Why a violin doesn't sound like a flute
Tonight's idea
A violin and a flute both play A = 440 Hz. A tuner says they match; your ear disagrees. Same pitch, different timbre.
A string of length \(L\) fixed at both ends vibrates only in shapes still at the ends: its modes; a still point is a node. One hump is half a wavelength, \(\lambda_1 = 2L\); two humps, \(\lambda_2 = L\); in general \(\lambda_n = 2L/n\).
Frequency is wave speed over wavelength, so \(f_n = v/\lambda_n = n \cdot v/(2L) = n f_1\): whole-number multiples of the lowest, the harmonics.
A real string does several modes at once. The lowest, \(f_1\), is the pitch; the rest colour it. A flute is nearly a pure sine, a violin is rich in upper harmonics, and a clarinet drops the even ones and sounds hollow.
Fourier (1807): a repeating sound of period \(T\) is a constant plus sines at \(1/T, 2/T, 3/T, \ldots\), each with an amplitude and a phase (a time shift). Amplitudes alone are not enough: every \(\sin(k\omega t)\) is zero at \(t = 0\), so no mix of them is \(\cos \omega t\), which is 1 there.
The arrow is \(e^{i\omega t}\) from Corner 7 and Corner 8. For a steady tone the amplitudes, the spectrum, are most of timbre; attack and fade matter too. Build one at mathdetours.com/demos/fourier.html.
Watch one: the harmonics of A = 220 Hz, and beats
A 220 Hz string also holds 440, 660, 880 and 1100 Hz: A, E, A, C♯. A major chord in one note.
Two flutes play 440 Hz and 444 Hz. With \(\sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}\):
\[\sin(2\pi\cdot 440 t) + \sin(2\pi\cdot 444 t) = 2\cos(2\pi\cdot 2t)\,\sin(2\pi\cdot 442 t).\]
You hear 442 Hz swell and fade with the slow cosine. Loudness ignores sign, so each cosine cycle gives two beats: 4 beats a second, the difference of the frequencies. Piano tuners tune by beats.
Your turn
- The low A string of a guitar is 65 cm long and sounds 110 Hz. List its first six harmonics, the wavelength of each of the first three modes, and the speed of waves on the string.
- Guitarists touch a string lightly to force a node there. Touch it at the midpoint: which harmonic sounds, at what frequency? At one third? At one quarter? Why does one quarter not give the second harmonic?
- You hold a 440 Hz tuning fork against a string and hear 3 beats per second. You tighten the string a little and the beats slow to 1 per second. Was the string at 437 Hz or 443 Hz before? Explain.
- Notes at 300 Hz and 303 Hz are played together. Write the sum of the two sines as a product, state the frequency you hear and the number of beats per second.
Stretch: a square wave, and a formula for \(\pi\)
Fourier's recipe for the square wave that jumps between \(+1\) and \(-1\) is \(\text{square}(x) = \frac{4}{\pi}\left(\sin x + \frac{\sin 3x}{3} + \frac{\sin 5x}{5} + \cdots\right)\). Put \(x = \pi/2\), where the square wave is 1 and every sine is \(\pm1\). Rearrange into a formula for \(\pi\) and add the first five terms. How close are you?
Answers
- 110, 220, 330, 440, 550, 660 Hz. Wavelengths \(2L/n\): 130 cm, 65 cm, 43.3 cm. Speed \(v = f\lambda = 110 \times 1.30 = 143\) m/s.
- Midpoint: 2nd harmonic, 220 Hz. One third: 3rd, 330 Hz. One quarter: 4th, 440 Hz. The 2nd mode's only interior still point is at \(L/2\); the lowest mode that is still at \(L/4\) is the 4th, and that is the one that sounds.
- 437 Hz. Tightening raises the pitch. A string at 443 would move further from 440 and the beats would speed up. At 437 it moves to 439, and \(440 - 439 = 1\) beat per second.
- \(\sin(2\pi\cdot300t) + \sin(2\pi\cdot303t) = 2\cos(2\pi\cdot1.5t)\sin(2\pi\cdot301.5t)\). You hear 301.5 Hz, and the envelope \(|\cos(2\pi\cdot 1.5t)|\) peaks 3 times a second: 3 beats per second.
Stretch. \(1 = \frac{4}{\pi}\left(1 - \frac13 + \frac15 - \frac17 + \cdots\right)\), so \(\pi = 4\left(1 - \frac13 + \frac15 - \cdots\right)\), Leibniz's series. Five terms: \(0.8349 \times 4 = 3.340\). Ten terms give 3.042. The error after \(n\) terms is about \(1/n\), so nobody computes \(\pi\) this way, but a sound wave knows about \(\pi\).