose one another.

Phase — position of a point within a wave cycle relative to another wave or reference.

In phase — waves having corresponding parts of their cycles aligned.

Out of phase — waves having different positions within their cycles.

Path difference — difference between distances travelled by two waves to reach a point.

Coherent waves — waves maintaining a predictable phase relationship.

Beat — periodic variation in amplitude caused by interference between waves with slightly different frequencies.

Beat frequency — number of amplitude or loudness cycles per second, equal to the absolute difference between the two frequencies.

Beat period — time between successive beat maxima.

Envelope — slowly varying outline of the amplitude of a beat waveform.

Sound intensity — rate of sound energy transfer per unit area.

Active noise cancellation — use of controlled destructive interference to reduce unwanted sound.

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Key Takeaways

• The principle of superposition states that overlapping wave displacements add algebraically.
• Constructive interference occurs when wave displacements reinforce each other.
• Destructive interference occurs when wave displacements oppose each other.
• Equal waves perfectly in phase can produce twice the individual amplitude.
• Equal waves 180° out of phase can completely cancel at a particular location.
• Wave intensity is proportional to amplitude squared.
• Path difference determines whether coherent waves arrive in or out of phase.
• For initially in-phase coherent waves, constructive interference occurs when Δd = nλ.
• Destructive interference occurs when Δd = (n + ½)λ.
• Beats result from interference between waves with slightly different frequencies.
• The changing phase relationship causes alternating constructive and destructive interference.
• Beat frequency is calculated using f_beat = |f₁ − f₂|.
• Beat period is T_beat = 1/f_beat.
• The approximate pitch of two nearby frequencies is related to their average frequency.
• Musicians can tune instruments by listening for beats.
• As two frequencies approach each other, the beats become slower.
• When the frequencies match, the beats disappear.
• Sound interference affects the amplitude and intensity heard at different locations.
• Active noise cancellation uses destructive interference to reduce unwanted sound.
• Acoustic engineers must consider interference when designing rooms and multi-speaker sound systems.
• Beats and spatial interference are related through superposition but should not be confused: beats vary mainly with time, while spatial interference patterns vary with position.

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Check Your Understanding

1. State the principle of superposition.
2. Explain the difference between constructive and destructive interference.
3. Two identical waves each have an amplitude of 3.0 cm and arrive perfectly in phase. What is their resultant amplitude?
4. Two waves with amplitudes of 7.0 cm and 5.0 cm arrive exactly opposite in phase. Calculate the magnitude of their resultant amplitude.
5. Explain why two equal-amplitude waves 180° out of phase can completely cancel.
6. Two tuning forks produce frequencies of 440 Hz and 446 Hz. Calculate the beat frequency.
7. Two sound sources produce 256 Hz and 259 Hz. Calculate both the beat frequency and beat period.
8. A musician hears 4 beats per second when comparing an unknown note with a 500 Hz reference. What are the two possible frequencies of the unknown note?
9. Explain how the musician could determine which of the two possible frequencies in Question 8 is correct.
10. Explain why the beat frequency decreases as a musical instrument approaches the correct tuning.
11. A sound wave has a wavelength of 0.80 m. Determine the smallest non-zero path difference that would produce constructive interference and the smallest path difference that would produce destructive interference for two initially in-phase waves.
12. Two equal-amplitude sound waves interfere constructively and produce twice the amplitude of either individual wave. How does the resulting intensity compare with the intensity of one wave?
13. Explain how active noise-cancelling headphones use destructive interference and why perfect cancellation is difficult in a real environment.
14. A student walks through a room containing two speakers producing the same frequency and hears alternating loud and quiet regions. Explain why this is an interference pattern rather than a beat pattern.
15. Two musical instruments produce frequencies of 392 Hz and 395 Hz. Analyze what a listener would hear, calculate the beat frequency and beat period, and explain how a musician could use the observed beats to tune the instruments to the same frequency.