Standing waves and resonance
🎯What you need to be able to do
- Explain how a standing wave forms, and contrast it with a travelling wave.
- Identify nodes and antinodes and use the spacing between them.
- Find the harmonics of a string fixed at both ends, and of open and closed pipes.
- Describe natural frequency, forced oscillation and resonance.
- Describe light, critical and heavy damping, and their effect on a resonance curve.
📚The physics
A standing wave forms when two waves of the same frequency and amplitude travel through each other in opposite directions — in practice, a wave and its own reflection. Superposition then produces points that never move at all (nodes) and points that oscillate with maximum amplitude (antinodes).
The contrast with a travelling wave is a favourite exam question, so learn it as a list. A travelling wave transfers energy along its direction; a standing wave stores energy and transfers none. In a travelling wave every point has the same amplitude; in a standing wave amplitude varies with position, from zero at nodes to maximum at antinodes. In a travelling wave phase varies continuously along the wave; in a standing wave every point between two adjacent nodes is in phase, and points on opposite sides of a node are exactly antiphase.
The spacing rule is the one to hold onto: adjacent nodes are half a wavelength apart, and so are adjacent antinodes. A node and its neighbouring antinode are a quarter wavelength apart. Most standing-wave calculations are this fact plus \( c = f\lambda \).
String fixed at both ends. Both ends must be nodes, so the length holds a whole number of half-wavelengths:
All integer harmonics are present: fundamental, then twice, three times, four times that frequency.
Pipe open at both ends behaves the same way arithmetically — antinodes at both ends instead of nodes, but still \( L = n\lambda/2 \) and all harmonics present.
Pipe closed at one end is the interesting case. The closed end must be a node and the open end an antinode, so the shortest fit is a quarter wavelength: \( L = n\lambda/4 \) with \(n\) odd only. The fundamental is an octave lower than an open pipe of the same length, and only odd harmonics exist. That missing set of even harmonics is why a clarinet, effectively a closed pipe, sounds different from a flute, effectively an open one, even at the same pitch.
Natural frequency is the frequency at which a system oscillates when displaced and released. Forced oscillation is what happens when a periodic driver is applied at some other frequency. Resonance occurs when the driving frequency matches the natural frequency: energy is transferred into the system most efficiently and the amplitude grows dramatically.
Damping removes energy from an oscillator. Light damping lets it oscillate many times with slowly decaying amplitude. Critical damping returns it to equilibrium in the shortest possible time without overshooting — the design target for car suspension and for the needle of an analogue meter. Heavy damping returns it slowly without oscillating at all.
On a resonance curve — amplitude against driving frequency — increasing the damping does two things: it lowers the peak and broadens it, and it shifts the peak very slightly below the natural frequency. Being able to sketch that family of curves is a standard question.
✏️Worked example
(a) Fundamental frequency. For a closed pipe the fundamental fits a quarter wavelength, so \( \lambda = 4L = 4 \times 0.68 = 2.72 \) m. Then \( f = c/\lambda = 340/2.72 = 125 \) Hz.
(b) Next two frequencies it will resonate at. Only odd harmonics exist, so the next are the third and fifth: \( 3 \times 125 = 375 \) Hz and \( 5 \times 125 = 625 \) Hz. There is nothing at 250 Hz.
(c) If the closed end is opened, what is the new fundamental? The pipe now fits a half wavelength, so \( \lambda = 2L = 1.36 \) m and \( f = 340/1.36 = 250 \) Hz — exactly an octave higher, and now all harmonics are available.
🔭See it happen
Blow across the top of a bottle, then add water and blow again. The pitch rises because you have shortened the air column. It is the closed-pipe formula, and you can hear yourself changing \(L\).
📝Practise
Work through these, then reveal the answer. Each question targets a different objective from the list above.
1. A string 0.65 m long is fixed at both ends, and waves travel along it at 240 m s\(^{-1}\). Find the fundamental frequency.
2. Find the frequency of the third harmonic of that string.
3. A pipe 0.50 m long is open at both ends. Take the speed of sound as 340 m s\(^{-1}\). Find the fundamental frequency.
4. The same 0.50 m pipe is now closed at one end. Find the fundamental frequency and the next frequency at which it resonates.
5. A standing wave of frequency 250 Hz forms in air where the speed of sound is 340 m s\(^{-1}\). Find the distance between adjacent nodes.
6. Describe how increasing the damping changes the shape and position of a resonance curve.
🔗Go deeper — other people’s work
These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.
- The Physics Classroom — Sound Waves and Music, standing wave patterns
- The Physics Hypertextbook — standing waves and resonance