The Doppler Effect

1. BASIC WAVE
EQUATION

2. 3 FACTORS TO BE
CONSIDERED IN ANY
WAVE:
1.Frequency
2.Wavelength
3.Velocity

3. SPEED OF A WAVE:
v = f λ
where: v – velocity (m/s)
f – frequency (Hz = 1/s)
λ – wavelength (m)

4. SAMPLE PROBLEMS:
1. If a dog whistle has a frequency of 30 000
Hz, what is its speed if the wavelength of the
sound emitted is 0.011 m?
Given:
f = 30 000 Hz
λ = 0.011 m

5. SAMPLE PROBLEMS:
2. Suppose a station broadcasts on two radio
frequencies: 630 Hz and 101.9 Hz on your radio dial.
The speed of radio wave in air is 3 x 108 m/s. Find
the wavelength for each frequency.
Given:
f1 = 630 Hz
f2 = 101.9 Hz
v = 3 x 108 m/s

6. SPEED OF SOUND

7. •Do you notice that several
seconds pass between the
time you see lightning and
hear thunder? How come you
see lightning first when
actually, lightning and thunder
occur at the same time?

8. SPEED OF SOUND AT
ANY TEMPERATURE
GIVEN:
v = 331m/s + (0.6 m/s)(T)
Where v is the speed of
sound in air and T is
temperature

9. WHY USE 331 M/S?
•The speed of sound in air at 0oC is
331 m/s. For every 1 Celsius degree
rise in temperature, the speed of
sound increases by 0.6 m/s.

10. SAMPLE PROBLEM:
•What is the speed of sound at an
altitude of 30.5 km where the
temperature is about -50oC?
Solution:
v = 331 m/s + (0.6 m/s)(T)
v = 331 m/s + (0.6 m/s)(-50)
v = 301 m/s

11. SAMPLE PROBLEMS:
1. What is the speed of sound at:
a. 20oC? b. -30oC
v = 331 m/s + (0.6 m/s)(T)
v = 331 m/s + (0.6 m/s)(20)
v = 343 m/s
c. 100oC?
v = 331 m/s + (0.6 m/s)(T)
v = 331 m/s + (0.6 m/s)(-30)
v = 301.6 m/s
v = 331 m/s + (0.6 m/s)(T)
v = 331 m/s + (0.6 m/s)(100)
v = 391 m/s

12. SAMPLE PROBLEMS:
2. The frequency of a certain sound is 440 Hz. What is
the wavelength of this sound when the temperature of
air is
a) 20 degrees Celsius b) -30 degrees Celsius
Given:
f = 440 Hz
T = 20
λ = ?
Solution:
v = 331 m/s + (0.6 m/s)(T)
v = 331 m/s + (0.6 m/s)(20)
v = 343 m/s

14. Imagine a bug
jiggling its legs and
bobbing up and
down in the middle
of a quiet puddle

15. What would happen
to the waves if the
bug began to swim
forward, toward B?

16. What does that do
to the frequency of
the waves, in front
of the bug and
behind the bug?
In front of the
bug…the frequency ↑
Behind the bug…the
frequency ↓

18. The Doppler
effect causes
the changing
pitch of a siren

19. Note: The change in
loudness is not the
Doppler Effect! It is
the shift in
frequency!

22. When a source
moves toward you,
do you measure an
increase or
decrease in wave
speed?
Neither! It is the
frequency of a
wave that
undergoes a
change, not the
wave speed.

23. How does the
apparent
frequency of
waves change as a
wave source
moves?
 The sound waves
become more
frequent
(compressed
together).
 The sound waves
become less
frequent (stretched
apart).

25.  The speed of sound is 340 m/s one day and the source
whose actual frequency is 700 Hz is moving 20 m/s away
from the observer who is not moving. What is the
perceived frequency of the sound by the observer?
fo = 700 Hz
v = 340 m/s
vo = 0
vs = 20 m/s
f = ?
f = fo ((v (+/-) vo)/ (v (-/+) vs))
source moving away and the observer not
moving makes the equation:
f = fo ((v / (v + vs))
f =700 ((340 / (340 + 20))
f = 661 Hz

26.  A police car drives towards a jogger at 28 m/s as she runs
in the direction of the police car at 2 m/s. What would
the 700 Hz siren sound like to the jogger? (The speed of
sound today is 343 m/s)
Givens:
Fo = 700 Hz
v = 343 m/s
vo = 2 m/s
vs = 28 m/s
f = ?
f = fo ((v (+/-) vo)/ (v (-/+) vs))
source moving towards the observer and the
observer not moving towards the source makes
the equation:
f = fo ((v + vo)/ (v - vs))
f = 700 ((343 + 2)/ (343 - 28)) = 767 Hz

27.  A humming bird’s wings give off a frequency of 80 Hz as it
flies toward you at 21 m/s. What frequency would you
hear on a day the temperature is 25°C?
Givens:
Fo = 80 Hz
v = 346 m/s
vo = 0 m/s
vs = 21 m/s
f = ?
Equation:
f = fo ((v (+/-) vo)/ (v (-/+) vs))
source moving towards and the observer not
moving makes the Doppler equation:
f = fo ((v)/ (v - vs))
Work:
f = 80 ((346)/ (346 - 21)) = 85.2 Hz