1. Sound and Hearing Physics 2 Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

2. Sound Waves
Sound waves are longitudinal pressure waves. Air is alternately compressed and rarefied along the direction of propagation. The mathematical description is the same as the transverse waves we saw before. However, the pressure is out-of-phase from the displacement by a quarter-cycle.
𝑦 𝑥,𝑡 =𝐴𝑐𝑜𝑠 𝑘𝑥−𝜔𝑡 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
𝑝 𝑥,𝑡 =𝐵𝑘𝐴𝑠𝑖𝑛 𝑘𝑥−𝜔𝑡 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 (𝐵=𝐵𝑢𝑙𝑘 𝑀𝑜𝑑𝑢𝑙𝑢𝑠)
We can use the pressure to determine the speed of a sound wave.
Bulk Modulus
𝑣 𝑠𝑜𝑢𝑛𝑑 = 𝐵 𝜌 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑜𝑢𝑛𝑑 𝑖𝑛 𝑎 𝑓𝑙𝑢𝑖𝑑
Young’s Modulus
𝑣 𝑠𝑜𝑢𝑛𝑑 = 𝑌 𝜌 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑜𝑢𝑛𝑑 𝑖𝑛 𝑎 𝑠𝑜𝑙𝑖𝑑

3. Sound Intensity and Decibel Level
Intensity is the power per unit area.
It can be computed from the expressions for pressure and velocity.
𝐼= 𝜌𝐵 𝜔 2 𝐴 2
The main thing to notice is that intensity increases as the square of the amplitude or frequency. So a wave with higher frequency or amplitude will be more intense (i.e. louder).
The Sound Intensity Level is defined as:
𝛽= 10𝑑𝐵 log 𝐼 𝐼 𝐼 0 = 10 −12 𝑊 𝑚 2
This is the decibel scale, which gives convenient values for typical sounds. The reference level, I0, is a sound that is just of the threshold of human hearing. This corresponds to 0 decibels.
A rule of thumb to use is that every factor of 10 increase in intensity will be an addition of 10 decibels to the sound level.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

4. Sound Intensity and Decibel Level
How to calculate decibels from intensity:
Suppose a sound has intensity I=10-3 W/m2.
To find the Intensity Level (in dB) we can use the definition:
How to calculate intensity from decibels:
Suppose a sound has a sound level of 73 dB.
To Find the intensity (in W/m2) we use the definition again:
Another rule of thumb we get from this is that a 3dB change means a factor of 2 change in intensity.
(73dB is twice the intensity of 70dB)
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

5. Example - A baby’s mouth is 30. 0 cm from her father’s ear and 3
Example - A baby’s mouth is 30.0 cm from her father’s ear and 3.0m from her mother’s ear. What is the difference of the sound intensity levels (in dB) heard by the father and mother?
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

6. Don’t forget to look at the units – 3.0m is 300cm.
Example - A baby’s mouth is 30.0 cm from her father’s ear and 3.0m from her mother’s ear. What is the difference of the sound intensity levels (in dB) heard by the father and mother?
The first step is to find the ratio of the distances. Just divide them to find that the mother is 10 times farther away.
Don’t forget to look at the units – 3.0m is 300cm.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

7. Don’t forget to look at the units – 3.0m is 300cm.
Example - A baby’s mouth is 30.0 cm from her father’s ear and 3.0m from her mother’s ear. What is the difference of the sound intensity levels (in dB) heard by the father and mother?
The first step is to find the ratio of the distances. Just divide them to find that the mother is 10 times farther away.
Don’t forget to look at the units – 3.0m is 300cm.
Now we can use this to find out what happens to the intensity.
Since sound intensity depends on 1/r2, the mother’s intensity is 1/100 times the father’s.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

8. Don’t forget to look at the units – 3.0m is 300cm.
Example - A baby’s mouth is 30.0 cm from her father’s ear and 3.0m from her mother’s ear. What is the difference of the sound intensity levels (in dB) heard by the father and mother?
The first step is to find the ratio of the distances. Just divide them to find that the mother is 10 times farther away.
Don’t forget to look at the units – 3.0m is 300cm.
Now we can use this to find out what happens to the intensity.
Since sound intensity depends on 1/r2, the mother’s intensity is 1/100 times the father’s.
This is 2 factors of 10.
For each factor of 10 change in intensity, you change the sound level by 10 decibels.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

9. Don’t forget to look at the units – 3.0m is 300cm.
Example - A baby’s mouth is 30.0 cm from her father’s ear and 3.0m from her mother’s ear. What is the difference of the sound intensity levels (in dB) heard by the father and mother?
The first step is to find the ratio of the distances. Just divide them to find that the mother is 10 times farther away.
Don’t forget to look at the units – 3.0m is 300cm.
Now we can use this to find out what happens to the intensity.
Since sound intensity depends on 1/r2, the mother’s intensity is 1/100 times the father’s.
This is 2 factors of 10.
For each factor of 10 change in intensity, you change the sound level by 10 decibels.
So the difference in sound intensity level is 20 dB.
For example, if the father hears an 80 dB sound, the mother will hear 60 dB.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

10. Don’t forget to look at the units – 3.0m is 300cm.
Example - A baby’s mouth is 30.0 cm from her father’s ear and 3.0m from her mother’s ear. What is the difference of the sound intensity levels (in dB) heard by the father and mother?
The first step is to find the ratio of the distances. Just divide them to find that the mother is 10 times farther away.
Don’t forget to look at the units – 3.0m is 300cm.
Now we can use this to find out what happens to the intensity.
Since sound intensity depends on 1/r2, the mother’s intensity is 1/100 times the father’s.
This is 2 factors of 10.
For each factor of 10 change in intensity, you change the sound level by 10 decibels.
So the difference in sound intensity level is 20 dB.
For example, if the father hears an 80 dB sound, the mother will hear 60 dB.
The more math-y version of the solution is below. It’s best if you can do it both ways.
r1=30cm=0.3m; r2=3m
Since Intensity is proportional to 1/r2, we can write:
Now we can write the equations for the decibel levels:
We need the difference, so subtract them (and simplify using log rules)
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

11. Doppler Effect
Basic Idea: Things sound different when objects are moving relative to each other.
If objects are moving toward each other, sounds seem to have higher frequencies.
If objects are moving away from each other, sounds have lower frequencies.
If the source of the sound is moving, the wavelength will change as well.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

12. Doppler Effect
Basic Idea: Things sound different when objects are moving relative to each other.
If objects are moving toward each other, sounds seem to have higher frequencies.
If objects are moving away from each other, sounds have lower frequencies.
If the source of the sound is moving, the wavelength will change as well.
Here is a quick explanation:
If you are standing on either side of a stationary car, you will hear the same sound because the sound waves hit you at the same frequency.
However, if you are near a moving car, the frequencies are different depending on where you stand.
In front of the car, the waves are bunched up, so the frequency is increased. The car is moving toward you.
motion toward = upward shift in frequency
Behind the car, the waves are spread out, so the frequency is decreased. The car is moving away from you.
motion away = downward shift in frequency
170 m/s
340 m/s
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

13. Doppler Effect
Basic Idea: Things sound different when objects are moving relative to each other.
If objects are moving toward each other, sounds seem to have higher frequencies.
If objects are moving away from each other, sounds have lower frequencies.
If the source of the sound is moving, the wavelength will change as well.
Here is a quick explanation:
If you are standing on either side of a stationary car, you will hear the same sound because the sound waves hit you at the same frequency.
However, if you are near a moving car, the frequencies are different depending on where you stand.
In front of the car, the waves are bunched up, so the frequency is increased. The car is moving toward you.
motion toward = upward shift in frequency
Behind the car, the waves are spread out, so the frequency is decreased. The car is moving away from you.
motion away = downward shift in frequency
170 m/s
340 m/s
Note that the bottom car is moving at the speed of sound (are we sure it’s not an airplane??), so the waves in front are so bunched up that they hit you all at once when the car drives by – that’s a sonic boom!
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

14. Doppler Effect Formula for Doppler Shift 𝑓 𝐿 = 𝑓 𝑠 𝑣± 𝑣 𝐿 𝑣± 𝑣 𝑠
𝑓 𝐿 = 𝑓 𝑠 𝑣± 𝑣 𝐿 𝑣± 𝑣 𝑠
In this formula v is the speed of the waves.
L is for Listener, and S is for Source.
Set up your coordinate system so that positive is away from the listener and toward the source. OR
Choose the plus/minus signs based on the situation:
If the motion is toward, the fraction has to come out larger than 1 (to get an upward shift in frequency)
If the motion is away, the fraction has to come out less than 1 (to get a downward shift in frequency)
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

15. Example: On a hot afternoon you hear an ice cream truck down the street. The truck’s music box is broken, so it just emits a continuous tone at exactly 1000 Hz.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

16. Example: On a hot afternoon you hear an ice cream truck down the street. The truck’s music box is broken, so it just emits a continuous tone at exactly 1000 Hz.
v=10m/s
Case 1: The ice cream truck is approaching your house at 10 m/s.
What is the frequency of the sound you hear?
972 Hz
1000 Hz
1030 Hz
who cares – give me the ice cream already!!
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

17. The correct answer is c)
Example: On a hot afternoon you hear an ice cream truck down the street. The truck’s music box is broken, so it just emits a continuous tone at exactly 1000 Hz.
v=10m/s
Case 1: The ice cream truck is approaching your house at 10 m/s.
What is the frequency of the sound you hear?
972 Hz
1000 Hz
1030 Hz
who cares – give me the ice cream already!!
The correct answer is c)
The truck is moving toward you, so there is an upward shift in frequency.
answer d) is also correct
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

18. Here is the calculation using our Doppler shift formula:
Example: On a hot afternoon you hear an ice cream truck down the street. The truck’s music box is broken, so it just emits a continuous tone at exactly 1000 Hz.
v=10m/s
Positive Direction
Case 1: The ice cream truck is approaching your house at 10 m/s.
What is the frequency of the sound you hear?
972 Hz
1000 Hz
1030 Hz
who cares – give me the ice cream already!!
Here is the calculation using our Doppler shift formula:
Negative Sign on bottom makes the fraction bigger than 1
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

19. Example: On a hot afternoon you hear an ice cream truck down the street. The truck’s music box is broken, so it just emits a continuous tone at exactly 1000 Hz.
v=10m/s
Case 2: The ice cream truck has passed by your house and is driving away at 10 m/s.
What is the frequency of the sound you hear?
972 Hz
1000 Hz
1030 Hz
you can’t hear anything over the sound of your sobbing – you missed the ice cream truck!!
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

20. The correct answer is a)
Example: On a hot afternoon you hear an ice cream truck down the street. The truck’s music box is broken, so it just emits a continuous tone at exactly 1000 Hz.
v=10m/s
Case 2: The ice cream truck has passed by your house and is driving away at 10 m/s.
What is the frequency of the sound you hear?
972 Hz
1000 Hz
1030 Hz
you can’t hear anything over the sound of your sobbing – you missed the ice cream truck!!
The correct answer is a)
The truck is moving away from you, so there is a downward shift in frequency.
answer d) is not correct – you can still catch the truck. hurry! run!!
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

21. Here is the calculation using our Doppler shift formula:
Example: On a hot afternoon you hear an ice cream truck down the street. The truck’s music box is broken, so it just emits a continuous tone at exactly 1000 Hz.
v=10m/s
Positive Direction
Case 2: The ice cream truck has passed by your house and is driving away at 10 m/s.
What is the frequency of the sound you hear?
972 Hz
1000 Hz
1030 Hz
you can’t hear anything over the sound of your sobbing – you missed the ice cream truck!!
Here is the calculation using our Doppler shift formula:
Positive Sign on bottom makes the fraction smaller than 1
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

22. Example: On a hot afternoon you hear an ice cream truck down the street. The truck’s music box is broken, so it just emits a continuous tone at exactly 1000 Hz.
v=15m/s
v=10m/s
Case 3: The ice cream truck has passed by your house and is driving away at 10 m/s. Your mom chases the truck at her top speed of 15 m/s. What frequency sound does she hear?
986 Hz
1000 Hz
1014 Hz
you can’t hear anything because you are screaming for ice cream!
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

23. Example: On a hot afternoon you hear an ice cream truck down the street. The truck’s music box is broken, so it just emits a continuous tone at exactly 1000 Hz.
v=15m/s
v=10m/s
Case 3: The ice cream truck has passed by your house and is driving away at 10 m/s. Your mom chases the truck at her top speed of 15 m/s. What frequency sound does she hear?
986 Hz
1000 Hz
1014 Hz
you can’t hear anything because you are screaming for ice cream!
The correct answer is c) – the truck is moving away from you, but you are chasing it. Since your speed is greater than the truck, your relative velocity is toward, so there is a net upward shift in the frequency.
answer d) may also be correct – it depends upon the availability of your favorite flavor.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

24. Here is the calculation using our Doppler shift formula:
Example: On a hot afternoon you hear an ice cream truck down the street. The truck’s music box is broken, so it just emits a continuous tone at exactly 1000 Hz.
v=15m/s
v=10m/s
Positive Direction
Case 3: The ice cream truck has passed by your house and is driving away at 10 m/s. Your mom chases the truck at her top speed of 15 m/s. What frequency sound does she hear?
986 Hz
1000 Hz
1014 Hz
you can’t hear anything because you are screaming for ice cream!
Here is the calculation using our Doppler shift formula:
Positive Sign on top makes the fraction bigger than 1
Positive Sign on bottom makes the fraction smaller than 1
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB

25. Example: Bats perform echolocation by sending out cries and gathering information about their surroundings based on the echoes that return back to them.
Suppose a bat is traveling toward the wall of a cave at a speed of 10 m/s. It sends out a cry with frequency 70kHz. What is the frequency of the returning echo?

26. Example: Bats perform echolocation by sending out cries and gathering information about their surroundings based on the echoes that return back to them.
Suppose a bat is traveling toward the wall of a cave at a speed of 10 m/s. It sends out a cry with frequency 70kHz. What is the frequency of the returning echo?
This will be a two step problem. The bat is the source of the initial cry, and the wall is the “listener”. Then the sound wave reflects from the wall, and the wall is the source of the returning echo, and the bat is the listener.

27. Example: Bats perform echolocation by sending out cries and gathering information about their surroundings based on the echoes that return back to them.
Suppose a bat is traveling toward the wall of a cave at a speed of 10 m/s. It sends out a cry with frequency 70kHz. What is the frequency of the returning echo?
This will be a two step problem. The bat is the source of the initial cry, and the wall is the “listener”. Then the sound wave reflects from the wall, and the wall is the source of the returning echo, and the bat is the listener.
vL=0 m/s
Step 1: Bat is source, Wall is listener.
vs=-10 m/s
f0=70kHz
𝑓 1 = 70𝑘𝐻𝑧 𝑚 𝑠 𝑚 𝑠 −10 𝑚 𝑠 =72.1𝑘𝐻𝑧
Positive Direction

28. Example: Bats perform echolocation by sending out cries and gathering information about their surroundings based on the echoes that return back to them.
Suppose a bat is traveling toward the wall of a cave at a speed of 10 m/s. It sends out a cry with frequency 70kHz. What is the frequency of the returning echo?
This will be a two step problem. The bat is the source of the initial cry, and the wall is the “listener”. Then the sound wave reflects from the wall, and the wall is the source of the returning echo, and the bat is the listener.
vL=0 m/s
Step 1: Bat is source, Wall is listener.
vs=-10 m/s
f0=70kHz
𝑓 1 = 70𝑘𝐻𝑧 𝑚 𝑠 𝑚 𝑠 −10 𝑚 𝑠 =72.1𝑘𝐻𝑧
Positive Direction
vs=0 m/s
Step 2: Wall is source, Bat is listener.
vL=10 m/s
f1=72.1kHz
𝑓 2 = 71.2𝑘𝐻𝑧 𝑚 𝑠 𝑚 𝑠 𝑚 𝑠 =74.2𝑘𝐻𝑧
Positive Direction