Physics Study Guide/Wave overtones

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Wave overtones

For resonance in a taut string, the first harmonic is determined for a wave form with one antinode and two nodes. That is, the two ends of the string are nodes because they do not vibrate while the middle of the string is an antinode because it experiences the greatest change in amplitude. This means that one half of a full wavelength is represented by the length of the resonating structure.

The frequency of the first harmonic is equal to wave speed divided by twice the length of the string. (Recall that wave speed is equal to wavelength times frequency.)

F1 = v/2L

The wavelength of the first harmonic is equal to double the length of the string.

λ1 = 2L

The "nth" wavelength is equal to the fundamental wavelength divided by n.

λn = λ1/n

Harmonics for a taut string*

  Harmonic number Overtone number F = λ =
F1 First harmonic --- F1 = v/2L λ1 = 2L
F2 Second harmonic First overtone F2 = 2F1 λ2 =λ1/2
F3 Third harmonic Second overtone F3 = 3F1 λ3 = λ1/3
Fn Nth harmonic (Nth - 1) overtone Fn = nF1 λn = λ1/n

* or any wave system with two identical ends, such as a pipe with two open or closed ends. In the case of a pipe with two open ends, there are two antinodes at the ends of the pipe and a single node in the middle of the pipe, but the mathematics work out identically.

Definition of terms

Frequency (F): Units: (1/s), hertz (Hz)

Fundamental frequency, first harmonic (F)1: The lowest frequency (longest wavelength) allowed for the system.

Length of string (L): (or pipe, etc.) Units: meters (m).

Wavelength (λ): Units: meters (m).

The first overtone is the first allowed harmonic above the fundamental frequency (F1).

In the case of a system with two different ends (as in the case of a tube open at one end), the closed end is a node and the open end is an antinode. The first resonant frequency has only a quarter of a wave in the tube. This means that the first harmonic is characterized by a wavelength four times the length of the tube.

F1 = v/4L

The wavelength of the first harmonic is equal to four times thelength of the string.

λ1 = 4L

The "nth" wavelength is equal to the fundamental wavelength divided by n.

λn = λ1/n

Note that "n" must be odd in this case as only odd harmonics will resonate in this situation.

Harmonics for a system with two different ends*

  Harmonic number Overtone number F = λ =
F1 First harmonic --- F1 = v/4L λ1 = 4L
F2 Third harmonic First overtone F2 = 3F1 λ2 =2λ1/3
F3 Fifth harmonic Second overtone F3 = 5F1 λ3 = 2λ1/5
Fn Nth harmonic (Nth - 1)/2 overtone F(n-1)/2 = nF1 λn = λ1/n

* such as a pipe with one end open and one end closed
In this case only the odd harmonics resonate, so n is an odd integer.

Vs: velocity of sound

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