The Mathematics of Sound Waves

McEwen, © 2025 | All Rights Reserved | Distribution Prohibited

version 1.0, 2025

This is a first version. It is a work in progress. Please report any errors or typos.

Many of the animations in this notebook are inspired by those found at Dan Russell's Acoustics and Vibration Animations website at Penn State University.

import ipywidgets as widgets
from ipywidgets import interact, IntSlider, FloatSlider
from IPython.display import HTML

import numpy as np
%matplotlib inline 
import matplotlib.pyplot as plt
from bokeh.plotting import figure,show
from bokeh.io import output_notebook
output_notebook()

from numpy import pi, cos, sin, exp

#%load_ext autoreload

#%autoreload 2

from soundwave_plots import traveling_soundwave_animation, sound_reflection_animation, sound_standingwave_animation
from soundwave_Fourier import soundwave_interactive, pressure_plot

Loading BokehJS ...

anim=traveling_soundwave_animation()
HTML(anim.to_jshtml())

Once Loop Reflect

Sound waves are an example of longitudinal waves. A longitudinal wave is a wave whose displacements are parallel to the direction of motion. This is in contrast to transverse waves whose displacements are perpendicular to the direction of motion, for example a traveling wave on a string. A slinky toy provides a good demo of longitudinal waves, see here. In the video do notice that although the wave pulse moves to the left, the slinky as a whole remains stationary. Sound waves are similar; they are the consequence of vibrations of the medium parallel to the direction of motion. Think of air molecules moving back and forth. As adjacent air molecules move back and forth a sound wave can travel over a distance. For example, a vibrating drum head will be heard a distance away because your ear is sensitive to the vibrations of the air. Sound can travel through any compressible medium (including solids). Sound cannot travel without a medium.

The animation above is a demo of a traveling sound wave. The top panel illustrates the vibrations of particles (black dots). On the left is a moving grey wall (perhaps a piston or diaphragm). The particles move back and forth such that their displacement from their equilibrium positions is given by $s(x,t)=A \sin(kx - \omega t + \phi_0)$, where $A$ is an amplitude, $k$ a wave number, $\omega$ a frequency, and $\phi_0$ a phase constant. As you watch the animation you will see a wave move from left to right. But, you should note that none of the particles actually move all the way to the right. Each particle oscillates about its equilibrium position. To help you see this I have randomly selected a few particles and colored them in red (and one in blue). You can see that these particles oscillate back and forth (they have an average displacement of zero, just like all the other particles do).

Problem 1: If each particle has mass $m$ and the displacement of the particle is $A \sin(kx - \omega t+ \phi_0)$ then the position of the particle is $x(t)=x_0 + A \sin(kx_0 - \omega t+ \phi_0)$. Does this make sense to you? What is the expression for the force on the particle? What is the expression for the velocity of the particle?

Solution

Does the position expression make sense?

Yes. The particle oscillates about its equilibrium position $x_0$. The displacement from equilibrium is $s = A \sin(kx_0 - \omega t+ \phi_0)$, so the actual position is the equilibrium position plus the displacement: $x(t) = x_0 + s(t) = x_0 + A \sin(kx_0 - \omega t+ \phi_0)$.

Expression for velocity:

The velocity is the time derivative of position:

$$v(t) = \frac{dx}{dt} = \frac{d}{dt}\left[x_0 + A \sin(kx_0 - \omega t+ \phi_0)\right] = -A\omega \cos(kx_0 - \omega t+ \phi_0)$$

Note that $x_0$, $k$, and $\phi_0$ are constants, so only the $\omega t$ term contributes to the time derivative.

Expression for force:

Using Newton's second law, the force is:

$$F(t) = m\frac{dv}{dt} = m\frac{d}{dt}\left[-A\omega \cos(kx_0 - \omega t+ \phi_0)\right] = -mA\omega^2 \sin(kx_0 - \omega t+ \phi_0)$$

Alternatively, we can write this in terms of position:

$$F(t) = -m\omega^2[x(t) - x_0]$$

This is the form of simple harmonic motion, where the force is proportional to the displacement from equilibrium.

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When you look at the animation of the particles (top panel), you can see two distinct regions: a region of high particle density and a region of low particle density. The region of high density is called compression. The region of low density is called rarefaction.

The middle panel animates the displacement, $s(x,t)=A \sin(kx - \omega t + \phi_0)$. Although the middle panel shows displacement in the vertical, we should be clear that the displacement is that of the particles and is in the horizontal direction. Note that the displacement, $s(x,t)=A \sin(kx - \omega t + \phi_0)$, is both a function of location and time.

The bottom panel animates the pressure. Pressure is defined as the force per area, $P=F/A$. A typical unit of pressure is the Pascal. One Pascal is equal to $9.869 \times 10^{-6}$ atmospheric pressure. You should observe that high pressure regions correspond to compression regions and low pressure regions correspond to rarefaction (be sure you observe this). The pressure oscillates as well and is described by an expression $P\propto \cos(k x - \omega t + \phi_0)$.

Problem 2: Given the expression for the pressure and the displacement what is the phase difference between the two? Is your answer congruent with the animation?

Solution

The displacement is given by: $$s(x,t) = A \sin(kx - \omega t + \phi_0)$$

The pressure is proportional to: $$P(x,t) \propto \cos(kx - \omega t + \phi_0)$$

Using the trigonometric identity $\cos(\theta) = \sin(\theta + \pi/2)$, we can rewrite the pressure as: $$P(x,t) \propto \sin(kx - \omega t + \phi_0 + \pi/2)$$

Comparing this to the displacement expression, we see that the pressure leads the displacement by a phase of $\pi/2$ radians (or 90°).

Is this congruent with the animation?

Yes! In the animation, when the displacement is zero and increasing (particles moving right), the pressure is at a maximum (compression). When the displacement reaches its maximum (particles furthest right), the pressure is zero and decreasing. This confirms a 90° phase difference, with pressure leading displacement by a quarter cycle.

Physically, this makes sense: maximum compression (pressure) occurs when particles are converging, which happens when displacement is zero but velocity is maximum. Maximum displacement occurs when particles momentarily stop, which is when pressure returns to equilibrium.

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If the pressure is high, the region is a compression region and has higher density. This makes sense; after all if I squeezed an object (apply pressure) I would increase the density. If pressure is low, I would expect an expansion region (rarefaction).

You will notice that I plot a blue dot at $x=4$ in the displacement animation (middle panel). This is the same blue dot in the top panel. You will also notice that I plotted two red dots at $x=3.8$ and $x=4.2$ in the pressure panel. The two red dots are just to the left and right of the blue dot in the displacement panel. I did this so we can make a force analysis (as a problem below).

Pressure gradients source changes in motion. If the pressure on the left side of me is greater than the right side, then I would feel a force to the right. It is a good thing that the pressure in my room is uniform, otherwise I would be tossed around the room; pressure differences can be mighty dangerous. We could write the force on a particle as the difference between the right and left pressure forces $F_{net}=F_{left} - F_{right} = AP_{left} - AP_{right} = A (P_{left} - P_{right})$, where $A$ is the cross sectional area of the object of interest.

Discussion problem 1: We are going to analyze the relationship between the displacement, pressure, velocity, and force. You can move the animation between frames by adjusting the controls below the figure. Let us start at the first frame, the control all the way to the left. Observe the blue dot on the displacement curve. This should be at displacement 0 (vertical axis) and position $x=4$ (horizontal axis). Remember this is the same blue dot as in the top panel. Observe the two red dots on the pressure curve. The two red dots should be at the same height. We will call this snapshot 1. Draw a small box that takes a snapshot of the blue dot and another that takes the snapshot of the two red dots. You don't have to draw the entire displacement or pressure curves, just take a window around the location of interest. Now write down descriptive values of the displacement, pressure difference, velocity, and force. To start you off, here is what I have for those values: displacement is zero (but particle is moving right), there is zero pressure difference, there is no net force, the velocity is positive. Now advance the animation forward about an eighth of a period. This is snapshot 2. Do all the drawings we did in snapshot 1 and record the same values as in snapshot 1. Repeat this advancing the animation about an eighth of a period until you come back to the starting position. Note that advancements of a quarter period are easy; they correspond to the equilibrium positions, the crest and trough. That should help you estimate advancements of an eighth of a period as half way between those of a quarter of a period. When you write down your values for displacement, velocity, pressure difference, force, use terms like zero, positive/negative, left/right, up/down.

Wave equation for the displacement

We will derive the equations describing a sound wave here. The math is advanced, but not any more difficult than what is in your book. The method presented here does not use the bulk modulus as that is something we do not cover in our physics sequence (your book does use the bulk modulus). You should follow along with a sheet of paper. When you get to a derivation you don't understand, write that down. Then ask the discussion board for clarification. Remember that derivations are important; they are generally the place where the physics is.

Pressure differences provide the force that vibrates the medium. These pressure differences are small and we can write the pressure (a function of location and time) as \begin{equation} P(x,t)=P_0 + \delta P (x,t), \end{equation} where $P_0$ is the equilibrium pressure and $\delta P$ is a small variation of the pressure from equilibrium. We should read this expression as the pressure at a particular location $x$ and particular time $t$ is the equilibrium pressure $P_0$ plus a small change $\delta P$. We note that the equilibrium pressure is uniform in time and space.

As the medium vibrates the density changes. In the animation above we have already seen this phenomenon; compression is a region of higher density, while rarefaction is a region of lower density. Again, the changes in density are small and we can write the density as \begin{equation} \rho(x,t) = \rho_0 + \delta \rho(x,t), \end{equation} where $\rho_0$ is the density of the material without vibrations and $\delta \rho$ is the small change in density from $\rho_0$.

The pressure and density are related (generally called an equation of state). If I squeeze the material (by applying a large pressure), I generally increase the density. Here are two examples of how pressure and density are related. For an ideal gas, the pressure and density are related by $P=\rho~ k_B/m T$, where $T$ is the temperature measured in Kelvin ($K$), $k_B=1.38 \times 10^{-23}~J/K$ is the Boltzmann constant and $m$ is the mass of the gas particle. You may recall this formula from your chemistry class. This also relates temperature and pressure; higher temperature leads to higher pressure. Another is to consider the pressure in a fluid at particular depth $h$, for example what is the pressure at the bottom of a swimming pool. In this case the pressure is $P=P_0 + \rho g h$, where $P_0$ is atmospheric pressure. The deeper the swimming pool the larger the pressure, which makes sense.

For our purposes we simply state the relationship between pressure and density as $P=P(\rho)$. We then have \begin{align} \begin{split} P_0 + \delta P & = P(\rho_0 + \delta \rho) \\ & \approx P(\rho_0) + \frac{dP}{d\rho}\Big|_{\rho=\rho_0} \delta \rho \\ & = P_0 + \kappa \delta \rho, \end{split} \end{align} where in the second line we made a first order Taylor expansion (dropped all terms after the first term) and in the third line we identified $P_0 = P(\rho_0)$, meaning the pressure at the unperturbed density is the equilibrium pressure, and defined $\kappa= dP/d\rho$ at $\rho=\rho_0$.

From the above derivation we can write down a relationship between changes in pressure and changes in density as \begin{equation} \delta P = \kappa \delta \rho. \end{equation} We will call the above relationship expression I. Expression I states that small changes in pressure relate to small changes in density by an as yet unknown factor $\kappa$. $\kappa$ is a Greek letter pronounced kappa.

Problem 3: What are the units on $\kappa$? Based on the units, can you take a guess at what $\kappa$ might be?

Solution

Units of κ:

From the equation $\delta P = \kappa \delta \rho$, we can solve for $\kappa$:

$$\kappa = \frac{\delta P}{\delta \rho}$$

The units of pressure are Pa (Pascals) = N/m² = kg/(m·s²)

The units of density are kg/m³

Therefore: $$[\kappa] = \frac{\text{kg/(m·s}^2\text{)}}{\text{kg/m}^3} = \frac{\text{kg}}{\text{m·s}^2} \cdot \frac{\text{m}^3}{\text{kg}} = \frac{\text{m}^2}{\text{s}^2}$$

What might κ be?

The units of $\kappa$ are m²/s², which are the units of velocity squared. This suggests that $\kappa$ is related to the square of the wave speed in the medium.

Indeed, $\kappa$ is related to the bulk modulus $B$ of the material (which measures how resistant the material is to compression), and the speed of sound in the medium is given by $v = \sqrt{B/\rho_0} = \sqrt{\kappa}$.

For sound in air at room temperature, $v \approx 343$ m/s, so $\kappa \approx v^2 \approx 1.2 \times 10^5$ m²/s².

We have made some progress by just asserting that density and pressure are related, a reasonable assertion. Now we consider what happens to a small portion of the medium as the pressure changes. The figure below illustrates the situation. We take a small cylindrical portion of the medium (a parcel) when the pressure is at equilibrium ($P_0$) and the density is at $\rho_0$. The left side of the cylinder is at location $x_L=x$ and the right side is at location $x_R=x + \Delta x$. The cylinder has cross sectional area $A$ and therefore the volume is $V=A\Delta x$. The mass of the parcel is $m=\rho_0 V$. At a later time, the pressure and the density change, and the parcel is displaced to the right. The left end of the cylinder is displaced by $s(x,t)$ so that the position is denoted as $x'_L= x + s(x,t)$, where we are using primes to indicate quantities at the later time. The right side is displaced by $s(x + \Delta x,t)$ so that the position is $x_R'=x +\Delta x + s(x + \Delta x, t)$. The mass of the parcel is given by $m=\rho V' = \rho A (\Delta x + \Delta s)$, where $\Delta s = s(x + \Delta x, t) - s(x,t)$. Be sure you see that the mass calculated in the second scenario uses the density $\rho$ not the density $\rho_0$. This makes sense, as the second scenario is a result of pressure changes, which results in density changes.

Now here is an important observation. The mass of the parcel does not change. How could it? Mass is not created or destroyed. The parcel does change shape, but the mass of the parcel is always the same. Therefore we have \begin{align} \begin{split} & m = m' \\ & \rho_0 V = \rho V' \\ & \rho_0 A \Delta x= (\rho_0 + \delta \rho) A \left( \Delta x + \Delta s \right) \\ & \rho_0 \Delta x = \rho_0 \Delta x + \rho_0 \Delta s + \delta \rho \Delta x + \delta \rho \Delta s. \end{split} \end{align} In the above we can drop the last term $\delta \rho \Delta s$ since this is the multiplication of two small terms, which is just something even smaller. We then have \begin{align} \begin{split} \delta \rho & = - \rho_0 \frac{ \Delta s}{\Delta x} \\ & = - \rho_0 \frac{ \partial s}{\partial x} \quad \text{in the limit as $\Delta x\to 0$}. \end{split} \end{align} We will call the above relationship expression II.

Discussion problem 1: Expression II makes sense. One, it shows that density variations are present when the displacement changes. Two, the sign is correct. If the change in displacement is positive the medium stretched and the density goes down. If the change in displacement is negative, the medium compressed and the density goes up. (Remember that $\rho= \rho_0 +\delta \rho$). Use expression II to write $\rho= \rho_0(1 - \partial s/\partial x)$. Draw a cartoon (your own idea) depicting positive/negative changes in $s$ and the corresponding density changes. Imagine yourself as the teacher; what would you draw in your class?

We now turn to an analysis based on Newton's second law $\vec{F}= m\vec{a}$. The figure above shows our cylindrical parcel with the pressure difference labeled. On the left we have pressure $P(x,t)$ while on the right we have pressure $P(x + \Delta x, t)$. The parcel feels a nonzero force when the two pressures are not the same (a pressure gradient). Noting that pressure is force per area $F/A$, we can write out Newton's second law for the parcel as \begin{align} \begin{split} & m a = \left(F_{\text{left}} - F_{\text{right}}\right) \\ & m a = A\left(P(x,t) - P(x + \Delta x,t)\right) \\ & a =- \frac{ A \Delta x}{m} \frac{P(x + \Delta x,t) - P(x,t)}{\Delta x} \\ & a =- \frac{1}{\rho_0} \frac{\partial P}{\partial x} \quad \text{in the limit as $\Delta x \to 0$.} \end{split} \end{align} Now we note that the acceleration is the second time derivative of the displacement and we have \begin{equation} \frac{\partial^2 s}{\partial t^2} = - \frac{1}{\rho_0} \frac{\partial P}{\partial x}. \end{equation} We will call the above relationship expression III.

Let's recap our methodology.

expression	eqn.	how
I	$$\delta P=\kappa \delta \rho$$	changes in pressure and density are related
II	$$\delta\rho=-\rho_0 \frac{\partial s}{\partial x}$$	consequence of mass conservation
III	$$\ddot{s} = - \rho_0^{-1} \frac{\partial P}{\partial x}$$	from Newton's second law

We shall now combine all three expressions, starting from expression III as \begin{align} \begin{split} \frac{\partial^2 s}{\partial t^2}& = - \frac{1}{\rho_0} \frac{\partial P}{\partial x} \\ & = -\frac{\kappa}{\rho_0} \frac{\partial \rho}{\partial x} \quad \text{using expression I} \\ & = -\frac{\kappa}{\rho_0} (-\rho_0)\frac{\partial^2 s}{\partial x^2} \quad \text{using expression II} \\ & = \kappa \frac{\partial^2 s}{\partial x^2}. \end{split} \end{align}

We now have a wave equation for the displacement and can identify $\kappa = v^2$, where $v$ is the wave speed as \begin{equation} \frac{\partial^2 s}{\partial t^2} = v^2 \frac{\partial^2 s}{\partial x^2}. \end{equation}

A solution is $s(x,t)= A \sin(kx - \omega t + \phi_0)$, where $A$ is the amplitude of displacement, $k$ is the wave number, $\omega$ is the frequency, and $\phi_0$ is the phase constant. This solution represents a traveling wave.

The velocity is given by $u = \partial s/\partial t= -\omega A \cos(kx - \omega t + \phi_0)$.

Substituting $\ddot{s} = - \omega^2 A \sin(kx - \omega t + \phi_0)$ into expression III we have \begin{align} \begin{split} & \frac{1}{\rho_0} \frac{\partial P}{\partial x} =\omega^2 A \sin(kx - \omega t + \phi_0) \\ &\frac{\partial P}{\partial x}= \rho_0 \omega^2 A \sin(kx - \omega t + \phi_0), \end{split} \end{align} which we can solve by inspection to get \begin{align} \begin{split} P & = -\frac{\rho_0 \omega^2 A}{k} \cos(kx - \omega t + \phi_0) \\ & = - \rho_0 v \omega A \cos(kx - \omega t + \phi_0). \end{split} \end{align}

Discussion problem 2: Is the velocity $u = \partial s/\partial t= -\omega A \cos(kx - \omega t + \phi_0)$ the same as the velocity $v$ in the above equation? Explain.

In your book you will find an expression for the sound speed in terms of the pressure and bulk modulus, given by $v = \sqrt{B/\rho_0}$, where $B$ is the bulk modulus and $\rho_0$ is the density (in these notes we use $\rho_0$ for the uniform density). We can recover the book's equation for pressure using the previously noted expression and the relation $v = \omega/k$ as \begin{align} \begin{split} P & = - \rho_0 v \omega A \cos(kx - \omega t + \phi_0) \\ & = - \rho_0 v^2 k A \cos(kx - \omega t + \phi_0) \\ & = - \rho_0 \sqrt{\frac{B}{\rho_0}} k A \cos(kx - \omega t + \phi_0) \\ & = - B k A \cos(kx - \omega t + \phi_0). \end{split} \end{align}

Power and Intensity

Power is defined as the rate at which energy is transferred or the rate at which work is done. Mathematically:

$$\text{Power} = \frac{\text{Energy}}{\text{Time}} = \frac{\text{Work}}{\text{Time}}$$

Power is measured in watts (W), where 1 W = 1 J/s.

For a force acting on an object, power can also be expressed as:

$$P = \vec{F} \cdot \vec{v}$$

where $\vec{F}$ is the force and $\vec{v}$ is the velocity. This makes sense because work is force times distance ($W = F \cdot d$), so power (work per time) is force times distance per time, which is force times velocity.

Intensity is defined as the time-averaged power per unit area:

$$I = \frac{\text{Power}}{\text{Area}} = \frac{P}{A}$$

Intensity is measured in W/m². For sound waves, intensity represents how much acoustic power passes through a unit area perpendicular to the direction of wave propagation.

Don't confuse power with pressure; I used the same capital P for both, will redo in newer version.

Problem 4: Show that the product of force and velocity has the same units as power. Finish the calculation below.

Solution

Units of force times velocity:

Force has units: N = kg·m/s²

Velocity has units: m/s

Therefore: $$[F \cdot v] = \frac{\text{kg·m}}{\text{s}^2} \cdot \frac{\text{m}}{\text{s}} = \frac{\text{kg·m}^2}{\text{s}^3} = \frac{\text{J}}{\text{s}} = \text{W}$$

This is indeed the unit of power! This makes physical sense: power is the rate of doing work, and work equals force times distance. When we multiply force by velocity (distance per time), we get work per time, which is power.

Finishing the intensity calculation:

Sound intensity is (time averaged): \begin{align} \begin{split} I(x) & =\frac{1}{T} \int_0^T P(t) v(t) dt\\ & =\frac{1}{T} \int_0^T\rho_0 v \omega^2 A^2 \cos^2(kx -\omega t +\phi_0)dt \end{split} \end{align}

To evaluate this integral, we use the fact that the time average of $\cos^2(\omega t)$ over one period is $\frac{1}{2}$:

$$\frac{1}{T}\int_0^T \cos^2(kx - \omega t + \phi_0) dt = \frac{1}{2}$$

Therefore: \begin{align} I(x) = \frac{1}{2}\rho_0 v \omega^2 A^2 \end{align}

This shows that the intensity is proportional to:

• The density of the medium ($\rho_0$)
• The wave speed ($v$)
• The square of the frequency ($\omega^2$)
• The square of the amplitude ($A^2$)

Decibel Scale

The human ear can detect sound intensities over an enormous range—from about $10^{-12}$ W/m² (threshold of hearing) to about $1$ W/m² (threshold of pain). Because of this huge range, we use a logarithmic scale called the decibel scale (dB).

The sound intensity level $\beta$ in decibels is defined as: \begin{equation} \beta = 10\log_{10}\left(\frac{I}{I_0}\right) \end{equation}

where $I$ is the intensity of the sound and $I_0 = 10^{-12}$ W/m² is the reference intensity (approximately the threshold of human hearing at 1000 Hz).

Examples:

• Threshold of hearing: $I = 10^{-12}$ W/m² → $\beta = 0$ dB
• Whisper: $I \approx 10^{-10}$ W/m² → $\beta \approx 20$ dB
• Normal conversation: $I \approx 10^{-6}$ W/m² → $\beta \approx 60$ dB
• Rock concert: $I \approx 10^{-2}$ W/m² → $\beta \approx 100$ dB
• Threshold of pain: $I \approx 1$ W/m² → $\beta \approx 120$ dB

Note that an increase of 10 dB corresponds to a 10-fold increase in intensity, and an increase of 20 dB corresponds to a 100-fold increase in intensity.

Sound Reflections

The animation below shows a sound pulse traveling from left to right, encountering a hard boundary on the right, reflecting, and then traveling to the left. We have learned that when a wave reflects off a hard boundary, the wave is inverted. Sound waves make this notion intuitive. When the air particles hit the boundary at the right, they bounce back (what else would they do?). Therefore, the displacement must be inverted (be sure you understand this). The reflection of the displacement is shown in the bottom panel of the animation.

anim=sound_reflection_animation()
HTML(anim.to_jshtml())

Once Loop Reflect

Standing Waves

Just as for strings, we can excite standing sound waves. Below are several animations of standing waves in a pipe. The first is a pipe closed at both ends, showing the 2nd harmonic. You should observe the boundary conditions. At the left and right ends, the displacement is zero. This is a sensible boundary condition, as the air particles cannot penetrate the closed end of the pipe and therefore the displacement must be zero (unless you were a quantum particle—oh, where did the modern physics course go? Complain to the administration!).

The second animation is for a pipe with an open left end and a closed right end, showing the 5th harmonic. Here we see that the pressure variations are zero at the open end. This also makes sense, as the pressure should match the surrounding pressure (usually atmospheric) at the open end.

The third animation shows a pipe with both ends open, displaying the 6th harmonic.

# closed - closed
# you can change the value n here
n=2
anim=sound_standingwave_animation(n,"CC")
HTML(anim.to_jshtml())

Once Loop Reflect

# open - closed
# you can change the value n here
n=5
anim=sound_standingwave_animation(n,"OC")
HTML(anim.to_jshtml())

Once Loop Reflect

# open - open
# you can change the value n here
n=6
anim=sound_standingwave_animation(n,"OO")
HTML(anim.to_jshtml())

Once Loop Reflect

Standing Waves in Pipes

Standing waves in pipes depend on the boundary conditions at each end.

For a closed end, air cannot penetrate the pipe end. The air molecules at the very end are therefore "fixed" and have no ability to displace. Thus, a closed end of the pipe is a displacement node. In order to keep the air from displacing, the closed pipe end has to exert a force on the molecules by means of pressure. This means the closed end is a pressure antinode. Mathematically, pressure and displacement are out of phase with one another: if displacement is at a node, then pressure must be at an antinode.

For an open end, the argument is swapped. The pipe is open to the air, usually air at atmospheric pressure, so there must be a pressure node at the open end. Since pressure and displacement are out of phase, the open end is also a displacement antinode. In reality, the air pressure in the standing wave doesn't instantly equalize with the background pressure at an open end; it sort of "bulges" out of the pipe a bit. The displacement antinode is therefore just outside the pipe end, not exactly at the pipe end. But for our course, an introductory course, we will ignore this subtlety and just state the open end is a displacement antinode.

Here are the configurations for each case we will study.

Pipe Configuration	Left End	Right End
Closed-Closed	Displacement node Pressure antinode	Displacement node Pressure antinode
Open-Open	Displacement antinode Pressure node	Displacement antinode Pressure node
Open-Closed	Displacement antinode Pressure node	Displacement node Pressure antinode

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Closed-Closed and Open-Open Pipes

For a pipe of length $L$ that is either open on both ends or closed on both ends, we have the following frequency and wavelength relationships.

The allowed wavelengths are: \begin{equation} \lambda_n = \frac{2L}{n}, \quad n = 1, 2, 3, \ldots \end{equation}

The frequency for each mode is: \begin{equation} \begin{split} & f_n = \frac{nv}{2L}, \quad n = 1, 2, 3, \ldots \\ & f_n = n f_1. \end{split} \end{equation}

Just as with standing waves on a string, all harmonics are allowed: the 1st, 2nd, 3rd, and so on.

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Open-Closed Pipe

For a pipe with one end open and the other end closed, we have the following frequency and wavelength relationships.

The allowed wavelengths are: \begin{equation} \lambda_n = \frac{4L}{n}, \quad n = 1, 3, 5, \ldots \end{equation}

The frequency for each mode is \begin{equation} \begin{split} & f_n = \frac{nv}{4L}, \quad n = 1, 3, 5, \ldots \\ & f_n = n f_1. \end{split} \end{equation}

Only odd harmonics are allowed: the 1st, 3rd, 5th, and so on.

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Recap Questions

Question 1: Describe the motion of particles in a sound wave. Are the waves transverse or longitudinal?

Solution

The particles oscillate about an equilibrium point. The waves are longitudinal, meaning the wave is a disturbance along the direction of motion.

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Question 2: How are pressure and displacement oscillations related?

Solution

Pressure and displacement oscillations are out of phase with one another. When displacement is at a node, pressure is at an antinode. When displacement is at an antinode, pressure is at a node.

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Question 3: What are the displacement and pressure boundary conditions at an open end of a pipe?

Solution

At an open end of a pipe, there is a displacement antinode and a pressure node. The particles can move freely with maximum amplitude, and the pressure remains approximately constant at atmospheric pressure.

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Question 4: What are the displacement and pressure boundary conditions at a closed end of a pipe?

Solution

At a closed end of a pipe, there is a displacement node and a pressure antinode. The particles cannot move, and the pressure oscillates with maximum amplitude.

Problem

A pipe is closed at one end and open at the other end. A tuning fork of frequency 512 Hz is positioned near the open end of the pipe. When struck, the tuning fork excites a standing wave in the pipe. It is known that the standing wave is in the 3rd harmonic. What is the length of the pipe? (Use $v = 340$ m/s for the speed of sound.)

Solution

For an open-closed pipe, only odd harmonics are allowed. The frequency of the $n$th harmonic is: \begin{equation} f_n = \frac{nv}{4L}, \quad n = 1, 3, 5, \ldots \end{equation}

We can rearrange this to solve for the length: \begin{equation} L = \frac{nv}{4f_n} \end{equation}

Given:

• $f_3 = 512$ Hz
• $n = 3$ (3rd harmonic)
• $v = 340$ m/s

Substituting: \begin{equation} L = \frac{3 \times 340}{4 \times 512} = \frac{1020}{2048} = 0.498 \text{ m} \approx 50 \text{ cm} \end{equation}

The length of the pipe is approximately 50 cm (or 0.5 m).