IB Physics Topic 3.4: Standing waves and resonance | Notes & Flashcards | Aimnova

IB Physics — Standing waves and resonance

Topic 3.4 of IB Physics covers Standing waves and resonance, which is part of Unit 3: Wave behaviour. Students explore key concepts including Standing waves: nodes, antinodes and superposition, Harmonics, resonance and wavelength from a standing-wave pattern. A strong understanding of standing waves and resonance is essential for IB Physics exams and builds the foundation for connected topics across the syllabus.

Key concepts in Standing waves and resonance

Key Idea: When two identical waves travel in opposite directions they superpose into a standing wave — a fixed pattern that does not move along and carries no net energy. It has nodes (points that never move) and antinodes (points that swing the most), and neighbouring nodes sit exactly half a wavelength apart. A string or air column only resonates at certain frequencies, its harmonics, where a standing wave fits the length. This is examined on both papers. Paper 1A is usually quick: spot how a standing wave is made, compare phase on a standing vs travelling wave, or read the wavelength condition for a string or pipe. Paper 2 is longer structured work: determine a wavelength from a measured node/antinode spacing and then a frequency from the given v = fλ, or calculate the harmonic frequencies of a string or pipe.

📋 Key formulas

Only the wave equation carries the data-booklet badge (look for it). The harmonic wavelength conditions and the node-spacing rule are not printed — they come from how the standing wave fits between the ends, so you memorise them.

v = f\lambda

The wave equation (given). Rearrange to f = v ÷ λ to turn a wavelength into a frequency — for a sound pipe v is the speed of sound, for microwaves v is the speed of light.

$v$: wave speed — how fast the wave travels (m s⁻¹)
$f$: frequency — waves per second (Hz)
$\lambda$: wavelength — length of one full wave (m)

\lambda = 2 \times d_{node}

Node-spacing rule (memorise — not in the booklet). Neighbouring nodes, and neighbouring antinodes, are HALF a wavelength apart, so double the measured spacing to get λ.

$\lambda$: wavelength of the standing wave (m)
$d_{node}$: distance between two neighbouring nodes (or two neighbouring antinodes) (m)

\lambda = \frac{2L}{n}, \quad n = 1, 2, 3, \dots

String fixed at both ends, or a pipe open at both ends (memorise — not in the booklet). n half-wavelengths fit into the length; ALL whole harmonics exist.

$\lambda$: wavelength of that harmonic (m)
$L$: length of the string or pipe (m)
$n$: harmonic number (1, 2, 3 …; for a closed pipe only odd: 1, 3, 5 …)

\lambda = \frac{4L}{n}, \quad n = 1, 3, 5, \dots

Pipe closed at one end (memorise — not in the booklet). A node at the closed end and an antinode at the open end mean only ODD harmonics exist; the next resonance after the fundamental is n = 3.

$\lambda$: wavelength of that harmonic (m)
$L$: length of the string or pipe (m)
$n$: harmonic number (1, 2, 3 …; for a closed pipe only odd: 1, 3, 5 …)

⚖️ Standing wave vs travelling wave

🎵 The three boundary types — which condition to use

A string or open pipe matches at its ends (both the same type), so a half-wave fits → 2L/n, all harmonics. A closed pipe is lopsided (node one end, antinode the other), so a quarter-wave fits → 4L/n, odd harmonics only — its 'second harmonic' is really n = 3, and the frequencies run 1 : 3 : 5, never 1 : 2 : 3.

✏️ Worked exam-style questions

IB-style question — wavelength and speed from the node spacing

A standing wave is set up on a stretched wire vibrating at 50 Hz. The distance from one node to the next node is measured as 0.18 m. (a) Find the wavelength of the wave. (b) Find the speed of the wave on the wire.

Solution:

(a) Neighbouring nodes are half a wavelength apart, so double the spacing:
$\lambda = 2 \times d_{node} = 2 \times 0.18$
(a) Keep the unit:
$\lambda = 0.36\ \text{m}$
(b) Now use the given wave equation:
$v = f\lambda = 50 \times 0.36$
(b) Work it out — keep the unit:
$v = 18\ \text{m s}^{-1}$

Final answer:

(a) λ = 0.36 m (twice the node-to-node spacing). (b) v = 18 m s⁻¹. The most common slip is reading the node spacing as a whole wavelength — it is HALF a wavelength, so always double it before using v = fλ.

IB-style question — fundamental of a string fixed at both ends

A guitar string of length 0.60 m is fixed at both ends. A wave travels along it at 300 m s⁻¹. (a) Find the wavelength of its fundamental (1st harmonic). (b) Find the fundamental frequency.

Solution:

(a) Both ends fixed → use λ = 2L/n with n = 1 (the fundamental):
$\lambda = \frac{2L}{n} = \frac{2 \times 0.60}{1}$
(a) So the wavelength is:
$\lambda = 1.2\ \text{m}$
(b) Rearrange the given wave equation for the frequency:
$v = f\lambda \;\Rightarrow\; f = \frac{v}{\lambda} = \frac{300}{1.2}$
(b) Work it out — keep the unit:
$f = 250\ \text{Hz}$

Final answer:

(a) λ = 1.2 m. (b) fundamental frequency f = 250 Hz. For a string every whole harmonic exists, so its harmonics would be 250, 500, 750 Hz, … (the ratio 1 : 2 : 3).

IB-style question — first two harmonics of a closed pipe

A pipe of length 0.25 m is closed at one end and open at the other. The speed of sound in the air inside is 340 m s⁻¹. Find the frequencies of its first two harmonics.

Solution:

Closed pipe → λ = 4L/n with odd n. Fundamental is n = 1:
$\lambda_{1} = \frac{4L}{1} = 4 \times 0.25 = 1.0\ \text{m}$
Its frequency from the given wave equation:
$f_{1} = \frac{v}{\lambda_{1}} = \frac{340}{1.0} = 340\ \text{Hz}$
The NEXT harmonic is n = 3 (there is no n = 2 for a closed pipe):
$\lambda_{3} = \frac{4L}{3} = \frac{1.0}{3} = 0.333\ \text{m}$
Its frequency — three times the fundamental:
$f_{3} = \frac{340}{0.333} = 1020\ \text{Hz}$

Final answer:

First two harmonics: f₁ = 340 Hz and f₃ = 1020 Hz (the ratio is 1 : 3, because a closed pipe has only odd harmonics). Watch the trap — the second resonance is n = 3, not n = 2.

IB-style question — microwave frequency from the melt spacing

A bar of chocolate is heated in a microwave oven with the turntable removed. An evenly spaced row of melted spots appears, with neighbouring spots 6.1 cm apart. Microwaves travel at c = 3.0 × 10⁸ m s⁻¹. (a) Find the wavelength of the microwaves. (b) Hence find their frequency.

Solution:

(a) Neighbouring melted spots are antinodes, half a wavelength apart, so double the spacing:
$\lambda = 2 \times d_{node} = 2 \times 0.061 = 0.122\ \text{m}$
(b) Rearrange the given wave equation, with c in place of v:
$c = f\lambda \;\Rightarrow\; f = \frac{c}{\lambda} = \frac{3.0\times10^{8}}{0.122}$
(b) Work it out — keep the unit:
$f = 2.5\times10^{9}\ \text{Hz} = 2.5\ \text{GHz}$

Final answer:

(a) λ = 0.122 m (about 12 cm). (b) f = 2.5 × 10⁹ Hz (2.5 GHz) — the working frequency of a kitchen microwave oven. The melted spots mark the antinodes, where the standing-wave field is strongest.

🧠 Quick self-check

Tap each card to reveal the answer.

🎯 Exam tips

Exam Tips

A standing wave needs two identical waves going opposite directions; nodes never move, antinodes swing the most, and the pattern carries no net energy along it.
Neighbouring nodes (or antinodes) are HALF a wavelength apart, so λ = 2 × the spacing. This is not in the data booklet — double the measured spacing, never report it as the wavelength itself.
String / open pipe: λ = 2L/n (all n). Closed pipe: λ = 4L/n (odd n only) — its second resonance is the THIRD harmonic, and the frequencies run 1 : 3 : 5.
Always finish by turning the wavelength into a frequency with the given v = fλ, rearranged f = v/λ (use the speed of sound for a pipe, the speed of light for microwaves).
On a standing wave points are only ever in phase or antiphase, unlike a travelling wave whose phase shifts smoothly — a classic Paper 1A comparison.
Convert every length to metres first (cm = 10⁻² m), and remember a string fixed at both ends in its nth harmonic has n loops and (n + 1) nodes, counting the two end nodes.

IB Physics — Standing waves and resonance

Key concepts in Standing waves and resonance

Key Idea: When two identical waves travel in opposite directions they superpose into a standing wave — a fixed pattern that does not move along and carries no net energy. It has nodes (points that never move) and antinodes (points that swing the most), and neighbouring nodes sit exactly half a wavelength apart. A string or air column only resonates at certain frequencies, its harmonics, where a standing wave fits the length. This is examined on both papers. Paper 1A is usually quick: spot how a standing wave is made, compare phase on a standing vs travelling wave, or read the wavelength condition for a string or pipe. Paper 2 is longer structured work: determine a wavelength from a measured node/antinode spacing and then a frequency from the given v = fλ, or calculate the harmonic frequencies of a string or pipe.

📋 Key formulas

v = f\lambda

The wave equation (given). Rearrange to f = v ÷ λ to turn a wavelength into a frequency — for a sound pipe v is the speed of sound, for microwaves v is the speed of light.

$v$: wave speed — how fast the wave travels (m s⁻¹)
$f$: frequency — waves per second (Hz)
$\lambda$: wavelength — length of one full wave (m)

\lambda = 2 \times d_{node}

Node-spacing rule (memorise — not in the booklet). Neighbouring nodes, and neighbouring antinodes, are HALF a wavelength apart, so double the measured spacing to get λ.

$\lambda$: wavelength of the standing wave (m)
$d_{node}$: distance between two neighbouring nodes (or two neighbouring antinodes) (m)

\lambda = \frac{2L}{n}, \quad n = 1, 2, 3, \dots

String fixed at both ends, or a pipe open at both ends (memorise — not in the booklet). n half-wavelengths fit into the length; ALL whole harmonics exist.

$\lambda$: wavelength of that harmonic (m)
$L$: length of the string or pipe (m)
$n$: harmonic number (1, 2, 3 …; for a closed pipe only odd: 1, 3, 5 …)

\lambda = \frac{4L}{n}, \quad n = 1, 3, 5, \dots

Pipe closed at one end (memorise — not in the booklet). A node at the closed end and an antinode at the open end mean only ODD harmonics exist; the next resonance after the fundamental is n = 3.

$\lambda$: wavelength of that harmonic (m)
$L$: length of the string or pipe (m)
$n$: harmonic number (1, 2, 3 …; for a closed pipe only odd: 1, 3, 5 …)

⚖️ Standing wave vs travelling wave

🎵 The three boundary types — which condition to use

A string or open pipe matches at its ends (both the same type), so a half-wave fits → 2L/n, all harmonics. A closed pipe is lopsided (node one end, antinode the other), so a quarter-wave fits → 4L/n, odd harmonics only — its 'second harmonic' is really n = 3, and the frequencies run 1 : 3 : 5, never 1 : 2 : 3.

✏️ Worked exam-style questions

IB-style question — wavelength and speed from the node spacing

Solution:

(a) Neighbouring nodes are half a wavelength apart, so double the spacing:
$\lambda = 2 \times d_{node} = 2 \times 0.18$
(a) Keep the unit:
$\lambda = 0.36\ \text{m}$
(b) Now use the given wave equation:
$v = f\lambda = 50 \times 0.36$
(b) Work it out — keep the unit:
$v = 18\ \text{m s}^{-1}$

Final answer:

IB-style question — fundamental of a string fixed at both ends

A guitar string of length 0.60 m is fixed at both ends. A wave travels along it at 300 m s⁻¹. (a) Find the wavelength of its fundamental (1st harmonic). (b) Find the fundamental frequency.

Solution:

(a) Both ends fixed → use λ = 2L/n with n = 1 (the fundamental):
$\lambda = \frac{2L}{n} = \frac{2 \times 0.60}{1}$
(a) So the wavelength is:
$\lambda = 1.2\ \text{m}$
(b) Rearrange the given wave equation for the frequency:
$v = f\lambda \;\Rightarrow\; f = \frac{v}{\lambda} = \frac{300}{1.2}$
(b) Work it out — keep the unit:
$f = 250\ \text{Hz}$

Final answer:

(a) λ = 1.2 m. (b) fundamental frequency f = 250 Hz. For a string every whole harmonic exists, so its harmonics would be 250, 500, 750 Hz, … (the ratio 1 : 2 : 3).

IB-style question — first two harmonics of a closed pipe

A pipe of length 0.25 m is closed at one end and open at the other. The speed of sound in the air inside is 340 m s⁻¹. Find the frequencies of its first two harmonics.

Solution:

Closed pipe → λ = 4L/n with odd n. Fundamental is n = 1:
$\lambda_{1} = \frac{4L}{1} = 4 \times 0.25 = 1.0\ \text{m}$
Its frequency from the given wave equation:
$f_{1} = \frac{v}{\lambda_{1}} = \frac{340}{1.0} = 340\ \text{Hz}$
The NEXT harmonic is n = 3 (there is no n = 2 for a closed pipe):
$\lambda_{3} = \frac{4L}{3} = \frac{1.0}{3} = 0.333\ \text{m}$
Its frequency — three times the fundamental:
$f_{3} = \frac{340}{0.333} = 1020\ \text{Hz}$

Final answer:

First two harmonics: f₁ = 340 Hz and f₃ = 1020 Hz (the ratio is 1 : 3, because a closed pipe has only odd harmonics). Watch the trap — the second resonance is n = 3, not n = 2.

IB-style question — microwave frequency from the melt spacing

Solution:

(a) Neighbouring melted spots are antinodes, half a wavelength apart, so double the spacing:
$\lambda = 2 \times d_{node} = 2 \times 0.061 = 0.122\ \text{m}$
(b) Rearrange the given wave equation, with c in place of v:
$c = f\lambda \;\Rightarrow\; f = \frac{c}{\lambda} = \frac{3.0\times10^{8}}{0.122}$
(b) Work it out — keep the unit:
$f = 2.5\times10^{9}\ \text{Hz} = 2.5\ \text{GHz}$

Final answer:

🧠 Quick self-check

Tap each card to reveal the answer.

🎯 Exam tips

Exam Tips

A standing wave needs two identical waves going opposite directions; nodes never move, antinodes swing the most, and the pattern carries no net energy along it.
Neighbouring nodes (or antinodes) are HALF a wavelength apart, so λ = 2 × the spacing. This is not in the data booklet — double the measured spacing, never report it as the wavelength itself.
String / open pipe: λ = 2L/n (all n). Closed pipe: λ = 4L/n (odd n only) — its second resonance is the THIRD harmonic, and the frequencies run 1 : 3 : 5.
Always finish by turning the wavelength into a frequency with the given v = fλ, rearranged f = v/λ (use the speed of sound for a pipe, the speed of light for microwaves).
On a standing wave points are only ever in phase or antiphase, unlike a travelling wave whose phase shifts smoothly — a classic Paper 1A comparison.
Convert every length to metres first (cm = 10⁻² m), and remember a string fixed at both ends in its nth harmonic has n loops and (n + 1) nodes, counting the two end nodes.

IB Physics — Standing waves and resonance

Key concepts in Standing waves and resonance

📋 Key formulas

⚖️ Standing wave vs travelling wave

🎵 The three boundary types — which condition to use

✏️ Worked exam-style questions

IB-style question — wavelength and speed from the node spacing

IB-style question — fundamental of a string fixed at both ends

IB-style question — first two harmonics of a closed pipe

IB-style question — microwave frequency from the melt spacing

🧠 Quick self-check

🎯 Exam tips

Exam Tips

What you'll learn in Topic 3.4

Study resources — 3.4 Standing waves and resonance

Standing waves: nodes, antinodes and superposition

Harmonics, resonance and wavelength from a standing-wave pattern

Ready to study Standing waves and resonance?

Ready to practice?

IB Physics — Standing waves and resonance

Key concepts in Standing waves and resonance

📋 Key formulas

⚖️ Standing wave vs travelling wave

🎵 The three boundary types — which condition to use

✏️ Worked exam-style questions

IB-style question — wavelength and speed from the node spacing

IB-style question — fundamental of a string fixed at both ends

IB-style question — first two harmonics of a closed pipe

IB-style question — microwave frequency from the melt spacing

🧠 Quick self-check

🎯 Exam tips

Exam Tips

What you'll learn in Topic 3.4

Study resources — 3.4 Standing waves and resonance

Standing waves: nodes, antinodes and superposition

Harmonics, resonance and wavelength from a standing-wave pattern

Ready to study Standing waves and resonance?

Ready to practice?