Standing waves in strings and pipes

Key ideas

A standing wave forms when a wave and its reflection superpose. The pattern stays in place: some points never move, others swing with maximum amplitude.
- Node — a point of zero amplitude (the two waves always cancel there).
- Antinode — a point of maximum amplitude (the two waves always reinforce there).
Adjacent nodes are half a wavelength apart: node-to-node distance $= \dfrac{\lambda}{2}$ .
A string or pipe only sustains the standing waves that fit its length. These allowed patterns are the harmonics, and each has its own natural frequency.
Resonance is what happens when the system is driven at one of those natural frequencies — energy is fed in step with the oscillation and a large-amplitude standing wave builds up.
Throughout, connect frequency and wavelength with $v = f\lambda$ , where $v$ is the wave speed in the string or the speed of sound in the pipe.

Strings (fixed at both ends)

Both ends must be nodes.
The fundamental (1st harmonic) fits half a wavelength: $\lambda_1 = 2L$ , so $f_1 = \dfrac{v}{2L}$ .
Every whole-number multiple fits: $f_n = n f_1$ for $n = 1, 2, 3, \dots$ — all harmonics are present.

Pipes

An open end is an antinode (air free to move); a closed end is a node (air can't move).
Open pipe (open both ends) — antinode at each end:
- $\lambda_1 = 2L$ , $f_1 = \dfrac{v}{2L}$ , and all harmonics occur ( $f_n = nf_1$ ).
Closed pipe (one end closed) — node at the closed end, antinode at the open end:
- only a quarter wavelength fits: $\lambda_1 = 4L$ , $f_1 = \dfrac{v}{4L}$
- only odd harmonics occur: $f_1, 3f_1, 5f_1, \dots$

A string fixed at both ends showing the fundamental as a single arc between two end nodes, and the second harmonic as two arcs with a node in the middle; nodes and antinodes are labelled and the node-to-node spacing is marked as half a wavelength.

Worked ExampleFundamental of a guitar string

A guitar string is $0.60$ m long, and waves travel along it at $120$ m s⁻¹. Find the fundamental frequency and the frequency of the 3rd harmonic.

Step 1 — Fundamental wavelength (both ends are nodes)

Half a wavelength fits the string, so $\lambda_1 = 2L = 2(0.60) = 1.2 \text{ m}$

Step 2 — Fundamental frequency from $v = f\lambda$

$f_1 = \frac{v}{\lambda_1} = \frac{120}{1.2} = 100 \text{ Hz}$

Step 3 — Strings carry every harmonic

$f_3 = 3f_1 = 300 \text{ Hz}$

Tips

Closed pipes have no even harmonics. If your working produces $2f_1$ for a closed pipe, you have made a mistake — check which end is a node.
The fundamental of a closed pipe is an octave below an open pipe of the same length ( $\tfrac{v}{4L}$ vs $\tfrac{v}{2L}$ ) — a favourite compare-and-explain question.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A string fixed at both ends is $0.80$ m long.

Show that the wavelength of its fundamental is $1.6$ m.

Merit

A closed pipe (one end closed) is $0.85$ m long. The speed of sound is $340$ m s⁻¹.

Calculate the fundamental frequency, and state which harmonic comes next.

Excellence

An open pipe and a closed pipe have the same length.

Compare their fundamental frequencies and their harmonic series, explaining the differences using nodes and antinodes.

Standing waves in strings and pipes

Covers AS91523

Key ideas

A standing wave forms when a wave and its reflection superpose. The pattern stays in place: some points never move, others swing with maximum amplitude.
- Node — a point of zero amplitude (the two waves always cancel there).
- Antinode — a point of maximum amplitude (the two waves always reinforce there).
Adjacent nodes are half a wavelength apart: node-to-node distance $= \dfrac{\lambda}{2}$ .
A string or pipe only sustains the standing waves that fit its length. These allowed patterns are the harmonics, and each has its own natural frequency.
Resonance is what happens when the system is driven at one of those natural frequencies — energy is fed in step with the oscillation and a large-amplitude standing wave builds up.
Throughout, connect frequency and wavelength with $v = f\lambda$ , where $v$ is the wave speed in the string or the speed of sound in the pipe.

Strings (fixed at both ends)

Both ends must be nodes.
The fundamental (1st harmonic) fits half a wavelength: $\lambda_1 = 2L$ , so $f_1 = \dfrac{v}{2L}$ .
Every whole-number multiple fits: $f_n = n f_1$ for $n = 1, 2, 3, \dots$ — all harmonics are present.

Pipes

An open end is an antinode (air free to move); a closed end is a node (air can't move).
Open pipe (open both ends) — antinode at each end:
- $\lambda_1 = 2L$ , $f_1 = \dfrac{v}{2L}$ , and all harmonics occur ( $f_n = nf_1$ ).
Closed pipe (one end closed) — node at the closed end, antinode at the open end:
- only a quarter wavelength fits: $\lambda_1 = 4L$ , $f_1 = \dfrac{v}{4L}$
- only odd harmonics occur: $f_1, 3f_1, 5f_1, \dots$

Worked ExampleFundamental of a guitar string

A guitar string is $0.60$ m long, and waves travel along it at $120$ m s⁻¹. Find the fundamental frequency and the frequency of the 3rd harmonic.

Step 1 — Fundamental wavelength (both ends are nodes)

Half a wavelength fits the string, so $\lambda_1 = 2L = 2(0.60) = 1.2 \text{ m}$

Step 2 — Fundamental frequency from $v = f\lambda$

$f_1 = \frac{v}{\lambda_1} = \frac{120}{1.2} = 100 \text{ Hz}$

Step 3 — Strings carry every harmonic

$f_3 = 3f_1 = 300 \text{ Hz}$

Tips

Closed pipes have no even harmonics. If your working produces $2f_1$ for a closed pipe, you have made a mistake — check which end is a node.
The fundamental of a closed pipe is an octave below an open pipe of the same length ( $\tfrac{v}{4L}$ vs $\tfrac{v}{2L}$ ) — a favourite compare-and-explain question.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A string fixed at both ends is $0.80$ m long.

Show that the wavelength of its fundamental is $1.6$ m.

Merit

A closed pipe (one end closed) is $0.85$ m long. The speed of sound is $340$ m s⁻¹.

Calculate the fundamental frequency, and state which harmonic comes next.

Excellence

An open pipe and a closed pipe have the same length.

Compare their fundamental frequencies and their harmonic series, explaining the differences using nodes and antinodes.

Standing waves in strings and pipes

Key ideas

Strings (fixed at both ends)

Pipes

Step 1 — Fundamental wavelength (both ends are nodes)

Step 2 — Fundamental frequency from v=fλv = f\lambdav=fλ

Step 3 — Strings carry every harmonic

Tips

Test yourself

Practice by grade

Standing waves in strings and pipes

Key ideas

Strings (fixed at both ends)

Pipes

Step 1 — Fundamental wavelength (both ends are nodes)

Step 2 — Fundamental frequency from v=fλv = f\lambdav=fλ

Step 3 — Strings carry every harmonic

Tips

Test yourself

Practice by grade

Step 2 — Fundamental frequency from $v = f\lambda$

Step 2 — Fundamental frequency from $v = f\lambda$