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3.2.4 Standing Waves

By the end of this lesson you can:

  • Explain how standing waves form
  • Identify nodes, antinodes and harmonics
  • Calculate frequencies for strings and pipes
Find your course: NSW Year 11: Wave behavioursNSW Year 12QLD Unit 2: WavesVIC Unit 4: Light, matter and relativityAP Physics 2: Unit 14IB: C.4 Standing waves and resonance
🎥 Watch📝 Practise🕹️ Explore🎯 Exam ready🍎 Teachers
🎥Watch
Learn the idea: a short explanation, then a check.

A "standing wave" may sound like a contradiction in terms, as waves are fundamentally defined by the transport of energy through space. However, a standing wave (or stationary wave) is actually a manifestation of wave interference and resonance.

When a continuous progressive wave encounters a boundary, it reflects back into the medium. The superposition of the incident wave and the reflected wave—moving in opposite directions with identical frequency and amplitude—creates a resultant wave pattern that appears trapped in place.

Key characteristics of standing waves include:

  • Nodes: Points of permanent zero displacement resulting from continuous destructive interference (governed by boundary conditions).
  • Antinodes: Points of maximum oscillation amplitude resulting from continuous constructive interference.
  • Energy Confinement: Unlike traveling waves, standing waves store kinetic and potential energy within fixed spatial regions rather than propagating energy along the medium.

This principle governs phenomena across classical and modern physics, from acoustic resonance in musical instruments and microwave cavities to standing electron probability waves in quantum mechanics.

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The places where the waves are at maximum displacement, or amplitudes are called the antinodes, and this is where are getting constructive interference
The places where there is no motion are called nodes and are the places where there is destructive interference .​
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I discuss the principles of a transverse standing wave using a guitar and a string to demonstrate and explain.

📝Practise
Worked examples first, then problems to try, then worksheets.
Sample Problem
We are now ready to try a sample problem
Below is a sample problem with a video that explain how to solve it. It is suggested you try the problem beforehand, as this actually aids understanding, even if you are unsure if you are correct.
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Practice Problems: Standing Waves on Strings

In each question the string is fixed at both ends, so a node sits at each end.

1. A guitar string is fixed at both ends and has a vibrating length of 0.65 m. Waves travel along the string at 260 m/s. Calculate (a) the frequency of the fundamental and (b) the frequency of the third harmonic.

Given: L = 0.65 m, v = 260 m/s

(a) Fundamental: The string forms one loop, so half a wavelength fits on the string: L = λ/2 → λ = 2L = 1.30 m

f1 = v / λ = 260 / 1.30 = 200 Hz

(b) Third harmonic: f3 = 3f1 = 3 × 200 = 600 Hz

Answer: (a) 200 Hz, (b) 600 Hz
2. A string vibrates with a fundamental frequency of 440 Hz when its vibrating length is 0.60 m. A guitarist presses the string against a fret so that the vibrating length becomes 0.45 m. The tension does not change. What is the new fundamental frequency?

Given: f1 = 440 Hz, L1 = 0.60 m, L2 = 0.45 m

Step 1 (Wave speed): The tension is unchanged, so the wave speed is unchanged. v = 2L1f1 = 2 × 0.60 × 440 = 528 m/s

Step 2 (New frequency): f2 = v / (2L2) = 528 / (2 × 0.45) = 528 / 0.90 ≈ 587 Hz

Check: The frequency is inversely proportional to the length, so f2 = 440 × (0.60 / 0.45) ≈ 587 Hz. A shorter string gives a higher pitch.

Answer: 587 Hz
3. On a violin, waves travel along a string at 396 m/s. What vibrating length of this string produces a fundamental frequency of 660 Hz?

Given: v = 396 m/s, f1 = 660 Hz

Formula: For a string fixed at both ends, f1 = v / (2L) → L = v / (2f1)

Calculation: L = 396 / (2 × 660) = 396 / 1320 = 0.30 m

Answer: 0.30 m (30 cm)
🕹️Explore
Test your predictions in a simulation.

Let's first explore why a standing wave is produced​.

  1. Set the type to pulse, have damping to zero, and fixed end
  2. Now send down a single pulse - You will note that it reflects and switches side, or changes phase
  3. If you send another pulse whilst the previous one is returning, they will interfere with each other.

This is key to standing waves, but this only occurs at set frequencies, Let's see this in action.
  1. Set the system to oscillate, damping zero and fixed end
  2. If you press play you will see that you might get a standing wave when the reflected wave interferes with the coming wave.
  3. But the frequency (1.5 Hz) is not quite right. It may behave as a standing wave for a short time, but eventually it will become chaotic.
  4. See if you can alter the frequency slightly to make a standing wave. Press restart when you do this

Hopefully you will see that you will get a large amplitude (its in resonance) but the position of these amplitude does to move to the left or right (Hance its standing wave)

Here is another Standing wave in string interactive developed in Geogebra by Tom Walsh. Again, you will notice only specific frequencies will result in a standing wave

🎥Watch
Learn the idea: a short explanation, then a check.

Standing waves can also form in longitudinal sound waves, where air itself serves as the oscillating medium. Many wind and brass instruments rely on the resonance of standing waves inside air columns to produce pitch.

A closed pipe (a tube sealed at one end and open at the other) creates specific acoustic boundary conditions:

  • Closed End: Air molecules are constrained from moving, forming a displacement node (pressure antinode).
  • Open End: Air molecules oscillate freely, forming a displacement antinode (pressure node).

Because a node must exist at the closed end and an antinode at the open end, closed pipes only support odd harmonics (n = 1, 3, 5, ...), given by the frequency relation fn = n v / (4L).

📝Practise
Worked examples first, then problems to try, then worksheets.
Sample Problem
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Practice Problems: Closed Pipes

Take the speed of sound in air as 340 m/s. Ignore end corrections.

1. A pipe closed at one end and open at the other is 0.85 m long. Take the speed of sound in air as 340 m/s. Find the frequencies of the fundamental and the next two resonances.

Given: L = 0.85 m, v = 340 m/s

Step 1 (Fundamental): There is a node at the closed end and an antinode at the open end, so a quarter of a wavelength fits in the pipe: L = λ/4 → λ = 4L = 3.4 m

f1 = v / λ = 340 / 3.4 = 100 Hz

Step 2 (Next resonances): A closed pipe only supports odd harmonics (1, 3, 5, ...), so the next two are f3 = 3 × 100 = 300 Hz and f5 = 5 × 100 = 500 Hz

Answer: 100 Hz, 300 Hz and 500 Hz
2. A tuning fork of frequency 500 Hz is held above a tube. The water level in the tube is lowered slowly, which changes the length of the air column. The speed of sound is 340 m/s. Find the first two lengths of air column that make the sound loud (resonance).

Given: f = 500 Hz, v = 340 m/s

Step 1 (Wavelength): λ = v / f = 340 / 500 = 0.68 m

Step 2 (First resonance): The air column is closed at the water and open at the top, so L1 = λ/4 = 0.68 / 4 = 0.17 m

Step 3 (Second resonance): L2 = 3λ/4 = 3 × 0.68 / 4 = 0.51 m

The resonances are half a wavelength (0.34 m) apart. End correction is ignored.

Answer: 0.17 m and 0.51 m
3. An organ pipe closed at one end has a fundamental frequency of 200 Hz. The speed of sound is 340 m/s. (a) What is the length of the pipe? (b) What length of pipe open at both ends would have the same fundamental frequency?

Given: f1 = 200 Hz, v = 340 m/s

(a) Closed pipe: f1 = v / (4L) → L = v / (4f1) = 340 / (4 × 200) = 0.425 m

(b) Open pipe: An open pipe has an antinode at both ends, so f1 = v / (2L) → L = 340 / (2 × 200) = 0.85 m

The open pipe has to be twice as long as the closed pipe to give the same fundamental.

Answer: (a) 0.425 m, (b) 0.85 m
🕹️Explore
Test your predictions in a simulation.

In this animation you can explore both the standing wave in longitudinal form as well as the transverse form used in representations
​​Standing wave in pipe interactive developed in Geogebra by Tom Walsh

Standing Waves in 2 dimensions - Chladni plates

In the previous examples we were looking at standing waves in one dimension. However standing waves can occur in two and three dimensions. Chladni plates provide a fascinating visual representation of standing waves in 2 dimensions​, this video will discuss the principles of how it works.
🍎For teachers
Ideas for using this lesson in class, and where to find the answers.

Experiment

These is an online interactive lab on standing waves that I produced.  Ideally you would do this with physical equipment. However, you may be limited in resources or maybe teaching this topic remotely so this may be useful.

Applications
As the above videos demonstrate, there are applications for standing waves, most obviously the production and use of musical instruments
There are other applications. It is worthwhile t o mention them in class as way of providing context. In fact, I start the lesson on standing waves by bringing in a guitar and pretend I can play. 

  • Microwave ovens use microwave standing waves to heat your food. This principle was used to measure the speed of light and is further discussed here.
  • Louis de Broglie used the concepts of standing waves to explain why electrons have a set energy of discrete orbits. This is discussed further here