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When two progressive waves of the same frequency and amplitude travel in opposite directions through the same region, their superposition produces a wholly different kind of pattern — a stationary wave (also called a standing wave). Rather than travelling through space, the wave appears to be "frozen" in place, with certain points (nodes) where the displacement is always zero and other points (antinodes) where the oscillation is maximum.
Stationary waves are the physics of musical instruments. Every string you have plucked, every pipe you have blown across, and every tuning fork you have struck produces sound by setting up stationary waves. In this lesson we introduce stationary waves in general, the distinction between nodes and antinodes, and the specific cases of stationary waves on strings (with both ends fixed) and in pipes (open and closed). The OCR A-Level specification expects quantitative treatment using:
f = (1/2L) √(T/μ) (fundamental frequency on a stretched string)
Consider a progressive wave travelling to the right along a string, meeting a fixed end. The wave is reflected, with a π phase shift (because a fixed end cannot move, the reflected wave must cancel the incoming displacement). The result is two waves of the same frequency and amplitude, travelling in opposite directions along the same string.
By the principle of superposition, the total displacement at any point is the sum of the two waves' displacements. Mathematically, the two opposing travelling waves combine to give:
y(x, t) = 2 A sin(kx) cos(ωt)
The spatial part sin(kx) is independent of time; the temporal part cos(ωt) applies equally to every point. At positions where sin(kx) = 0 — called nodes — the displacement is always zero. At positions where sin(kx) = ±1 — called antinodes — the amplitude of oscillation is maximum (2A, twice that of the constituent waves).
This is the defining feature of a stationary wave: the pattern is stationary in space — nodes remain nodes and antinodes remain antinodes — while each point between them oscillates in simple harmonic motion.
A node is a point on a stationary wave where the displacement is always zero. Nodes form at positions where the two travelling waves permanently cancel.
An antinode is a point where the amplitude of oscillation is maximum. Antinodes form at positions where the two travelling waves permanently reinforce.
Key facts about nodes and antinodes:
This is worth memorising — it is often tested.
A string of length L, fixed at both ends (like a guitar or violin string) can support stationary waves only if the pattern has nodes at both ends. This is because the fixed ends cannot move.
The only stationary waves that fit this condition are those with a whole number of half-wavelengths between the ends:
L = n λ / 2 where n = 1, 2, 3, …
So the allowed wavelengths are:
λ_n = 2 L / n
And the corresponding frequencies, using v = fλ:
f_n = n v / (2 L) where n = 1, 2, 3, …
These are called the harmonics of the string.
The fundamental frequency (first harmonic) is the lowest allowed frequency:
f_1 = v / (2 L)
It corresponds to n = 1 — a single "half-wavelength loop" with nodes at each end and one antinode in the middle.
flowchart TB
subgraph H1["Fundamental: n=1"]
A1[Node at x=0]
B1[Antinode in middle]
C1[Node at x=L]
end
subgraph H2["2nd harmonic: n=2"]
D2[Node-Antinode-Node-Antinode-Node]
end
subgraph H3["3rd harmonic: n=3"]
E3[Three antinodes, four nodes]
end
H1 -. doubles .- H2
H1 -. triples .- H3
The overtones of a string are the harmonics above the fundamental. In music theory, the second harmonic is the first overtone, the third harmonic is the second overtone, etc.
The speed of a transverse wave on a stretched string is:
v = √(T / μ)
where:
Substituting this into f_1 = v / (2L), the fundamental frequency of a stretched string becomes:
f_1 = (1 / 2L) √(T / μ)
This is the formula you must know. It tells you that the pitch of the string rises:
This is the full physics of how guitars, violins, pianos and harps are tuned and played.
Q. A guitar string has length 0.65 m, mass per unit length 1.0 × 10⁻³ kg m⁻¹, and is under tension 100 N. Calculate the fundamental frequency.
A.
v = √(T/μ) = √(100 / 1.0 × 10⁻³) = √(1.00 × 10⁵) = 316 m s⁻¹ f_1 = v / (2L) = 316 / (2 × 0.65) = 243 Hz
This is slightly lower than the 247 Hz of B3 on the standard guitar — plausibly, a B string that needs tuning up a little.
Q. For the same guitar string, calculate the frequency of the second harmonic.
A. f_2 = 2 f_1 = 2 × 243 = 486 Hz.
Q. A string of length 0.50 m and μ = 2.0 × 10⁻³ kg m⁻¹ is to produce a fundamental of 300 Hz. Calculate the required tension.
A. Using f_1 = (1/(2L)) √(T/μ):
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