Sound waves in a tube

We now examine sound waves in a tube, that is, in one dimension. Consider a tube filled with gas with cross sectional area $A$ , pressure $P_0$ , and density $\rho$ . We will look at one region of this tube, from $x$ to $x+a$ .

Define $\psi(x,t):=$ the displacement of air molecules. This means that the particles whose equilibrium position is $x$ , will at time $t$ be at $x+\psi(x,t)$ .

The region of air moved and because $\psi(x)$ may not equal $\psi(x+a)$ , its volume and pressure may have changed. We can write the change in volume as

\begin{align*} \Delta V &= V_\text{after} - V_\text{before} = A(a + \psi(x+a,t) - \psi(x,t)) - Aa \\ &= A(\psi(x+a,t)-\psi(x,t)) \cdot \frac{a}{a} \\ &= A \left(\frac{\psi(x+a,t) - \psi(x,t)}{a}\right) \cdot a. \end{align*}

Taking the limit as $a \to 0$

\Delta V = Aa \frac{\partial \psi}{\partial x}.

From the ideal gas law, $PV=nRT$ . We now assume that the compression in the cylinder is adiabatic, i.e., that energy transfer in the gas is slow relative to compression. This gives us $PV^\gamma =$ constant, where $\gamma$ is the adiabatic index, which depends on the molecular structure of the gas. For a diatomic gas $\gamma = \frac75$ .

We also assume small perturbations, so $\Delta V \ll V$ . Using these assumptions we can say

\begin{align*} ☺︎ &:= P_0 V_0^\gamma \\ ☺︎ &= (P_0 + \Delta P)(V_0 + \Delta V)^\gamma \\ &= (P_0 + \Delta P) V_0^\gamma \left(1 + \frac{\gamma \Delta v}{V_0}\right) \\ &= \underbrace{P_0 V_0^\gamma}_{\text{This is just ☺︎}} + P_0 \gamma \Delta V V_0^{\gamma-1} + \Delta P V_0^\gamma + \underbrace{\Delta P V_0^{\gamma-1} \gamma \Delta V}_{\text{Insignificant; has two small terms}} \\ 0 &= P_0 \gamma \Delta V V_0^{\gamma - 1} + \Delta P V_0^\gamma \\ \Delta P &= -\frac{P_0 \gamma \Delta V}{V_0}. \end{align*}

Plugging in $\Delta V$ , we get $\Delta P = -\gamma P_0 \frac{\partial \psi}{\partial x}$ .

Consider the two graphs below. The first shows $\psi(x)$ , while the second shows the corresponding $\Delta P(x)$ .

Note that the change in pressure is proportional to the negative derivative of displacement. Consider why this makes sense physically. If $\psi' < 0$ , then $\psi(x) > \psi(x+a)$ , which means that the region is compressed, and so the pressure increases.

Now let us consider the force acting on the gas in the tube. $F = P \cdot A$ , so

\begin{align*} F(x,t) &= (P_\text{left} - P_\text{right}) A \\ &= (P_0 + \Delta P(x,t))A - (P_0 + \Delta P(x+a,t))A \\ &= A(\Delta P(x,t) - \Delta P(x+a,t)) \cdot \frac{a}{a} \\ &= -Aa \frac{\Delta P(x+a,t) - \Delta P(x,t)}{a}. \end{align*}

Again we take the limt as $a \to 0$ to find $F = -A \frac{\partial \Delta p}{\partial x} a = m \ddot \psi$ . We know that the mass of the region is $\rho a A$ , and we have already found $\Delta P = - \gamma P_0 \frac{\partial \psi}{\partial x}$ . We plug these in to find

\begin{align*} \rho a A \ddot \psi &= A a \gamma P_0 \frac{\partial^2\psi}{\partial x^2} \\ \frac{\partial^2 \psi}{\partial t^2} &= \frac{\gamma P_0}{\rho} \frac{\partial^2\psi}{\partial x^2}. \tag{1} \end{align*}

This is the wave equation! We’ve already solved it, so we can just write down the solution.

\psi(x,t) = \sum_{m=1}^\infty A_m \sin(\mathcal K_m x + \alpha_m) \sin(\omega_m t + \beta_m).

Where $v_p^2 = \frac{\gamma P_0}{\rho}$ .

Boundary conditions

We incorporate our boundary conditions the same as in any wave. An open open end of the tube is equivalent to a massive string attached to a ring on a pole, and a closed end is equivalent to a fixed wall.