Sound Notes
Level 3: Characteristics of Sound
from Knapp Notes
Sound is a longitudinal mechanical wave. For most of the time, what we will be talking about is a wave that travels
through air. Sound can travel through other mediums - water, other liquids, solids, and gases. We can hear sounds that
travel through other mediums than air – put your ear to the wall and hear the sounds on the other side. You hear
sounds when you are under water - although not as well as in air. This is because our ears are set up for listening to
sounds that move through the atmosphere.
Rarefaction
Compression
v
Wavelength
The disturbance which travels through air is the compression of air molecules – they are squeezed together and
pulled apart. Sound is a series of traveling high pressure and low pressure fronts.
Sound waves are frequently graphed with pressure on the y axis and time on the x axis. This makes the wave look like a
transverse wave - a sine wave shape on the graph. But in this depiction, changes in pressure are being plotted Vs time
and it is not a depiction of the disturbance itself, which is longitudinal. So please, the Physics Kahuna begs of you, don't
be confused about the thing.
Pressure
Time
Here we see a graph of pressure Vs time. The compressions are regions where the air pressure is greater than the
ambient pressure of the air. The rarefactions are areas of lower pressure. These high and low pressure ridges travel
outward in an expanding sphere from the sound source.
The sound source is simply something that vibrates. It can be the clangor on an alarm clock, a window shade
flapping in the wind, or your vocal cords vibrating because air is passing through them. The vibrating sound
source collides with air molecules, causing them to scrunch together and pull apart. These scrunches travel
through the air. But the air molecules do not physically travel across the room. They are excited by the sound
source and gain kinetic energy. They move outward and have elastic collisions with other air molecules, which
then gain energy, and so on.
Sound waves
propagate as
expanding sphere
Sound source
Air molecule gets excited, gains energy,
collides with another molecule and transfers
energy to it and so on.
The damping of a sound wave (decrease in amplitude) as it travels is called
attenuation. Attenuation depends on the medium and the frequency of
the sound. Low frequency sounds are attenuated less than high frequency
sounds. Whales make very low frequency sounds (in the neighborhood of
1 - 10 Hz) which can travel hundreds of miles through the ocean. It has
recently been found that elephants also employ similar low frequency
sounds to communicate. In air, these sounds can travel many miles.
Practice Problem
A vibrating tuning fork sends sound waves into the air surrounding it.
During the time in which the tuning fork makes one complete vibration,
the emitted wave travels
(A) one wavelength
(B) about 340 meters
(C) a distance directly proportional to the frequency of the vibration
(D) a distance directly proportional to the square root of the air density
(E) a distance inversely proportional to the square root of the pressure
Solution
The time to make 1 cycle, is also the time it takes the wave to travel 1 λ.
Audible Spectrum:
The human ear is not the world's best sound receptor, although it does all right (if you take
care of them). A typical person can hear sounds whose frequency ranges from 20 Hz to about 20 000 Hz. This is known
as the audio spectrum. Sounds with a higher frequency are called ultrasonic sounds and sounds of a lower frequency
are called infrasonic sounds.
Other animals have different hearing spectrums. Dogs can easily hear sounds up to 45 000 Hz. Whales and elephants
hear very low frequency sounds (below 10 Hz).
Supplemental: Infrasound Check out this Wikipedia article on infrasound. I’ve checked most of this out
myself and it is spot on. Ghosts, vibrating, eyeballs… it is worth a look if you have the time.
Sound Characteristics
There is an overlap in the type of words we use to describe
sounds and physics terms. Let’s try to translate these.
frequency  pitch You probably could have guessed this
one without me. A high pitched voice has a high frequency.
loudness  intensity If you could visualize sound as a
transverse wave (which it is not) the “loudness” would be the amplitude. This would also depend on where it was in our
hearing range. We measure “loudness” in decibels (see the supplemental below if you are interested.
octave  You hear this term used a lot in music. If you tell someone to sing an octave higher, you are telling them to
double their frequency. So if someone was singing at 256 Hz and they increase the frequency to 512 Hz, they have gone
up an octave.
Forced Vibrations
When the big piece of cardboard was brought out and the speaker was pressed onto it, the speaker forced the
cardboard to vibrate. Since it was mechanically connected to the cardboard, the vibrations were easily transferred. The
cardboard had a very large area in contact with the air, so it was more efficient at sending the sound into the air. This is
why the sound suddenly sounded so much better when the cardboard made its appearance.
We call this phenomenon forced vibration.
Forced vibration  The vibration of an object that is made to vibrate by another
vibrating object in contact.
A tuning fork makes a very weak sound - you can barely hear the thing. Place a vibrating tuning fork against a
window or desktop, however, and the sound will become much louder.
This is actually how a music box work. The inner mechanism of a music box is actually quite quiet. But when
you give it something large to force a vibration onto (such as a piece of wood) it gets much louder.
Supplemental: Easy Ipod Speaker Think this is interesting? This guy shows you how to use forced vibrations
to create ipod speakers using a toilet paper roll.
The Speed of Sound
from Princeton Review
You do not need to have a deep, mathematical understanding of it. Just read it to get the general idea. You don’t need
to actually calculate the speed of sound on the AP test, but you do need to have a qualitative understanding of how it is
calculated.
What isn’t covered in these notes is the fact that when it comes to air, the less dense the air is, the slower the sound will
travel. This is because the air molecules are farther apart and don’t interact as easily. Because of this, when the weather
is warmer, sound will travel a little slower. If you’ve ever been outside on a quiet winter morning and realized the
sounds were clear and crisp, you already know this.
You need to memorize the speed of sound in air at room temperature as 343 m/s. Not memorizing this will make every
problem a lot harder.
Practice Problem
As sound travels from steel into air, both its speed and its:
A) wavelength increase B) wavelength decrease C) frequency increase
D) frequency decrease E) frequency remain unchanged
Solution
When sound travels into less dense medium, its speed decreases (unlike light) … however, like
all waves when traveling between two mediums, the frequency remains constant. Based on
v = f λ, if the speed decreases and the frequency is constant then the λ must decrease also
Beats
from Princeton Review
How beats are made… notice the difference in the waves (50-40=10 Hz).
Practice Problem
Solution
Doppler Effect
The Doppler Effect:
Imagine a water bug floating motionless on the surface of a calm pond on a lovely summer
day. The bug, bored out of its little bug brain, is tapping the water with a pair of its little segmented legs, making a
series of waves that radiate outward on the surface
Wave Pattern Created By A Stationary
Bug wiggling its Legs in the Water
- like the ripples on a pond (actually, they are ripples on a pond, ain’t they? Curious, what?). The bug is unwittingly
producing a traveling wave. It would look like this:
The waves spread out in all directions. The distance between the wave crests is the wavelength, . This wavelength is
the same in all directions. Now, imagine that the bug starts swimming in one direction, but it still makes its little
periodic vibrations with its legs at a constant frequency. We would see a different wave pattern.
Notice that the waves in front of the bug are pushed closer together. Behind the bug, the waves are stretched further
apart. The waves in front have a shorter wavelength, the waves to the rear have a longer wavelength. Since the speed
of the wave is a constant and equal to the wavelength multiplied by the frequency, this means that the frequency of the
waves traveling in front of the bug is higher. The waves behind the bug are lower in frequency. We call this frequency
change the Doppler shift or the Doppler effect.
This happens because the bug makes a wave and then swims after it. So that, when he makes the next wave, it will start
out closer to the first wave and so on. As the wave travels to the rear, it is already further away from the first wave, so
the wavelength is longer and the frequency shorter. All the waves travel at the same speed so they can't make up the
difference.
Sound, like all waves, undergoes this Doppler shift. A car moving towards you
pushes its sound waves closer together in front so the sound you actually hear
has a higher frequency. When it moves away from you its frequency is lower.
You can hear this change when you are near traffic. You can listen to a car
and tell from the change in pitch when it stops coming toward you and starts
moving away.
In order to experience the Doppler shift, there must be relative motion
between the sound source and the listener.
Doppler Equation- Memorize this
You are going to see that people write this equation lots of ways. The do lots of algebraic gymnastics writing this in
different formats. Just be aware of that when you are checking the solutions below.
Practice Problem
A train is traveling at 125 km/h. It has a 550.0 Hz train whistle. What is frequency heard by a stationary listener in front
of train?
Solution
First, convert the train’s speed to meters per second:
125
km  1 h  1 000 m 
m


  34.72
h  3 600 s  1 km 
s
Then plug the data into the equation which you will have derived (as above). Make sure to use the proper sign. In this
case the train is closing on the listener, so the negative sign is selected.
m


345


 v 
s
f ' f 

550.0
Hz

 

m
m
v

v
s 

 345  34.72 
s
s 

Practice Problem
Practice Problem
612 Hz
Supplemental: Big Bang Theory Now you can fully appreciate these jokes.
Level 4: Closed and Open Pipes
Resonance
This is one of the cooler parts of physics. Objects have something called a natural frequency. We won’t get too into the
crazy chemistry involved in this, but basically, when knocked or hit, objects tend to vibrate at a set frequency. This exists
all around you. Tap a piece of metal or the side of a water bottle. If you happen to have a wine glass in your bag, take it
out. Then put it away because that is probably not school appropriate. When you get home, take it out and tap it again,
you will hear a nice, clear, high pitched sound. That is the glass’s natural frequency. When we tap objects, they sing at
their natural frequency. This is also why you can get a sound out of a glass when you rub your finger along the top. What
determines the object’s natural frequency? The shape and makeup of the object. Do you have a natural frequency? Yes,
you probably do. And if you are lucky I will never use it against you. Mwahahaha. Wow, it is getting late… are these
notes done yet? Wow, okay onward…
Imagine you had two glasses with the same natural frequency. You tapped one- what would happen to the other? You
guessed it- it would start to vibrate as well.
Now, let’s get dark and creepy. What happens if we play an object its own natural frequency? We call this resonance.
This is where you are putting energy into an object at its natural frequency, it will begin to vibrate. If you keep putting
more an more energy in, the object will eventually shatter. Check out the example below.
Supplemental: Resonance in action Check out this video. Start it at 5:24.
Resonance in Pipes
Probably one of the most useful and beautiful things we can do with resonance is create music. Music is created with we
use the natural frequencies (sometimes called the resonant frequencies) of
objects to make beautiful clear sounds. The easiest way to visualize this is to think
about guitar strings. When you pluck a guitar string, it will start to sing at its
resonant frequency. If you want to change the sound it makes you will need to
change its natural frequency, so you will need to alter the string physically. You
can do this by changing its length (pushing your finger down on the string,
essentially shortening it) or by changing its tension (tightening or tuning a string).
Notice, that for a string, first harmonic tells us the number of antinodes a wave has. So,
for a guitar string, the first harmonic would have half a wavelength vibrating along the
length of the string. Each harmonic tells us how many ½ wavelengths exist in the string.
Notice that for a sr
That takes care of a lot of instruments- harps, guitars, banjos, even timpanis work this
way. What about the rest of the instruments- the tubas, clarinets, flutes? How do they
work?
Well, columns of air also have their own resonant frequencies, and if you happen to put
energy into them (maybe by blowing on them with the right speed) you can cause them
to vibrate and make sound. Think about blowing into the top of a bottle. You formed a
standing wave in the bottle, causing the bottle to vibrate. We are going to explore this in a lot more detail below.
Close Pipes
Let’s start with pipes that are closed at one end. When you blow into them, you create a standing
longitudinal wave in the tube. At the closed end, the air becomes trapped, builds up pressure, and
comes to a relative stop. It is fairly still- creating a node. The other end is capable of vibrating back
and forth, so you have an antinode here. This is true for all closed pipes. A node at the closed endand antinode at the other end.
Even though this all has to do with longitudinal wave, this can be difficult to picture longitudinally,
so we often show this using a transverse wave. There is an example of this on the right. Notice that
a quarter of a wavelength currently exists in the closed pipe. So at the first harmonic, we have
standing wave with ¼ a wavelength. When we discussed strings, we could figure out the harmonics
by counting the antinodes (1st harmonic- 1 antinode, 2nd harmonic- 2 antinodes…). This gets a little
more complex with pipes. Let me walk you through it.
from Classroom Physics
Don’t worry about it cutting off there… I had to use this because that had the absolute best description of this, but I
don’t want to go into the next section.
Notice that there is a pattern. No matter what, we can find the wavelength in the tube as long as we keep it a multiple of
¼ of a wavelength.
𝜆𝑛 =
4𝐿
𝑓𝑜𝑟
𝑛
𝑎𝑛𝑦 𝑜𝑑𝑑 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 Memorize this
Where L is the length of the pipe and n is the number of the harmonic.
Using 𝑣 = 𝑓𝜆 we can also say
𝑓𝑛 =
𝑛𝑣
4𝐿
𝑓𝑜𝑟 𝑎𝑛𝑦 𝑜𝑑𝑑 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 Memorize this
where v is the speed of sound in the medium (often 343 m/s) and f is the frequency and L is the length
You can start to see how we create music. If we can control the length of the pipe of the velocity in the medium, we can
control the frequency. Ever see anyone play a song by blowing into bottles full of different amounts of water? They were
using the water to change the height of the column of air in the tube.
Supplemental: Watch this kid play closed PVC pipes like a champ. Great use of closed pipes.
Practice Problem
Determine the length of an closed-end air column that produces a fundamental frequency (1st harmonic) of
480 Hz. The speed of waves in air is known to be 340 m/s.
Solution
The solution to the problem begins by first identifying known information, listing the desired quantity,
and constructing a diagram of the situation.
Given:
v = 340 m/s
Find:
L = ??
Diagram:
f1 = 480 Hz
The problem statement asks us to determine the length of the air column. From the graphic above,
the only means of finding the length of the air column is from knowledge of the wavelength. But the
wavelength is not known. However, the frequency and speed are given, so one can use the wave
equation (speed = frequency • wavelength) and knowledge of the speed and frequency to determine
the wavelength. This calculation is shown below.
speed = frequency • wavelength
wavelength = speed / frequency
wavelength = (340 m/s) / (480 Hz)
wavelength = 0.708 m
Now that the wavelength is found, the length of the air column can be calculated. For the first
harmonic, the length is one-fourth the wavelength. This relationship is derived from the diagram of
the standing wave pattern (see table above); it may also be evident to you by looking at the standing
wave diagram drawn above. This relationship between wavelength and length, which works only for
the first harmonic of a closed-end air column, is used to calculate the wavelength for this standing
wave.
Length = (1/4) • Wavelength
Length = (1/4) • Wavelength
Length = 0.177 m
Practice Problem
A pipe is closed at one end and is 1.50 m in length. If the sound speed is 345 m/s, what are the frequencies of the first
three harmonics that would be produced?
Use the close ended pipe formula to find the first harmonic (the fundamental frequency):
fn  n
v
4L
f1 
v
4L
f1  345

m
1

 
s  41.50 m  
57.5 Hz
Recall that close ended pipes only have the odd harmonics, so the next two would be the third and fifth harmonics:
f3  n  f1   3  57.5 Hz  
172 Hz
f5  n  f1   5  57.5 Hz  
288 Hz
Open Pipes
Open pipes use the same idea, but are kind of the opposite of closed pipes. In open pipes, you create standing waves as
well, but here, both ends are open and able to move back and forth. This means that for an open pipe, you get
antinodes the open ends.
from the Physics Classroom
We can use an equation to keep track of these. Notice that there is a pattern. No matter what, we can find the
wavelength in the tube as long as we keep it a multiple of 1/2 of a wavelength.
𝜆𝑛 =
2𝐿
𝑓𝑜𝑟
𝑛
𝑎𝑛𝑦 𝑖𝑛𝑡𝑒𝑔𝑒𝑟
Memorize this
Where L is the length of the pipe and n is the number of the harmonic.
Using 𝑣 = 𝑓𝜆 we can also say
𝑓𝑛 =
𝑛𝑣
2𝐿
𝑓𝑜𝑟 𝑎𝑛𝑦 𝑖𝑛𝑡𝑒𝑔𝑒𝑟
Memorize this
where v is the speed of sound in the medium (often 343 m/s) and f is the frequency and L is the length
Practice Problem
The speed of sound waves in air is found to be 340 m/s. Determine the fundamental frequency (1st harmonic)
of an open-end air column that has a length of 67.5 cm.
Solution
The solution to the problem begins by first identifying known information, listing the desired quantity,
and constructing a diagram of the situation.
Given:
v = 340 m/s
Find:
f1 = ??
Diagram:
L = 67.5 cm = 0.675 m
The problem statement asks us to determine the frequency (f) value. From the graphic above, the
only means of finding the frequency is to use the wave equation (speed = frequency • wavelength)
and knowledge of the speed and wavelength. The speed is given, but wavelength is not known. If
the wavelength could be found then the frequency could be easily calculated. In this problem (and
any problem), knowledge of the length and the harmonic number allows one to determine the
wavelength of the wave. For the first harmonic, the wavelength is twice the length. This relationship
is derived from the diagram of the standing wave pattern (see table above). The relationship, which
works only for the first harmonic of an open-end air column, is used to calculate the wavelength for
this standing wave.
Wavelength = 2 • Length
Wavelength = 2 • 0.675 m
Wavelength = 1.35 m
Now that wavelength is known, it can be combined with the given value of the speed to calculate the
frequency of the first harmonic for this open-end air column. This calculation is shown below.
speed = frequency • wavelength
frequency = speed / wavelength
frequency = (340 m/s) / (1.35 m)
frequency = 252 Hz
The end, where
Supplemental: Some of the most amazing uses of sound out there. Fast cool, very visual. Worth your time.
Supplemental: Not going to lie. This video on how microphones work is pretty swag. You should be able to
following all of it using stuff you learned in AP.