2. Basic Concepts & Key Equation
frequency of the source :
frequency of the receiver :
speed of sound in air :
speed of the receiver :
speed of the source :
fr =
v ± vr
v ∓ vs
fs
fr
fs
v
vr
vs
3. Different Cases with Different Equations
➢ Case 1: Stationary Source & Stationary Receiver
In this case, Vr = Vs = 0.
Therefore,
➢ Case 2: Stationary Source & Moving Receiver
In this case, Vr ≠ 0 & Vs = 0.
If the receiver is moving toward to the source, then
If the receiver is moving away from the source, then
fr =
v + 0
v + 0
fs = fs
fr =
v + vr
v + 0
fs =
v + vr
v
fs
fr =
v − vr
v + 0
fs =
v − vr
v
fs
4. ➢ Case 3: Moving Source & Stationary Receiver
In this case, Vr = 0 & Vs ≠ 0.
If the source is moving toward to the receiver, then
If the source is moving away from the receiver, then
➢ Case 4: Moving Source & Moving Receiver
In this case, Vr ≠ 0 & Vs ≠ 0.
It depends on detailed subcases:
Different Cases with Different Equations
fr =
v + 0
v − vs
fs =
v
v − vs
fs
fr =
v + 0
v + vs
fs =
v
v + vs
fs
fr =
v ± vr
v ∓ vs
fs
5. Simple Clicker Questions
Q1:
Person A is riding a bike at 20m/s toward to Person B who is
standing still. The sound of bike in frequency is 640Hz. What is the
frequency at the Person B?
(The speed of sound in air is 340m/s)
A. 604.44Hz
B. 677.64Hz
C. 680Hz
D. 602.35Hz
E. 640Hz
6. Answer to Q1:
The answer is C.
In this question, the bike is the source, and the
standing Person B is the receiver.
So the case is moving source with stationary receiver.
Therefore, according to the key equation,
fPersonB =
v
v − vs
fbike =
340m / s
340m / s − 20m / s
× 640Hz = 680Hz
7. Simple Clicker Questions
Q2:
The bell of a still bike is ringing, and the sound in frequency is
680Hz. A person is running away from the bike at 10m/s. What is
the frequency at that person?
(The speed of sound in air is 340m/s)
A. 680Hz
B. 700Hz
C. 700.6Hz
D. 660.5Hz
E. 660Hz
8. Answer to Q2:
The answer is E.
In this question, the still bike is the source, and the
running Person is the receiver.
So the case is stationary source with moving receiver.
Therefore, according to the key equation,
fPerson =
v − vr
v
fbike =
340m / s −10m / s
340m / s
× 680Hz = 660Hz
9. Moving source with moving receiver
Q3:
Person A is driving a bus at 72km/h with the horn ringing in the frequency
of 640Hz toward to the Person B who is the receiver and riding a bike at
18km/h toward to A. What is the frequency at the Person B?
( The speed of sound in air is 340m/s)
🚲 🚌
72km/h18km/h
10. In this case, the person who is riding a bike is the receiver, and
the person who is driving a bus is the source, and both the source
and receiver are moving.
Since they are moving toward to each other, according to the
key equation, we will use the following format.
fr =
v + vr
v − vs
fs
fr =
340m / s + 5m / s
340m / s − 20m / s
× 640Hz = 690Hz
First, we have to convert
the unit of the giving speed.
72km / h =
72km ×1000
3600s
= 20m / s 18km / h =
18km ×1000
3600s
= 5m / s
11. Challenge Question
As shown in the picture, R is a sound reflector; S is a tuning fork which
can produce a sound of 612Hz in frequency; O is a person as receiver.
Define the speed of sound in air is 340m/s.
(1) If R and O are both still, S is moving toward to O at the speed of 20m/s,
what is the frequency of sound that O directly hears from S and the echo
returned by R?
(2) Now S is still, R is moving rightward at v=2m/s and O is moving rightward
at u=10m/s, what is the frequency of sound that O directly hears from S and
the echo returned by R?
(3) Now R and S are both still, O is moving rightward at 14m/s, and the wind is
blowing from O to R at 7m/s. What is the frequency of echo that O hears?
R S O
12. Answers to the Challenge Question
(1)
For the sound that O directly hear from the tuning fork, we can consider
the fork S is the moving source and O is the stationary receiver. And S
is moving toward to O.
So the frequency of direct sound is:
For the echo returned by the reflector R, we can consider the reflector
as a stationary RECEIVER, and the fork S as the source is moving away
from the receiver. Since both reflector R and the person O are stationary,
the frequency that R “hears” from the fork S is the same as the frequency
that O hears from R. Note that the sound that O hears from R is the echo.
So the frequency of echo is:
fdirect =
v
v − vs
fS =
340m / s
340m / s − 20m / s
× 612Hz = 650.25Hz
fecho = fR<−S =
v
v + vs
fS =
340m / s
340m / s + 20m / s
× 612Hz = 578Hz
13. (2)
For the sound that O directly hear from the tuning fork, we can consider the
fork S is the stationary source and O is the moving receiver. And the receiver
O is moving away from the source S.
So the frequency of direct sound is:
For the echo returned by the reflector R, we first consider the reflector as a
moving RECEIVER, and the fork S as the source is stationary. So now the
receiver R is moving toward to the source S.
Hence the frequency of sound that R hears from S is:
Now let’s consider the sound that O hear from R (that’s the echo sound we
want). So now the reflector R is source and the person O is the receiver.
Since the reflector R is moving toward to the receiver O, but the receiver O is
also moving away from the reflector R.
Therefore, the frequency of echo is:
fdirect =
v − vr
v
fS =
340m / s −10m / s
340m / s
× 612Hz = 594Hz
fR<−S =
v + vr
v
fS =
340m / s + 2m / s
340m / s
× 612Hz = 615.6Hz
fecho =
v − vreceiver
v − vreflector
freflector ( freflector = fR<−S = 615.6Hz)
fecho =
340m / s −10m / s
340m / s − 2m / s
× 615.6Hz = 601Hz
14. (3)
Recall from the textbook, the definition of Vr and Vs from the equation is:
Vr and Vs are the speed of the receiver and the source, respectively, relative
to the air.
So what will happen if the speed of air exists?
To conquer this question, let’s first figure out how to deal with the air speed.
Now we only consider the reflector R and the source S.
They are both stationary and the speed of air is 7m/s.
So what’s the frequency of sound that R hears from S?
Well, probably we can quickly answer that the frequency is still 612Hz, because
this is the case that both RECEIVER and SOURCE are STATIONARY. Or to be
more precise, both RECEIVER and SOURCE are relatively stationary to each other.
But now, let’s change our perspective on the frequency that R hears from S.
R S
Air blowing direction
7m/s
15. Now let’s take another look at the air speed.
Supposed the air is stationary, in order to reframe the same situation as when
the air speed is 7m/s, both R and S have to move leftward at the speed of 7m/s.
We have already known that the frequency from case 1 is 612Hz.
So let’s verify whether the frequency from case 2 is the same as case 1.
Basically, case 2 is moving source with moving receiver.
And the source is moving toward to the receiver, however, the receiver is
moving away from the source.
Therefore, the frequency of the sound that R hears from S in case 2 is:
R S
Air direction
7m/s
R S
7m/s
7m/s
no air speed
case 1 case 2
fR =
v − vr
v − vs
fS =
340m / s − 7m / s
340m / s − 7m / s
× 612Hz = 612Hz
16. So now, let’s begin to solve this problem!!!
We know that the frequency of the sound that R hears from S is 612Hz.
And now considering O as the receiver, R as the source, we know that R as the
source is stationary, and O as the receiver is moving away from the source.
So this is the case that stationary source with moving receiver.
Therefore, the frequency of the echo returned by R is:
R S
Air direction
7m/s
R S
7m/s
7m/s
no air speed
O O
7m/s
14m/s
fecho =
v − vr
v
fR =
(340m / s − 7m / s)−14m / s
(340m / s − 7m / s)
× 612Hz = 586.3Hz
