Applications: Musical Instruments

Learning Objectives

By the end of this section, you will be able to:

1. Derive the formula $f_n = \frac{n}{2L}\sqrt{T/\mu}$ for the natural frequencies of a vibrating string directly from the 1D wave equation
2. Predict how a guitarist's fingers, a violinist's bow pressure, and a piano tuner's wrench each change pitch
3. Distinguish open–open and closed–open pipe modes and explain why a clarinet plays an octave lower than a flute of the same length
4. Compute drum-head frequencies using Bessel-function zeros and prove that drums are inharmonic
5. Relate Fourier coefficients of the pluck shape to the timbre (the "color") of an instrument's sound
6. Implement all three instrument families in plain Python, then use PyTorch's autograd to solve the inverse problem: what tension produces a given pitch?

The Big Picture: How Music Lives Inside the Wave Equation

"Every musical instrument is a boundary-value problem with a pretty face. The boundary conditions decide the scale; the shape decides the timbre."

For the last five sections we have built a deep theory of waves — the wave equation $u_{tt} = c^2 \nabla^2 u$, d'Alembert's traveling-wave solution, separation of variables for standing waves, Bessel functions for circular geometry, finite-difference schemes for numerical experiments. We have been climbing a tall ladder. This section is the view from the top.

Everything you have heard in your life — every guitar, every flute, every drum, the sound of your own voice — is a particular solution of the wave equation, sculpted by the geometry of the instrument. Pitch comes from the eigenvalues. Loudness comes from the amplitudes. Timbre comes from the relative weights of the modes. Even the difference between an oboe and a clarinet — both wooden tubes — is nothing more than a difference in boundary conditions.

Why a Whole Section on Instruments?

Three reasons. First, instruments are the most honest teachers of the wave equation: their boundary conditions are physically visible, their eigenmodes audible, their inharmonicities measurable in cents on a tuner app. Second, the same three problem types — string, pipe, drum — generalize to every wave application you will meet later: antennas (electromagnetic strings), waveguides (electromagnetic pipes), and quantum confinement (probability-amplitude membranes). Third, every one of these problems can be solved forward (geometry → pitch) and inverse (desired pitch → geometry), and the inverse direction is exactly where modern ML and physics meet.

The Vibrating String: Guitars, Violins, and Pianos

Consider a string of length $L$, tension $T$, and linear mass density $\mu$ (kilograms per metre). Both ends are clamped — the bridge at $x = 0$ and the nut (or your finger on the fret) at $x = L$. The vertical displacement $y(x, t)$ obeys the 1D wave equation

$\dfrac{\partial^2 y}{\partial t^2} = c^2 \dfrac{\partial^2 y}{\partial x^2}, \quad c = \sqrt{T/\mu}$

with boundary conditions $y(0,t) = y(L,t) = 0$. We derived this in Section 1 of this chapter from Newton's second law applied to a tiny segment of string. We solved it in Sections 2–3. Now we read off the consequences for music.

The intuition before the algebra

Pluck the string and let it go. The string wants to return to its straight rest shape; tension pulls it down, but its inertia overshoots, and a wave bounces back and forth between the bridge and the nut. Because both ends are pinned, only shapes that vanish at both ends can persist — the same way only certain notes fit between two walls. Those allowed shapes are the eigenfunctions $\sin(n\pi x / L)$. The time they take to complete one oscillation is the period, and its reciprocal is the frequency you hear.

From the Wave Equation to Pitch in Three Steps

Step 1. Separate the variables

Write $y(x, t) = X(x)\,\Theta(t)$ and plug into the wave equation. Dividing by $c^2 X \Theta$:

$\dfrac{\Theta''(t)}{c^2 \Theta(t)} = \dfrac{X''(x)}{X(x)} = -\lambda$

The left side depends only on $t$, the right only on $x$. They can be equal only if both equal a constant; we write the constant as $-\lambda$ (negative, so the time solution oscillates rather than blows up).

Step 2. Apply the boundary conditions

The spatial ODE $X'' + \lambda X = 0$ with$X(0) = X(L) = 0$ only has nonzero solutions when

$\lambda_n = \left(\dfrac{n\pi}{L}\right)^2, \quad X_n(x) = \sin\left(\dfrac{n\pi x}{L}\right), \quad n = 1, 2, 3, \dots$

Those are the standing waves. Any other value of $\lambda$ forces $X(x) \equiv 0$ — silence.

Step 3. Read the frequency off the time equation

With $\lambda_n$ known, the time equation $\Theta'' + c^2 \lambda_n \Theta = 0$ has angular frequency

$\omega_n = c\sqrt{\lambda_n} = \dfrac{n\pi c}{L} = \dfrac{n\pi}{L}\sqrt{T/\mu}$

Converting to hertz, $f_n = \omega_n / 2\pi$, we get the master formula for stringed instruments:

$f_n = \dfrac{n}{2L}\sqrt{T/\mu}, \quad n = 1, 2, 3, \dots$

Interactive Guitar String Lab

Move the sliders below to feel the formula. Watch the composite vibration (white) and the individual harmonics (colored). The bar chart is the harmonic spectrum — the recipe of the sound.

Worked Example: Tuning a Guitar's High-E String

Let's do one full calculation by hand. A standard tuned high-E guitar string vibrates at $f_1 = 329.63$ Hz. A typical scale length is $L = 0.65$ m and a typical mass density is $\mu = 0.00040$ kg/m (a 0.010-inch steel string). What tension do we need?

The Harmonic Series and Timbre

Why does a violin sound different from a piano playing the same note? Both have a fundamental at, say, 440 Hz. The difference is in the amplitudes of the higher harmonics — the Fourier coefficients $b_n$ in

$y(x, t) = \sum_{n=1}^{\infty} b_n \sin\!\left(\dfrac{n \pi x}{L}\right) \cos(\omega_n t)$

For a triangular pluck at position $x = pL$, $b_n \propto \sin(n\pi p) / n^2$. Two facts follow immediately:

1. The $1/n^2$ decay means high harmonics are quiet — exactly why a pluck sounds rich but not screechy.
2. If $p = 1/n$ for some integer $n$, then $\sin(n\pi p) = 0$ and the $n$-th harmonic is silent. Pluck a string exactly in the middle ($p = 1/2$) and the 2nd, 4th, 6th harmonics all vanish — the sound becomes hollow and flute-like. This is why guitarists pluck near the bridge for a bright sound and over the fretboard for a mellow one.

Wind Instruments: Sound in a Tube of Air

Replace the transverse displacement of a string with the longitudinal pressure of air inside a tube and you get a wind instrument. Air is compressible; the wave equation now governs pressure fluctuations$p(x, t)$:

$\dfrac{\partial^2 p}{\partial t^2} = c^2 \dfrac{\partial^2 p}{\partial x^2}, \quad c \approx 343 \text{ m/s}$

The wave speed $c$ is no longer $\sqrt{T/\mu}$ — it's the speed of sound in air, set by temperature, not by the instrument. The geometry only enters through boundary conditions:

• Open end (the bell of a flute, the open tone hole): pressure equals atmospheric pressure outside, so the standing-wave pressure has a node there.
• Closed end (the player's lips on a clarinet mouthpiece, the cap on an organ pipe): air cannot move past the cap, so the velocity is zero and the pressure swings maximally — a pressure antinode.

Open vs. Closed Pipes — Why a Clarinet Sounds Different

Open–open pipe (flute, recorder, open organ pipe)

Pressure node at both ends. The allowed spatial pattern is $p_n(x) = \sin(n\pi x / L)$ — identical to the string! So the frequencies are

$f_n = \dfrac{n c}{2L}, \quad n = 1, 2, 3, \dots$

Every integer multiple of the fundamental is allowed. The full harmonic series is present — flutes sound bright and full.

Closed–open pipe (clarinet, panpipe, stopped organ pipe)

Now we have an antinode at $x = 0$ (the mouthpiece) and a node at $x = L$ (the open bell). The allowed pattern is $p_n(x) = \cos((2n-1)\pi x / (2L))$ and the frequencies are

$f_n = \dfrac{(2n-1)c}{4L}, \quad n = 1, 2, 3, \dots$

Only odd multiples of the fundamental. This has three consequences musicians know in their bones:

1. A closed–open pipe of length $L$ plays half the fundamental frequency of an open–open pipe of the same length — one octave lower. A clarinet body the same length as a flute body plays an octave deeper.
2. The lowest overtone of a clarinet is the third harmonic (twelfth above the fundamental, not an octave). Clarinetists have a technique called "overblowing the twelfth" for exactly this reason.
3. Missing even harmonics gives the "hollow, reedy" clarinet timbre. Open–open pipes have a rounder, more vocal sound because every harmonic is present.

Interactive Pipe Acoustics Lab

Switch between flute (open–open) and clarinet (closed–open), drag the length slider, and step through the mode index. Pink curve = pressure standing wave; cyan dots = air particles displaced by the wave.

Worked Example: Tuning a Clarinet to Middle C

How long must a clarinet (closed–open pipe) be to sound middle C ($f_1 = 261.63$ Hz) at room temperature ($c = 343$ m/s)?

Drums: When Calculus Goes Inharmonic

For a circular drumhead of radius $a$, surface tension $T$ (N/m), and areal density $\sigma$ (kg/m²), the 2D wave equation in polar coordinates gives modes

$u_{mn}(r, \theta, t) = J_m\!\left(\dfrac{j_{mn}\, r}{a}\right) \cos(m \theta) \cos(\omega_{mn} t)$

with frequencies

$\omega_{mn} = \dfrac{c\, j_{mn}}{a}, \quad c = \sqrt{T/\sigma}$

Here $j_{mn}$ is the $n$-th positive zero of the Bessel function $J_m$. The crucial thing about these zeros is that they are not equally spaced.

Compare with a string, whose ratios are 1, 2, 3, 4, 5, … — every integer. The drum's ratios are 1, 1.594, 2.136, 2.296, … — irrational and clustered. That single fact is why drums have indefinite pitch. Your ear can't lock onto a fundamental because the overtones don't reinforce it. Drums are tones with body, not pitches.

Interactive Drum Membrane Lab

Step through the modes and watch the nodal lines appear. Dashed white circles and lines are zero-displacement curves — sprinkle sand on a real drumhead vibrating in this mode and it will settle along exactly these lines.

Worked Example: A Timpani Tuned to A₂

A concert timpani has radius $a = 0.30$ m and an areal density of about $\sigma = 0.25$ kg/m². The orchestra needs it tuned to A₂ ≈ 110 Hz. What surface tension $T$ do we need on the (1, 1) mode, which dominates the perceived pitch?

Python and PyTorch Implementation

Putting everything together: closed-form formulas for the three instrument families, plus a PyTorch routine that inverts the string equation to solve a tuning problem. Plain Python first to build intuition, then PyTorch to show how the same idea scales to learnable systems.

np.sqrt(80 / 0.0007)  → 338.06   # wave speed in m/s

string_frequency(1, L=0.65, T=80, mu=0.0007) → 260.05 Hz (about middle C)

Steel guitar string:  c = √(80 / 0.0007) ≈ 338 m/s
Nylon classical:      c = √(80 / 0.005)  ≈ 126 m/s

n = 1: ω₁ = π · 338 / 0.65 ≈ 1633 rad/s

f = 1633 / (2π) ≈ 260 Hz

Pluck at x = 0.2 L → b₁ ≈ 0.81, b₂ ≈ 0.20, b₃ ≈ 0.09, …
(amplitudes fall like 1/n²)

pipe_frequencies(0.6, 'open-open')  → [285.8, 571.7, 857.5, 1143.3, 1429.2]
pipe_frequencies(0.6, 'closed-open') → [142.9, 428.8, 714.6, 1000.4, 1286.3]

n=1 → harmonic # 1
n=2 → harmonic # 3
n=3 → harmonic # 5
(even harmonics missing)

jn_zeros(0, 3) → [2.4048, 5.5201, 8.6537]
jn_zeros(1, 3) → [3.8317, 7.0156, 10.1735]

drum_frequency(0, 1, a=0.30, T=2500, sigma=0.25) ≈ 40.4 Hz (a deep timpani)

Mode (0,1): j ≈ 2.405,  ratio = 1.000 (fundamental)
Mode (1,1): j ≈ 3.832,  ratio = 1.594
Mode (2,1): j ≈ 5.136,  ratio = 2.136

T = torch.tensor(50.0, requires_grad=True)
f1 = torch.sqrt(T / mu) / (2 * L)
f1.requires_grad   # → True (gradient flowed through sqrt and division)

f1 = √(T/μ) / (2L)
loss = (f1 − target)²
∂loss/∂T = 2(f1 − target) · ∂f1/∂T
         = 2(f1 − target) · 1 / (4Lμ · √(T/μ))

1import numpy as np
2import torch
3
4# ---------------------------------------------------------------
5# 1. Vibrating string  -- y_n(x, t) = A_n sin(n*pi*x/L) cos(omega_n t)
6#    omega_n = n * pi / L * sqrt(T / mu)
7# ---------------------------------------------------------------
8def string_frequency(n, L, T, mu):
9    """Return the n-th natural frequency (Hz) of a vibrating string."""
10    c = np.sqrt(T / mu)        # wave speed
11    omega_n = n * np.pi * c / L
12    return omega_n / (2 * np.pi)
13
14
15def string_motion(x, t, L, T, mu, n_modes=10, pluck=0.2):
16    """Time evolution of a plucked string using Fourier series."""
17    c = np.sqrt(T / mu)
18    y = np.zeros_like(x)
19    for n in range(1, n_modes + 1):
20        # Fourier coefficient for triangular pluck at x = pluck * L
21        b_n = (2.0 / (n * np.pi) ** 2) * np.sin(n * np.pi * pluck) / (
22            pluck * (1.0 - pluck)
23        )
24        omega_n = n * np.pi * c / L
25        y += b_n * np.sin(n * np.pi * x / L) * np.cos(omega_n * t)
26    return y
27
28
29# ---------------------------------------------------------------
30# 2. Pipes  --  open-open:  f_n = n*c/(2L)
31#               closed-open: f_n = (2n-1)*c/(4L)
32# ---------------------------------------------------------------
33def pipe_frequencies(L, kind="open-open", n_modes=5, c=343.0):
34    """Return the first n_modes resonant frequencies (Hz) of a pipe."""
35    if kind == "open-open":
36        return np.array([n * c / (2 * L) for n in range(1, n_modes + 1)])
37    elif kind == "closed-open":
38        return np.array([(2 * n - 1) * c / (4 * L) for n in range(1, n_modes + 1)])
39    else:
40        raise ValueError(f"Unknown pipe kind: {kind}")
41
42
43# ---------------------------------------------------------------
44# 3. Circular membrane -- u_mn = J_m(j_mn r/a) cos(m*theta) cos(omega_mn t)
45#    omega_mn = c * j_mn / a, where j_mn = n-th zero of J_m.
46# ---------------------------------------------------------------
47from scipy.special import jn_zeros
48
49def drum_frequency(m, n, a, T, sigma):
50    """Return the (m, n) mode frequency (Hz) of a circular drum."""
51    c = np.sqrt(T / sigma)        # wave speed on the membrane
52    j_mn = jn_zeros(m, n)[-1]     # n-th zero of J_m
53    omega_mn = c * j_mn / a
54    return omega_mn / (2 * np.pi)
55
56
57# ---------------------------------------------------------------
58# 4. PyTorch demo: fit string tension to a target pitch
59# ---------------------------------------------------------------
60def fit_tension_to_pitch(target_hz, L, mu, lr=1.0, steps=400):
61    """Use autograd to find the tension T that produces target_hz on mode 1."""
62    # Start with a guess; require_grad lets PyTorch track derivatives.
63    T = torch.tensor(50.0, requires_grad=True)
64    optim = torch.optim.Adam([T], lr=lr)
65    for step in range(steps):
66        optim.zero_grad()
67        c = torch.sqrt(T / mu)
68        f1 = c / (2 * L)
69        loss = (f1 - target_hz) ** 2
70        loss.backward()
71        optim.step()
72    return T.detach().item()
73
74
75if __name__ == "__main__":
76    # Guitar high E (frequency ~ 329.63 Hz)
77    L, mu = 0.65, 0.0007
78    f1 = string_frequency(1, L=L, T=80.0, mu=mu)
79    print(f"Guitar E1 with T=80 N : {f1:6.2f} Hz")
80
81    # Find tension that hits exactly 329.63 Hz
82    T_best = fit_tension_to_pitch(329.63, L, mu)
83    print(f"Tension to hit 329.63 Hz: {T_best:6.2f} N")
84
85    # Clarinet length to play middle C (f1 ~ 261.63 Hz, closed-open pipe)
86    L_clarinet = 343.0 / (4 * 261.63)
87    print(f"Closed-open pipe length : {L_clarinet:6.3f} m")
88    print(f"Overtones:               {pipe_frequencies(L_clarinet, 'closed-open')}")
89
90    # Timpani (drum) fundamental for a 30 cm radius head
91    f_drum = drum_frequency(m=0, n=1, a=0.30, T=2500.0, sigma=0.25)
92    print(f"Timpani fundamental f01 : {f_drum:6.2f} Hz")

Common Pitfalls

Summary

The three classical instrument families — strings, pipes, drums — are the wave equation looked at through three different boundary windows. One PDE, three geometries, an entire orchestra.

Key formulas

Key takeaways

1. Pitch is an eigenvalue. The geometry of the instrument decides which eigenvalues are allowed; everything else is consequence.
2. Timbre is a Fourier coefficient. The way you excite the instrument (pluck, strike, bow, blow) decides the relative amplitudes of the modes — and that decides what the note sounds like.
3. Boundary conditions matter more than material. A flute and a clarinet of the same length play an octave apart purely because one end is closed.
4. Drums are mathematical proof that not every wave is harmonic. Bessel-zero ratios are irrational, which is why percussion exists as a separate musical category.
5. Inverse problems are the future. PyTorch's autograd lets us pose questions like "what tension produces this pitch?" without a closed-form inverse — the same trick used for shape optimization, antenna design, and acoustic metamaterials.

Coming Next: In Section 27.7 we leave the orchestra for the laboratory and apply the same machinery to electromagnetic waves — Maxwell's equations reduce to a vector wave equation whose boundary problems describe antennas, waveguides, and the colors of light.