A six is hit at a packed cricket stadium, and fifty thousand people roar at once. The noise is so loud that the players cannot hear each other shout. A television microphone at the boundary picks up every bit of it. The roar seems enormous. Yet the power in it is surprisingly small.
Intensity
A sound spreads its energy over an area. A loudspeaker sends the same energy each second across a small window near it and across a large wall far away. What the ear or a microphone feels is the energy that arrives each second on each square metre. This is the intensity of the sound.
Intensity is the objective, physical measure of the power of a sound spread over an area. It is the power that crosses a surface at right angles to the direction of travel, divided by the area of the surface: \(I = \text{Power}/\text{Area}\). Its unit is the watt per square metre, W/m².
Intensity is a physical quantity that an instrument can measure. It does not depend on who listens. How loud a sound seems to a listener is a different matter, which lesson 4 takes up.
Energy density times speed
Lesson 1 found the energy that a sound holds in each cubic metre of air. Intensity asks how fast that energy crosses a surface. The link is the wave speed. Consider an area \(A\) across a tube of air, with sound travelling to the right at speed \(v\).
\[ dE = \langle u\rangle\,(A\,v\,dt) \]
Why. the energy in the shaded slab of Fig 2.1 is the average energy density times the volume of the slab.
\[ P = \dfrac{dE}{dt} = \langle u\rangle\,A\,v \]
Why. all this energy crosses the area \(A\) in the time \(dt\), so this is the power through \(A\).
\[ I = \dfrac{P}{A} = \langle u\rangle\,v \]
Read it. divide by the area. The intensity is the average energy density times the wave speed.
This is the twin of a result for a string. There the average power was the average energy per unit length times the wave speed. Here the intensity is the average energy per unit volume times the wave speed. Each second the wave carries \(v\) metres of its energy past every square metre of the surface.
The loud sound of the example in lesson 1 has an average energy density of \(5.9 \times 10^{-6}\,\text{J/m}^3\) and travels at \(340\,\text{m/s}\). What is its intensity, in mW/m²?
The intensity of a harmonic sound
Put the average energy density of lesson 1, \(\tfrac12\rho\omega^2 s_{max}^2\), into \(I = \langle u\rangle v\). This gives the intensity in terms of the displacement amplitude. It is the twin of \(\tfrac12\mu v\omega^2A^2\) for a string.
The displacement amplitude is hard to measure, because it is so small. The pressure amplitude \(\Delta P_{max}\) is easier. The chapter Propagation of Sound Waves found \(\Delta P_{max} = Bk\,s_{max}\). Two facts turn this into a simpler form: \(B = \rho v^2\) and \(k = \omega/v\).
\[ I = \tfrac12\,\rho\,\omega^2 s_{max}^2 \cdot v \]
Why. the average energy density of lesson 1 times the wave speed.
\[ \Delta P_{max} = Bk\,s_{max} = (\rho v^2)\left(\dfrac{\omega}{v}\right)s_{max} = \rho v\,\omega\,s_{max} \]
Why. the pressure amplitude, with \(B = \rho v^2\) and \(k = \omega/v\).
\[ \omega\,s_{max} = \dfrac{\Delta P_{max}}{\rho v} \]
Why. solve step 2 for the product \(\omega s_{max}\), which appears squared in step 1.
\[ I = \tfrac12\,\rho v\left(\dfrac{\Delta P_{max}}{\rho v}\right)^2 \]
Why. put step 3 into step 1.
\[ I = \dfrac{(\Delta P_{max})^2}{2\rho v} \]
Read it. one factor \(\rho v\) cancels. The intensity grows as the square of the pressure amplitude.
For air with \(\rho = 1.2\,\text{kg/m}^3\) and \(v = 340\,\text{m/s}\), the product \(2\rho v\) is \(816\) in SI units. So in air \(I = (\Delta P_{max})^2/816\), with \(\Delta P_{max}\) in pascals. Twice the pressure amplitude gives four times the intensity.
The same intensity follows from force times velocity, as the power of a wave on a string did. The air to the left of a surface pushes on the air to the right with the extra pressure \(\Delta P\). That surface moves at \(\partial s/\partial t\). So the power per unit area is \(\Delta P\,\partial s/\partial t\). Lesson 1 showed \(\Delta P = \rho v\,\partial s/\partial t\), so this power is \((\Delta P)^2/(\rho v)\). It is never negative, and its average is \((\Delta P_{max})^2/(2\rho v)\). The product \(\rho v\) is called the acoustic impedance of the medium, the twin of \(\sqrt{T\mu}\) for a string.
Normal conversation delivers an intensity of about \(1.0 \times 10^{-6}\,\text{W/m}^2\) to a listener's ear. Take air with \(\rho = 1.2\,\text{kg/m}^3\) and \(v = 340\,\text{m/s}\). Find the pressure amplitude of this sound, and its displacement amplitude at a frequency of \(1\,\text{kHz}\).
Known: \(I = 1.0 \times 10^{-6}\,\text{W/m}^2\), \(\rho = 1.2\,\text{kg/m}^3\), \(v = 340\,\text{m/s}\) and \(f = 1000\,\text{Hz}\). So \(2\rho v = 816\) and \(\rho v = 408\) in SI units.
From \(I = (\Delta P_{max})^2/(2\rho v)\), \(\Delta P_{max} = \sqrt{2\rho v I}\). From \(\Delta P_{max} = \rho v\,\omega\,s_{max}\), \(s_{max} = \Delta P_{max}/(\rho v\,\omega)\).
\(\Delta P_{max} = \sqrt{816 \times 1.0 \times 10^{-6}} = 2.9 \times 10^{-2}\,\text{Pa}\). Then \(\omega = 2\pi \times 1000 = 6283\,\text{rad/s}\), so \(s_{max} = 0.0286/(408 \times 6283) = 1.1 \times 10^{-8}\,\text{m}\).
The pressure swings by only about \(0.03\,\text{Pa}\) about the atmospheric pressure of \(1.0 \times 10^5\,\text{Pa}\). The air moves to and fro by about \(10^{-8}\,\text{m}\), about a hundred times the size of an atom.
A sound in air has a pressure amplitude of \(0.20\,\text{Pa}\). Take \(\rho = 1.2\,\text{kg/m}^3\) and \(v = 340\,\text{m/s}\). What is its intensity, in μW/m²?
The pressure amplitude of a sound is doubled, and the medium stays the same. How does its intensity change?
An intensity of \(1.0\,\text{W/m}^2\) hurts the ear. Take \(\rho = 1.2\,\text{kg/m}^3\) and \(v = 340\,\text{m/s}\). What pressure amplitude is this, in pascals?
Try the slider in the figure above. At \(0.010\,\text{Pa}\) the intensity is \(0.12\,\mu\text{W/m}^2\). At \(0.020\,\text{Pa}\) it is \(0.49\,\mu\text{W/m}^2\), four times as much. At \(0.029\,\text{Pa}\) it is about \(1.0\,\mu\text{W/m}^2\), normal conversation. The displacement readout is only about \(11\,\text{nm}\) there.
Now compare two sounds of the same intensity, one at \(100\,\text{Hz}\) and one at \(1000\,\text{Hz}\). The pressure amplitude \(\Delta P_{max} = \sqrt{2\rho v I}\) contains no frequency, so it is the same for both. The displacement amplitude \(s_{max} = \Delta P_{max}/(\rho v\,\omega)\) falls as the frequency rises. At \(100\,\text{Hz}\) it is \(1.1 \times 10^{-7}\,\text{m}\), ten times the value at \(1\,\text{kHz}\).
So the pressure amplitude is the natural measure of a sound. A microphone has a thin diaphragm, and the extra pressure of the sound pushes it in and out. The eardrum is a diaphragm too. Both respond to \(\Delta P\), which tells the intensity directly, whatever the frequency. The displacement of the air is far too small to see, and it depends on the frequency as well.
The ear is astonishingly sensitive. At the faintest sound a young ear can hear, \(1.0 \times 10^{-12}\,\text{W/m}^2\), the pressure amplitude is about \(3 \times 10^{-5}\,\text{Pa}\). At \(1\,\text{kHz}\) the air then moves by about \(1 \times 10^{-11}\,\text{m}\). That is a tenth of the size of an atom. The eardrum senses a pressure, not this tiny motion.
Two sounds in air have the same intensity. One has a frequency of \(100\,\text{Hz}\) and the other \(1000\,\text{Hz}\). Which statement is true?
Explain why a microphone is built to respond to the pressure of a sound, and not to the displacement of the air.
A microphone's diaphragm is pushed in and out by the extra pressure of the sound, so it responds to the pressure amplitude. For a sound in a given medium the intensity is \((\Delta P_{max})^2/(2\rho v)\), which does not depend on the frequency. So one pressure reading gives the intensity at any frequency. The displacement amplitude is tiny, about \(10^{-8}\,\text{m}\) for normal conversation. It also falls as \(1/f\) at a fixed intensity, so it would give different readings for equally intense sounds.
A loud shout carries about a thousandth of a watt of sound power. Estimate the total sound power of \(50\,000\) people roaring together in a stadium, in watts.
In 1917 Edward Wente, an engineer at the Western Electric company in the United States, described a new condenser microphone. A thin stretched diaphragm formed one plate of a capacitor. The pressure of a sound moved the diaphragm and changed the capacitance. Wente showed that it responded almost equally to sounds of every frequency in the range of speech. So it could measure the intensity of a sound from its pressure. Microphones for measuring sound are still built this way.
Now return to the roaring stadium. The fans pour out only about \(50\,\text{W}\) of sound, less than a ceiling fan uses. Yet each square metre near the boundary receives far more than the \(10^{-6}\,\text{W/m}^2\) of a conversation. The boundary microphone does not track the motion of the air, which is tiny. It reads the pressure of the roar, which tells its intensity directly.
- Treating intensity as power.Intensity is power per unit area, in W/m². The same power gives a smaller intensity when it is spread over a larger area.
- Taking the intensity to be proportional to \(\Delta P_{max}\).\(I = (\Delta P_{max})^2/(2\rho v)\). Twice the pressure amplitude gives four times the intensity.
- Writing \(\Delta P_{max} = \rho v\,s_{max}\).\(\Delta P_{max} = \rho v\,\omega\,s_{max}\). The angular frequency is part of it, since \(\omega s_{max}\) is the largest particle velocity.
- Thinking two equally intense sounds swing the air equally.At a fixed intensity \(s_{max}\) falls as \(1/f\); only \(\Delta P_{max}\) is the same.