The medium, not the source
One of the most basic facts about the speed of a wave is this: the speed belongs to the medium through which the wave travels. It does not depend on the source of the sound. It does not depend on the loudness, the frequency, or the way the sound was made.
To make this idea concrete, consider a row of dominoes. A ball strikes the first domino in a long line and starts a chain reaction. Each part of the picture has a partner in the physics of sound:
- the source: the ball, which provides the first energy;
- the medium: the line of dominoes, which stands for the particles;
- the wave: the chain reaction, which stands for the wave itself.
A ball strikes the first domino of a long row, once gently and once hard. Far down the row, how does the speed of the toppling front compare in the two runs?
A common question is this: if the ball hits harder, will the wave not travel faster? The first few dominoes may topple more violently. But the speed along the long line quickly settles into a steady speed. That speed is decided not by the first push but by the dominoes themselves: their mass, their height and their spacing. The source starts the wave, but the medium sets the speed limit.
In the figure above, the motion is slowed ten times. After a hard strike, the front starts at \(2.20\,\text{m/s}\) and slows. After a gentle strike, it starts at \(0.40\,\text{m/s}\) and speeds up. Within a few dominoes, both readouts show \(1.00\,\text{m/s}\). In this model that speed belongs to the row of dominoes and to nothing else.
This principle holds for every wave, and one formula captures it:
\[ v = \sqrt{\frac{\text{elastic property}}{\text{inertial property}}} \]
For a fluid, the elastic property is its resistance to compression, the bulk modulus \(B\). The inertial property is its density \(\rho\). So the speed of sound in a fluid should be \(v = \sqrt{B/\rho}\). We now derive this step by step from Newton's second law. The derivation is the twin of the one for a string in the chapter Propagating Waves on a String.
A stiff medium springs back hard, so each layer hands the push on quickly. A heavy medium responds slowly, because each layer has more inertia. The speed grows with the stiffness and falls with the density. On a string the same pair is the tension and the mass per unit length.
A drummer strikes a drum twice as hard. In the same air, what happens to the speed of the sound?
A thin slab of fluid
Let us see how Newton's second law leads straight to the speed of sound in a fluid. Imagine an extremely thin cylinder of fluid, a tiny slice. Its cross-sectional area is \(A\) and its thickness is \(\Delta x\). A sound wave is about to pass through it.
The mass of the slice is its volume times its density, \(m = \rho V = \rho A\,\Delta x\). As the sound wave passes, the pressure on the two faces of the slice differs slightly. It is higher on one face and lower on the other. The difference gives a net force. Fig 3.2 shows the forces on the thin layer.
\[ F_{\text{left}} = P(x)A \]
Why. pressure is force per unit area, so the fluid on the left pushes the left face to the right with \(P(x)A\).
\[ F_{\text{right}} = P(x + \Delta x)A \]
Why. the fluid on the right pushes the right face to the left with the pressure there times the same area.
\[ F_{net} = \left[P(x) - P(x + \Delta x)\right]A \]
Why. take the right as positive. The left push counts plus and the right push counts minus.
\[ P(x + \Delta x) - P(x) = \dfrac{\partial P}{\partial x}\Delta x \]
Why. for a very thin slice, the change of pressure across it is its rate of change along \(x\) times the thickness.
\[ F_{net} = -\dfrac{\partial P}{\partial x}A\,\Delta x \]
Read it. put step 4 into step 3. The bracket is minus the change of pressure.
The sign makes sense. If the pressure falls towards the right, \(\partial P/\partial x\) is negative, and the net force points to the right. Fluid is always pushed from high pressure towards low pressure.
Newton's second law for the slice
By Newton's second law, the net force equals mass times acceleration. The acceleration of the slice is the second time derivative of its displacement \(s(x, t)\), so \(a = \partial^2 s/\partial t^2\). To link pressure with displacement we use the elastic property of the medium, the bulk modulus \(B\). From Lesson 2, the pressure variation is \(\Delta P = -B\,\partial s/\partial x\), and \(P = P_0 + \Delta P\).
\[ -\dfrac{\partial P}{\partial x}A\,\Delta x = \rho A\,\Delta x\,\dfrac{\partial^2 s}{\partial t^2} \]
Why. Newton's second law, \(F_{net} = ma\), with the net force just found and \(m = \rho A\,\Delta x\).
\[ -\dfrac{\partial P}{\partial x} = \rho\,\dfrac{\partial^2 s}{\partial t^2} \]
Why. \(A\,\Delta x\) appears on both sides and cancels. What is left links pressure and displacement directly.
\[ \dfrac{\partial P}{\partial x} = \dfrac{\partial (\Delta P)}{\partial x} \]
Why. \(P = P_0 + \Delta P\), and the ambient pressure \(P_0\) is the same everywhere, so its derivative is zero.
\[ \dfrac{\partial (\Delta P)}{\partial x} = \dfrac{\partial}{\partial x}\left(-B\,\dfrac{\partial s}{\partial x}\right) = -B\,\dfrac{\partial^2 s}{\partial x^2} \]
Why. put in \(\Delta P = -B\,\partial s/\partial x\). The bulk modulus is a constant of the medium, so it comes outside.
\[ -\left(-B\,\dfrac{\partial^2 s}{\partial x^2}\right) = \rho\,\dfrac{\partial^2 s}{\partial t^2} \]
Why. put this pressure gradient back into step 2.
\[ \dfrac{\partial^2 s}{\partial x^2} = \dfrac{\rho}{B}\,\dfrac{\partial^2 s}{\partial t^2} \]
Read it. the two minus signs cancel. Divide both sides by \(B\).
This is the classic one-dimensional wave equation. Compare it with the general form found in the chapter Introduction to Waves, \(\partial^2 y/\partial x^2 = (1/v^2)\,\partial^2 y/\partial t^2\). The two match term by term, so we can read off the wave speed.
\[ \dfrac{1}{v^2} = \dfrac{\rho}{B} \]
Why. the factors in front of the second time derivative must be equal in the two matching equations.
\[ v = \sqrt{\dfrac{B}{\rho}} \]
Read it. turn both sides upside down and take the positive square root, since a speed is positive.
So, by walking through these physical steps, we find what the speed of sound in a fluid depends on. It depends only on how strongly the medium resists compression, its bulk modulus \(B\). It also depends on how much inertia the medium has, its density \(\rho\). The celebrated formula is rooted in Newton's laws, the physics of pressure and the elasticity of the medium.
In the derivation, the factor \(A\,\Delta x\) cancels from both sides. Why does it appear on both sides?
Speed in liquids and solids
For bulk materials, the formula \(v = \sqrt{B/\rho}\) works well.
- In liquids, the elastic property is the bulk modulus \(B\).
- In solids, sound can travel both as longitudinal (compression) waves and as transverse (shear) waves. For longitudinal waves in a thin rod, the elastic property is Young's modulus \(Y\). The speed is then \(v_{\text{solid}} = \sqrt{Y/\rho}\).
Water has a bulk modulus of \(2.2 \times 10^9\,\text{Pa}\) and a density of \(1000\,\text{kg/m}^3\). Find the speed of sound in water, in m/s.
A thin steel rail has a Young's modulus of \(2.0 \times 10^{11}\,\text{Pa}\) and a density of \(7800\,\text{kg/m}^3\). Find the speed of a longitudinal wave along it, in m/s.
Children in a village near a railway line know an old trick. With an ear pressed to the rail, they hear a train long before its sound arrives through the air. Sound runs along steel at about \(5060\,\text{m/s}\), and through air at about \(343\,\text{m/s}\). Over \(1\,\text{km}\) of track, the steel brings the sound in about \(0.2\,\text{s}\), while the air takes about \(2.9\,\text{s}\). Never try this on a real line: the train arrives soon after its sound.
- Thinking that a harder strike or a louder source makes sound travel faster.The speed is \(\sqrt{B/\rho}\), a property of the medium. The source changes the amplitude, not the speed.
- Using the bulk modulus for a longitudinal wave in a thin rod.In a thin rod the rod bulges sideways as it is squeezed, so its stiffness is Young's modulus: \(v = \sqrt{Y/\rho}\).
- Losing the minus sign of the net force.The slice is pushed from high pressure towards low pressure, so \(F_{net} = -(\partial P/\partial x)A\,\Delta x\).