Standing Waves Explorer

See how standing waves form from two interfering waves. Adjust the harmonic number to see overtones.

Starting values — Standing Waves Explorer

String/pipe length (m)	1
Wave speed (m/s)	343
Harmonic n	1

Change any control above and the results recompute from these.

Loading interactive simulation...

Only whole half-wavelengths fit, which is why strings sound musical

A string clamped at both ends cannot vibrate at just any frequency. The ends are forced to stay still, so the only patterns that survive are those fitting a whole number of half-wavelengths between them: λ = 2L/n. Since frequency is speed divided by wavelength, the allowed frequencies come out as exact integer multiples of the lowest one, fₙ = n · f₁. This tool’s 1 m string at 343 m/s reads 171.5 Hz at n = 1 and exactly 514.5 Hz at n = 3 — three times over, not merely nearby. That integer relationship is what the ear hears as a single pitch with a timbre rather than as a chord or a noise, and it is why a guitar and a flute can be in tune with each other at all.

Two waves that trap themselves

A standing wave isn't really going anywhere. It appears when a moving wave bounces off a boundary and overlaps its own reflection: the two travel in opposite directions and lock into a pattern that stays put, with the nodes never budging. Only wavelengths that fit a whole number of half-loops between the ends survive. Takeaway: shorten the string or pipe and those surviving wavelengths shrink, so the pitch climbs.

Your microwave can measure light's speed

A microwave oven fills its cavity with a standing electromagnetic wave, so energy piles up at fixed antinodes while the nodes stay cold — which is exactly why the food needs a turntable. Remove it, melt a flat bar of chocolate, and the molten patches mark the antinodes about 6 cm apart. Multiply that half-wavelength by the 2.45 GHz printed on the back and you recover c ≈ 3 × 10⁸ m/s on your kitchen counter.

STANDING WAVES — WHICH MODE, AND WHAT DO THE ENDS ALLOW?

Which Standing-Wave Case Are You In?

A standing wave is the one shape that fits: a wave whose reflections line up with itself. The length of the medium picks the wavelengths that can survive, λ = 2L/n, and the speed of the wave turns those into frequencies, f = v/λ. What the ends do — held still, or free to move — decides where the nodes sit, not which wavelengths are allowed.

The fundamental — the longest wave the length will hold λ = 2L, f = v/λ

A higher harmonic — n half-wavelengths, and n times the frequency λ = 2L/n, f = n·f₁

Open ends — the pattern flips, the wavelengths do not L = nλ/2

A real instrument — the length is the thing you change f ∝ 1/L

The fundamental — the longest wave the length will hold

What you know: n = 1: half a wavelength spans the whole medium, so λ = 2L. This is the lowest frequency that fits, and it is the pitch you hear.

Relation: λ = 2L, f = v/λ

Worked example: L = 1 m with v = 343 m/s → λ = 2 m and f = 343/2 = 171.5 Hz

Open this case: Fundamental

The fundamental — the longest wave the length will hold. One arch across the whole length: the longest wave that still fits. n = 1: half a wavelength spans the whole medium, so λ = 2L. This is the lowest frequency that fits, and it is the pitch you hear. — One arch across the whole length: the longest wave that still fits.

A higher harmonic — n half-wavelengths, and n times the frequency

What you know: For mode n the medium holds n half-wavelengths: λ = 2L/n and f = n·f₁. The nodes sit at every multiple of L/n along the length.

Relation: λ = 2L/n, f = n·f₁

Worked example: The same 1 m at n = 3 → λ = 0.667 m and f = 514.5 Hz, exactly three times the fundamental

Open this case: 3rd harmonic

A higher harmonic — n half-wavelengths, and n times the frequency. Three half-wavelengths fit in the same length, with fixed points between them. For mode n the medium holds n half-wavelengths: λ = 2L/n and f = n·f₁. The nodes sit at every multiple of L/n along the length. — Three half-wavelengths fit in the same length, with fixed points between them.

Open ends — the pattern flips, the wavelengths do not

What you know: With both ends open the air is free to move there, so the ends are antinodes rather than nodes. The allowed wavelengths are the same λ = 2L/n; only the shape is inverted.

Relation: L = nλ/2

Worked example: An open pipe of L = 0.5 m at n = 2 → λ = 0.5 m and f = 686 Hz, the same arithmetic as a string of the same length

Open this case: Open pipe

Open ends — the pattern flips, the wavelengths do not. Both ends are free to move, so the wave has antinodes where the string had nodes. With both ends open the air is free to move there, so the ends are antinodes rather than nodes. The allowed wavelengths are the same λ = 2L/n; only the shape is inverted. — Both ends are free to move, so the wave has antinodes where the string had nodes.

A real instrument — the length is the thing you change

What you know: A guitar string is fixed at both ends, so the same λ = 2L/n applies. Fretting shortens L, and since f is proportional to 1/L, the pitch rises.

Relation: f ∝ 1/L

Worked example: L = 0.65 m at n = 2 → λ = 0.65 m and f = 527.7 Hz; the fundamental of that string is 263.8 Hz

Open this case: Guitar 2nd

A real instrument — the length is the thing you change. The playing length sets the pitch; every fret is a new L in the same formula. A guitar string is fixed at both ends, so the same λ = 2L/n applies. Fretting shortens L, and since f is proportional to 1/L, the pitch rises. — The playing length sets the pitch; every fret is a new L in the same formula.

Problem solved in full

Lowest note of a 1 m string carrying waves at 343 m/s 5 steps
A 1 m string, clamped at both ends, carrying waves at 343 m/s. Find its lowest note — then work backwards to the tension that produces that wave speed, and to the pitch no steel string of this length can ever reach. This is Fixed-Fixed with L = 1 m, v = 343 m/s and n = 1.
1. Both ends are held still, so every pattern that survives must place a node there. That admits a whole number of half-wavelengths and nothing in between, which is why the string has a list of allowed notes rather than a range.
  
  $L=n\frac{\lambda_n}{2}\ \Longrightarrow\ \lambda_n=\frac{2L}{n}=\frac{2\times1}{1}=2.0000\ \mathrm{m}$
2. Frequency is speed over wavelength, and neither factor is a new fact here. The boundary picked the wavelength and the medium supplies the speed.
  
  $f_1=\frac{v}{\lambda_1}=\frac{343}{2.0000}=171.50\ \mathrm{Hz}$
3. The period is the reciprocal of the frequency, but it is also 2L/v — the time one pulse takes to run to the far end and back. A standing wave is a round trip that returns in phase with itself, so its period is the length of that trip.
  
  $T=\frac{1}{f_1}=5.831\times10^{-3}\ \mathrm{s}=\frac{2L}{v}=\frac{2}{343}$
4. Wave speed on a string is not something you set directly; it comes from tension over mass per unit length. For a 0.30 mm steel wire the linear density follows from the density and the cross-section, and inverting v = √(F_T/μ) puts 65 N on the wire — 6.7 kilograms of pull, which is the right order for a plain steel guitar string.
  
  $\mu=\rho\frac{\pi d^{2}}{4}=7850\times\frac{\pi(3.0\times10^{-4})^{2}}{4}=5.549\times10^{-4}\ \mathrm{kg/m}, \qquad v=\sqrt{\frac{F_T}{\mu}}\ \Longrightarrow\ F_T=\mu v^{2}=65.3\ \mathrm{N}$
5. Now divide that tension by the cross-section to get stress, and watch the diameter leave. Both μ and A carry d², so at a given wave speed the stress is ρv² and depends on the material alone. High-carbon music wire fails near 2.2 × 10⁹ Pa, and a ceiling on stress is therefore a ceiling on speed.
  
  $\sigma=\frac{F_T}{A}=\frac{\rho A v^{2}}{A}=\rho v^{2}=9.24\times10^{8}\ \mathrm{Pa}; \quad \sigma_{\max}\approx2.2\times10^{9}\ \Longrightarrow\ v_{\max}=\sqrt{\frac{\sigma_{\max}}{\rho}}=529\ \mathrm{m/s}, \quad f_{\max}=\frac{v_{\max}}{2L}=265\ \mathrm{Hz}$
Answer

The tool prints λ₁ = 2.0000 m, f₁ = 171.50 Hz and a period of 5.831 × 10⁻³ s. Step 5 is the one to keep: because the stress needed for a given wave speed is ρv², the gauge cancels, and a 1 m steel string cannot be tuned above about 265 Hz however thick or thin it is made. Instruments stay well under that — piano wire is worked at roughly half its breaking stress, which drops the usable speed to 374 m/s and makes 0.72 m the longest steel string that can sound middle C. A grand piano uses close to that length there, which is a fair check on the argument. It is also why the bass strings are wound with copper rather than made from thicker steel: the winding adds mass per unit length without adding cross-section that has to carry the tension, so it lowers v while leaving σ alone.

$\sigma=\rho v^{2}\ \ \forall d\quad\Longrightarrow\quad f_{\max}=\frac{1}{2L}\sqrt{\frac{\sigma_{\max}}{\rho}}=265\ \mathrm{Hz}$

Learning path

From one spring to a standing wave

References (1)

Insight block 3 claims a microwave oven can measure the speed of light. It can, and here is the method: C. Spence, "How to Measure the Speed of Light with a Microwave Oven and a Bar of Chocolate", in Lightspeed: The Ghostly Aether and the Race to Measure the Speed of Light. Oxford University Press, 2019.