The harmonic series
Equation (13) gives the wavelengths that a string fixed at both ends allows, \(\lambda_n = 2L/n\). A frequency follows from each wavelength through \(f = v/\lambda\), where \(v\) is the speed of waves on the string. The derivation below puts the two together.
\[ \lambda_n = \dfrac{2L}{n}, \qquad n = 1, 2, 3, \dots \]
Why. solve equation (13), \(L = n(\lambda_n/2)\), for \(\lambda_n\).
\[ f_n = \dfrac{v}{\lambda_n} \]
Why. for any wave, speed = frequency × wavelength. The speed \(v\) is the same for every mode, since it depends only on the string.
\[ f_n = n\left(\dfrac{v}{2L}\right) \]
Why. put step 1 into step 2.
\[ f_n = nf_1, \qquad f_1 = \dfrac{v}{2L} \]
Read it. name the lowest frequency, with \(n = 1\), as \(f_1\). This is equation (14).
The modes have names.
- \(n = 1\): the fundamental frequency, also called the first harmonic, \(f_1 = v/2L\).
- \(n = 2\): the second harmonic, or the first overtone, \(f_2 = 2f_1\).
- \(n = 3\): the third harmonic, or the second overtone, \(f_3 = 3f_1\).
Mode \(n\) has \(n\) loops and wavelength \(\lambda_n = 2L/n\). The overtones count only the frequencies above the fundamental, so the \(n\)th harmonic is the \((n - 1)\)th overtone.
For a string fixed at both ends, the resonant frequencies are all integer multiples of the fundamental frequency. This whole set is the string's harmonic series. Choose the chips of Fig 4.1 in turn and press play. Mode \(n\) oscillates \(n\) times as fast as mode 1, and its readout shows \(n \times 220\,\text{Hz}\). Now press Listen on each chip. The second harmonic, \(440\,\text{Hz}\), sounds like the same note one octave higher. The ear hears a doubled frequency as the same note, raised.
A string fixed at one end and free at the other
Now let the end at \(x = L\) be free. The chapter Superposition and Reflection made a free end with a massless ring that slides on a smooth vertical rod. The ring feels no vertical force, so the string there always has zero slope, and the ring swings with the largest displacement. So the rules change. The end at \(x = 0\) is a node, and the end at \(x = L\) is free to move, so it must be an antinode.
For a standing wave to exist here, its shape function, \(2A\sin(kx)\), must be zero at \(x = 0\) and must have its largest size at \(x = L\). The derivation below applies the second condition.
\[ 2A\sin(k \cdot 0) = 0 \]
Why. the node at the fixed end, \(x = 0\). It holds for every \(k\).
\[ |\sin(kL)| = 1 \]
Why. the antinode at the free end: \(|2A\sin(kL)| = 2A\). This is the boundary condition.
\[ kL = \left(n + \dfrac12\right)\pi, \qquad n = 0, 1, 2, \dots \]
Why. the sine is \(\pm 1\) only at odd multiples of \(\pi/2\). Here \(n = 0\) is allowed, since it gives \(kL = \pi/2\), not a flat string.
\[ \dfrac{2\pi}{\lambda_n}\,L = \dfrac{(2n + 1)\pi}{2} \]
Why. substitute \(k = 2\pi/\lambda\), and write \(n + \tfrac12\) as \((2n + 1)/2\).
\[ L = (2n + 1)\left(\dfrac{\lambda_n}{4}\right) \]
Read it. solve step 4 for \(L\). This is equation (15).
Equation (15) is equivalent to saying that the length must be an odd whole number of quarter wavelengths: \(L = m(\lambda/4)\) for \(m = 1, 3, 5, \dots\). A quarter wavelength runs from a node to the next antinode. So the string holds one, three, five or more such quarters.
The free-end rule of the chapter Superposition and Reflection gives the same result. At a free end the slope is zero. The slope of equation (11) is \(\partial y/\partial x = 2Ak\cos(kx)\cos(\omega t)\). For it to vanish at \(x = L\) at every instant, \(\cos(kL) = 0\). Since \(\sin^2 + \cos^2 = 1\), that is the same as \(|\sin(kL)| = 1\). An antinode is exactly where the string is level.
The odd harmonic series
The allowed wavelengths come from equation (15), \(\lambda_n = 4L/(2n + 1)\). With \(f = v/\lambda\), the allowed frequencies are equation (16):
\[ f = (2n + 1)\left(\frac{v}{4L}\right), \qquad n = 0, 1, 2, \dots \]
- \(n = 0\): the fundamental frequency, \(f_1 = v/4L\).
- \(n = 1\): the next mode, \(f_3 = 3(v/4L) = 3f_1\). This is the third harmonic.
- \(n = 2\): the next mode, \(f_5 = 5(v/4L) = 5f_1\). This is the fifth harmonic.
For a string with one free end, only the odd harmonics, \(f_1, 3f_1, 5f_1, \dots\), are resonant. The even harmonics cannot satisfy the boundary conditions, and they are absent. In Fig 4.3 every chip keeps a node at the wall and the ring at an antinode. An even harmonic would need a node at the ring too, and the free ring cannot be held still.
Two numbers are in play, and they are easy to mix up. In equation (16), \(n\) counts the modes from 0. The harmonic number is \(2n + 1\), and it names the frequency as a multiple of \(f_1\). So \(n = 1\) is the third harmonic, \(f_3\), not the second.
A cello string of length \(70.0\,\text{cm}\) is tuned to a fundamental frequency of \(220\,\text{Hz}\). Find (a) the speed of the transverse waves on this string and (b) the frequencies of the next two harmonics.
(a) The cello string is fixed at both ends, so the fundamental frequency follows equation (14): \(f_1 = v/(2L)\). So \(v = 2Lf_1 = 2(0.700\,\text{m})(220\,\text{Hz}) = 308\,\text{m/s}\).
(b) For a string fixed at both ends, all the harmonics are present. The next two are the second and the third harmonics. \(f_2 = 2f_1 = 2(220\,\text{Hz}) = 440\,\text{Hz}\), and \(f_3 = 3f_1 = 3(220\,\text{Hz}) = 660\,\text{Hz}\).
These are the first three tones that Listen plays in Fig 4.1.
Key ideas of resonance on a string
Four ideas carry this lesson and the last.
- Resonance occurs when waves reflect and return in phase with the driving source.
- For a string fixed at both ends, all the harmonics, \(f_1, 2f_1, 3f_1, \dots\), are present.
- For a string with one free end, only the odd harmonics, \(f_1, 3f_1, 5f_1, \dots\), are present.
- The fundamental frequency depends on the length of the string, its tension and its mass density, the mass per unit length.
The last idea is the subject of the next lesson.
A string of length \(0.8\,\text{m}\), fixed at both ends, resonates in its second harmonic. What is the wavelength of this mode?
A string is fixed at both ends. If the fundamental frequency is \(100\,\text{Hz}\), what is the frequency of the third harmonic?
A rope \(2.0\,\text{m}\) long is tied to a wall at one end, and its other end is a light ring on a smooth pole. Waves travel on it at \(40\,\text{m/s}\). Find the frequency of its third harmonic, in Hz.
A string fixed at one end and free at the other has a fundamental frequency of \(50\,\text{Hz}\). Which of these frequencies can it not resonate at?
Why does a string with one free end have no even harmonics?
The fixed end must be a node and the free end must be an antinode. From a node to an antinode is an odd number of quarter wavelengths, so \(L = (2n + 1)(\lambda/4)\). The frequencies are then odd multiples of \(v/4L\). An even multiple would fit a whole number of half wavelengths, with a node at the free end, which the free end does not allow.
The French mathematician Joseph Sauveur could not speak until the age of seven, and he never had a good ear for music. Yet he founded the science of musical sound. In 1701 he reported to the Academy of Sciences in Paris that a vibrating string divides into equal parts. He called the still points nodes and the swelling parts between them bellies, which English books call loops. He saw that a string sounds a set of higher tones with it, at two, three and four times its lowest frequency. He called the lowest tone the fundamental and the others harmonics. His names are the ones in this lesson.
A guitarist can make a string ring an octave higher without pressing it down. She touches the string lightly at its exact middle, over the twelfth fret, and plucks it. The finger forces a node at the middle. Every mode with an antinode there, the fundamental and the other odd harmonics, is stopped. The modes with a node at the middle, \(2f_1, 4f_1, \dots\), keep ringing. The note is the second harmonic, one octave above the open string, with a clear, bell-like sound.
- Using \(f_1 = v/2L\) for a string with one free end.A free end is an antinode, so the string holds a quarter wavelength in its fundamental: \(f_1 = v/4L\).
- Expecting a second harmonic on a string with one free end.With one free end only odd harmonics exist. The mode after \(f_1\) is \(3f_1\).
- Calling \(2f_1\) the second overtone.Overtones count from the first frequency above the fundamental. \(2f_1\) is the second harmonic and the first overtone.