Speak into the microphone of your phone. A thin sheet inside it, the diaphragm, moves as the air presses on it, and the phone turns that motion into a signal. The diaphragm responds to the pressure of the air. But the air particles near it are also moving to and fro. Are the pressure and the motion of the air two names for the same thing?
Two perspectives on sound
Sound has a double identity. We can describe it by how far the individual particles move, which is the displacement. Or we can describe it by how the local pressure varies. This lesson explores both views, and then joins them.
1. The displacement wave, s(x, t)
The displacement wave tells how far each particle has shifted from its equilibrium position as the wave passes. For a pure sound of a single frequency, its motion is captured by
\[ s(x, t) = s_{\text{max}}\cos(kx - \omega t) \]
Notice that the motion, forward and backward, is along the same line on which the wave travels. The animation of the piston in Lesson 1 used an equation of the same type.
Suppose we capture the positions of all the particles at one instant. Beside them we mark their mean positions, where they would sit if there were no sound. Fig 2.1 shows the result.
Notice how the compressions and rarefactions appear. At each compression C, the particles on both sides have moved towards C. This crowding makes a region of higher density, and so of higher pressure. At each rarefaction R, the particles on both sides have moved away from R. This spreading makes a region of lower density and lower pressure.
Fig 2.1 also shows two graphs under the particles. The upper one is the displacement wave \(s\), and the lower one is the pressure variation \(\Delta P\). The next section builds an exact mathematical link between them.
2. The pressure wave, \(\Delta P(x, t)\)
As the particles move, they squeeze or release the spaces between them. This causes tiny changes of pressure. With a sound wave present, the total pressure at \(x\) and \(t\) is
\[ P(x, t) = \underbrace{P_0}_{\text{ambient pressure}} + \underbrace{\Delta P(x, t)}_{\text{caused by sound}} \]
Here \(P_0\) is the ambient pressure of the still air, and \(\Delta P\) is the pressure variation caused by the sound.
To connect displacement and pressure, imagine a narrow tube of air of cross-section \(A\). At each position \(x\) and time \(t\), the particles are displaced by \(s(x, t)\). Over a tiny span \(dx\), the particles at \(x\) and at \(x + dx\) may move by different amounts. That difference changes the volume of the thin layer between them.
The change of volume makes a change of pressure. How large a change depends on the bulk modulus \(B\) of the medium. The bulk modulus measures how hard it is to compress the medium. A small fractional squeeze, \(\Delta V/V\), raises the pressure by \(-B\,\Delta V/V\).
\[ V = A\,dx \]
Why. at rest, the layer between \(x\) and \(x + dx\) has cross-section \(A\) and thickness \(dx\).
\[ \Delta V = A\left(s(x + dx, t) - s(x, t)\right) \]
Why. the right face moves by \(s(x + dx, t)\) and the left face by \(s(x, t)\). The thickness changes by their difference.
\[ \Delta P = -B\,\dfrac{\Delta V}{A\,dx} \]
Why. the bulk modulus gives the pressure change for a fractional change of volume \(\Delta V/V\). The minus sign says that a squeeze, \(\Delta V < 0\), raises the pressure.
\[ \Delta P = -B\,\dfrac{s(x + dx, t) - s(x, t)}{dx} \]
Why. put step 2 into step 3. The area \(A\) cancels.
\[ \Delta P = -B\,\dfrac{\partial s}{\partial x} \]
Why. as \(dx\) becomes very small, the change in \(s\) per unit length, at a fixed \(t\), is the partial derivative \(\partial s/\partial x\).
\[ \dfrac{\partial s}{\partial x} = -k\,s_{\text{max}}\sin(kx - \omega t) \]
Why. differentiate \(s = s_{\text{max}}\cos(kx - \omega t)\) with \(t\) held fixed. The derivative of \(\cos\) is \(-\sin\), and the chain rule brings out \(k\).
\[ \Delta P(x, t) = Bk\,s_{\text{max}}\sin(kx - \omega t) \]
Read it. put step 6 into step 5. The two minus signs give a plus.
So the change of volume per unit volume is just the slope of the displacement, \(\partial s/\partial x\). A cosine displacement wave gives a sine pressure wave. Its amplitude is the pressure amplitude, \(\Delta P_{\text{max}} = Bk\,s_{\text{max}}\).
A sound wave in air has a displacement amplitude of \(1.0 \times 10^{-7}\,\text{m}\) and a wave number of \(4\pi\,\text{rad/m}\). The bulk modulus of air is \(1.42 \times 10^5\,\text{Pa}\). What is the pressure amplitude, in Pa?
The phase relation
Compare the two equations side by side:
- the displacement wave is \(s(x, t) = s_{\text{max}}\cos(kx - \omega t)\);
- the pressure wave is \(\Delta P(x, t) = \Delta P_{\text{max}}\sin(kx - \omega t)\).
The displacement is a cosine, while the pressure is a sine. They are out of phase by 90°, that is, by \(\pi/2\) radians.
Let us read the same displacement graph once again, this time for the pressure. Fig 2.2 marks three points on it.
At the point P, the centre of a compression. The displacement \(s\) is zero there. The slope \(\partial s/\partial x\) is most negative. A negative slope means that the particles to the left of P have moved right, towards P. The particles to the right of P have moved left, also towards P. Particles from both sides pile up at P, so P is a region of maximum pressure.
At the point Q, the centre of a rarefaction. The displacement is again zero. The slope \(\partial s/\partial x\) is most positive. A positive slope means that the particles to the left of Q have moved left, away from Q. The particles to the right of Q have moved right, also away from Q. Particles leave Q on both sides, so Q is a region of minimum pressure.
At a point of maximum displacement, such as the crest. The displacement \(s\) is greatest there. The slope \(\partial s/\partial x\) is zero. So all the particles close to the crest have moved by the same largest amount; they have shifted together as a block. Nothing squeezes or stretches them relative to one another. So the pressure variation \(\Delta P\) at the crest must be zero.
This confirms the 90° phase shift. The points of maximum and minimum pressure are points of zero displacement. The points of maximum displacement are points of zero pressure variation. In Fig 2.2 each pair stays lined up as the waves travel.
Do not look for the highest pressure where the air has moved the most. Pressure follows the slope of the displacement graph, not its height. Where the particles have moved the most, they have all moved together, and the air is neither squeezed nor stretched.
In a sound wave, the displacement and the pressure variation are out of phase by
A sound wave travels through air at room temperature. Its displacement is \(s(x, t) = 2.0 \times 10^{-8}\cos\left[2\pi(2.0x - 686t)\right]\), with \(s\) and \(x\) in metres and \(t\) in seconds. The bulk modulus of air is \(B = 1.42 \times 10^5\,\text{Pa}\). What is the equation for the pressure variation \(\Delta P(x, t)\)?
Read the wave: \(s_{\text{max}} = 2.0 \times 10^{-8}\,\text{m}\), \(k = 2\pi \times 2.0\,\text{rad/m}\) and \(\omega = 2\pi \times 686\,\text{rad/s}\). So \(\lambda = 0.5\,\text{m}\) and \(f = 686\,\text{Hz}\), and \(v = \lambda f = 343\,\text{m/s}\), the speed of sound in air.
Use the general relation \(\Delta P(x, t) = -B\,\partial s/\partial x\). Differentiate with \(t\) held fixed: \(\dfrac{\partial s}{\partial x} = -2.0 \times 10^{-8} \cdot 2\pi \cdot 2.0\,\sin\left[2\pi(2.0x - 686t)\right]\).
Substitute: \(\Delta P = -1.42 \times 10^5\left[-2.0 \times 10^{-8} \cdot 2\pi \cdot 2.0\,\sin\left[2\pi(2.0x - 686t)\right]\right]\). This is \((1.42 \times 10^5)(2.0 \times 10^{-8})(2\pi \cdot 2.0)\sin\left[2\pi(2.0x - 686t)\right] = 0.0357\sin\left[2\pi(2.0x - 686t)\right]\,\text{Pa}\).
Final answer: \(\Delta P(x, t) = 0.0357\sin\left[2\pi(2.0x - 686t)\right]\,\text{Pa}\). The pressure amplitude \(Bk\,s_{\text{max}}\) is only \(0.0357\,\text{Pa}\), a few parts in ten million of the air pressure.
Where is the pressure variation greatest in a sound wave?
In a snapshot of a sound wave, the graph of \(s\) against \(x\) crosses zero at a point with a positive slope. What is the air doing there?
The diaphragm of a microphone responds to the pressure wave, \(\Delta P\), and not to the displacement. The pressures are tiny. In the example above, the pressure amplitude is \(0.0357\,\text{Pa}\). The air around it is at about \(1.013 \times 10^5\,\text{Pa}\), roughly three million times larger. A good microphone detects changes of a few parts in ten million of the air pressure.
Now return to the question of the microphone. The pressure and the motion of the air are not two names for one thing. The diaphragm feels the greatest push where the air particles pass through their mean positions, at the compressions. Where the air has moved the most, the pressure is just the ambient pressure. The two waves travel together, a quarter wavelength apart.
- Placing the pressure maximum at the displacement maximum.The pressure follows the slope, \(-B\,\partial s/\partial x\). Its extremes are at the zeros of \(s\).
- Dropping the minus sign in \(\Delta P = -B\,\partial s/\partial x\).A negative slope means particles converge, which raises the pressure. The minus sign says exactly that.
- Forgetting the factor \(k\) in the pressure amplitude.\(\Delta P_{\text{max}} = Bk\,s_{\text{max}}\). The chain rule brings out \(k\) when \(\cos(kx - \omega t)\) is differentiated.