What sets the note of a string
The last lesson ended with a claim: the fundamental frequency of a string depends on its length, its tension and its mass density. Equation (14) shows the length directly, \(f_1 = v/2L\). The other two hide inside the speed \(v\). The chapter Propagating Waves on a String found that speed from Newton's second law: \(v = \sqrt{T/\mu}\). Here \(T\) is the tension and \(\mu\) is the mass per unit length of the string.
\[ f_1 = \dfrac{v}{2L} \]
Why. equation (14) with \(n = 1\), for a string fixed at both ends.
\[ v = \sqrt{\dfrac{T}{\mu}} \]
Why. the speed of a transverse wave on a stretched string, from the chapter Propagating Waves on a String.
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Your turn. put step 2 into step 1. Which line is it?
\[ f_n = \dfrac{n}{2L}\sqrt{\dfrac{T}{\mu}} \]
Read it. every harmonic is \(n\) times the fundamental.
A steel wire of mass per unit length \(1.0\,\text{g/m}\) is stretched between two bridges \(0.60\,\text{m}\) apart with a tension of \(90\,\text{N}\). Find its fundamental frequency, in Hz.
Three laws in one formula
The formula \(f_1 = (1/2L)\sqrt{T/\mu}\) contains three separate laws. Each law holds one quantity at a time, with the other two kept fixed.
For a stretched string fixed at both ends:
The law of mass is often used for wires. A wire of radius \(r\) made of a material of density \(\rho\) has a mass \(\rho\pi r^2\) in each metre of its length. So \(\mu = \pi r^2\rho\), and
\[ f_1 = \frac{1}{2Lr}\sqrt{\frac{T}{\pi\rho}}. \]
Two wires of the same material, length and tension differ only in radius. The thicker wire has the lower note, with \(f_1 \propto 1/r\).
A string's vibrating length is halved, with its tension and its mass per unit length unchanged. What happens to its fundamental frequency?
A steel wire has a fundamental of \(400\,\text{Hz}\). It is replaced by a steel wire of twice the radius, with the same length and tension. Find the new fundamental, in Hz.
A sitar player uses all three laws. To play a melody, she presses the main string against a curved fret. This shortens the vibrating length, so the note rises: the law of length. Before playing, she turns the pegs to tune each string. This changes the tension: the law of tension. The low bass strings are thicker than the main string, so their larger mass per unit length gives lower notes: the law of mass.
The thinnest string of a guitar is \(0.65\,\text{m}\) long between its ends and sounds \(330\,\text{Hz}\). Its mass per unit length is about \(0.4\,\text{g/m}\). Estimate its tension, in newtons.
The sonometer
A sonometer tests the three laws in the laboratory. It is a hollow wooden box with a thin wire along its top. One end of the wire is tied to a peg. The wire passes over two sharp bridges and then over a pulley, where a hanger of mass \(M\) pulls it tight, so \(T = Mg\). One bridge is fixed and the other slides, so the vibrating length \(L\) between them can be set. The hollow box vibrates with the wire and makes its sound louder.
To compare the wire with a tuning fork, a small V of paper, the paper rider, is placed on the wire midway between the bridges. The fork is struck and its stem is pressed on the box. The box carries the fork's vibration to the wire. The bridge is then moved until the wire's fundamental matches the fork. At that length the wire resonates, as Lesson 3 explained. Its middle is an antinode and swings widely, and the rider jumps off.
In the figure above start with \(M = 4\,\text{kg}\), so \(T = 40\,\text{N}\) and \(v = 200\,\text{m/s}\). At \(L = 0.50\,\text{m}\) the wire gives \(200\,\text{Hz}\), and the rider stays. Slide the bridge in. At \(L = 0.39\,\text{m}\) the wire gives \(256.4\,\text{Hz}\), and the rider jumps. Press Listen near that length. As the two tones match, they merge into one nearly steady note. Now raise the mass to \(5\,\text{kg}\). The rider settles again, and it jumps next at \(L = 0.44\,\text{m}\), where \(f = 254.1\,\text{Hz}\).
With the tension fixed at \(40\,\text{N}\), a set of forks gives a set of resonant lengths. Each length is \(L = v/2f = 100/f\) metres. The product \(fL\) stays the same, which is the law of length.
The other two laws are tested the same way. With one fork, change the tension and find the new resonant length each time. Since \(f_1\) stays fixed, \(L \propto \sqrt{T}\), and \(L/\sqrt{T}\) stays constant. With one fork and one tension, wires of different mass per unit length give \(L \propto 1/\sqrt{\mu}\).
A sonometer wire resonates with a tuning fork when its vibrating length is \(0.40\,\text{m}\). The tension is made four times as large. Find the new vibrating length that resonates with the same fork, in m.
Melde's experiment
In 1859 the German physicist Franz Melde made standing waves on a thread in a new way. In Melde's experiment, one end of a light string is tied to a prong of an electrically maintained tuning fork. Such a fork keeps vibrating steadily. The string runs horizontally over a smooth pulley and carries a hanging mass, which sets the tension. The prong moves so little that its end of the string is nearly a node. The pulley end is a node too. So the string of length \(L\) settles into \(p\) loops, with \(L = p(\lambda/2)\).
The fork can be set in two ways.
- Prongs across the string. The fork lies so that its prongs vibrate at right angles to the string. Each vibration of the prong pushes the string sideways once. The string vibrates at the fork's frequency \(f\), so \(f = (p/2L)\sqrt{T/\mu}\).
- Prongs along the string. The fork stands so that its prongs vibrate along the length of the string. The prong no longer pushes the string sideways. It only tightens and slackens the string, once in each of its vibrations. The string then vibrates at half the fork's frequency, so \(f/2 = (p/2L)\sqrt{T/\mu}\).
Why half? A tightened string pulls hardest towards its rest line when it is displaced. The prong moves away from the pulley, and tightens the string, once in each of its own vibrations. For the string to be pulled back at its top and again at its bottom, it needs two such tugs in each of its oscillations. So the prong makes two vibrations while the string makes one.
A string of mass per unit length \(1.0\,\text{g/m}\) is tied to a prong of a \(100\,\text{Hz}\) fork. It is \(1.2\,\text{m}\) long up to the pulley, and the tension is \(3.6\,\text{N}\). Find the number of loops (a) with the prongs across the string and (b) with the prongs along it. Then find (c) the tension that gives 4 loops in the second arrangement.
The speed on the string is \(v = \sqrt{T/\mu} = \sqrt{3.6/10^{-3}} = 60\,\text{m/s}\). Both ends are nodes, so \(p = 2L/\lambda\).
Across: the string vibrates at \(100\,\text{Hz}\), so \(\lambda = v/f = 60/100 = 0.6\,\text{m}\). Along: it vibrates at \(50\,\text{Hz}\), so \(\lambda = 60/50 = 1.2\,\text{m}\).
(a) \(p = 2 \times 1.2/0.6 = 4\) loops. (b) \(p = 2 \times 1.2/1.2 = 2\) loops. (c) Four loops need \(\lambda = 0.6\,\text{m}\) at \(50\,\text{Hz}\), so \(v = 30\,\text{m/s}\) and \(T = \mu v^2 = 10^{-3} \times 900 = 0.9\,\text{N}\).
Turning the fork from across to along halves the loops at the same tension. To win them back, the tension must fall to a quarter, from \(3.6\,\text{N}\) to \(0.9\,\text{N}\).
For one fork, one string and one arrangement, the formula gives \(p\sqrt{T} = 2Lf\sqrt{\mu}\), a constant. So more loops need less tension: twice the loops need a quarter of the tension. The same formula finds the frequency of a fork from a count of loops.
In Melde's experiment the fork vibrates at \(f\), and its prongs move along the length of the string. At what frequency does the string vibrate?
In Melde's experiment with the prongs across the string, a string forms 4 loops at a tension of \(3.6\,\text{N}\). What tension makes it form 2 loops, in N?
The French friar Marin Mersenne set out the laws of a vibrating string in his book Harmonie universelle of 1636. He could not count the vibrations of a short, high string, which are far too fast for the eye. So he used very long strings, whose slow vibrations he could count. Then he used the law of length to scale his counts to short strings. He found that the frequency falls as the length grows and rises as the square root of the tension. It falls as the square root of the mass per unit length grows. These are still called Mersenne's laws.
- Using the mass per unit length in g/m inside \(\sqrt{T/\mu}\).Convert first: \(1\,\text{g/m} = 10^{-3}\,\text{kg/m}\). Then \(T\) in newtons gives \(v\) in m/s.
- Doubling the tension and expecting double the frequency.The frequency follows \(\sqrt{T}\). Double the frequency needs four times the tension.
- Assuming the string in Melde's experiment always vibrates at the fork's frequency.That holds only with the prongs across the string. With the prongs along the string it vibrates at half the fork's frequency.