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

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Wavelength of the nth Harmonic

Rearranging:

L = nλₙ/2

gives:

λₙ = 2L/n

As harmonic number increases, wavelength decreases.

For example:

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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.

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Harmonic Patterns

For a string fixed at both ends:

The two fixed ends are included when counting nodes.

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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.

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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

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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

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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

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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.

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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

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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

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Comparing the Three Factors

The complete relationship is:

f₁ = (1/2L)√(T/μ)

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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.

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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

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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

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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.

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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.

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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

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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

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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

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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.

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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.

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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.

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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

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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.

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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.

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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.

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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.

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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.