This document discusses the Doppler effect through examples of waves emitted from a source towards both a stationary and moving observer. It begins by defining the Doppler effect as a difference between the frequency of emitted waves and the observed frequency when the source and observer are in relative motion. Examples are then given of various scenarios where the observer or source is moving towards or away from the other. Equations for calculating the observed frequency are provided for each scenario. The document concludes by posing example problems applying the Doppler effect concepts and walking through the steps to solve them.

1. Doppler Effect by Vivian Tsang 14153143

2. What is this?
 This phenomenon occurs when the frequency emitted
by the source of the waves and the frequency at which
the observer notes the waves are different!
 The Doppler Effect occurs when the observer and the
source of the waves are moving relative to each other!

3. Let’s break it down
 First consider when the source and the observer are
both at rest. Let f = the frequency of the source and f’
be the frequency observed by the blue man
 Here, there is no Doppler shift: f = f’

4. Now let’s consider four examples
 Now let’s imagine the blue man running towards the source.
 Here, the Doppler effect comes into play as there is a difference in their
relative speeds
 f’=f(v+vo)/v
 v0 refers to the observer’s speed while v refers to the speed of the wave
 The (+) sign is used because the observer is going towards the source

5. Example 2
 Now let’s imagine the blue man running away from the source.
 Here, the Doppler effect comes into play as there is a difference in their
relative speeds
 f’=f(v-vo)/v
 v0 refers to the observer’s speed while v refers to the speed of the wave
 The (-) sign is used because the observer is going away from the source

6. Example 3
 Now let’s imagine the blue man stationary but the source of the wave moving closer to
the man.
 Here, the Doppler effect comes into play as there is a difference in their relative speeds
 f’=fv/(v-vs)
 vs refers to the source speed while v refers to the speed of the wave
 The (-) sign is used because the source is coming closer to the observer
 Note: the green wave has a smaller wavelength than the red wave

7. Example 4
 Now let’s imagine the blue man stationary but the source of the wave moving farther
from the man
 Here, the Doppler effect comes into play as there is a difference in their relative speeds
 f’=fv/(v+vs)
 vs refers to the source speed while v refers to the speed of the wave
 The (+) sign is used because the source is moving further from the observer
 Note: the green wave has a larger wavelength than the red wave

8. So what if everything’s moving all
at the same time?!?!?!
 Now that we know the basics, there are multiple combinations possible.
 Let’s imagine the blue man running away from the source and the source of the wave
moving farther from the man
 Here, the Doppler effect comes into play as there is a difference in their relative speeds
 f’=f(v-v0)/(v+vs)
 vs refers to the source speed while v refers to the speed of the wave
 v0 refers to the observer’s speed while v refers to the speed of the wave
 The (-) sign is used in the numerator because the object is moving further from the
observer
 The (+) sign is used in the denominator because the source is moving further from the
observer

9. Let’s apply this! Question 1
 A brand new ferrari is
moving along a straight
section of a road with its
engine in constant full
blast away from personA
and towards personB. If
fA is the frequency
observed by observer A
and fB is the frequency
observed by observer B…
Q: Which person
will observe a
larger frequency?

10. Hint
 Try visualizing the question!
Person A Person B

11. Answer
 That’s right!
 Person A will observe a lower frequency than person B
 This is because the wave fronts in the direction of the
car’s travel is going to be more compressed than the
wave fronts opposite to the car’s travelling direction!

12. Further applications! Question 2
 This baby dolphin sends out underwater sound waves
at 70 kHz to try and find its mom
 The baby dolphin is swimming at 3m/s and the mom
who is looking for her baby is waiting anxiously!
 What is the frequency of the echo that is heard by the
baby dolphin?

13. Hints
 Let’s assume that mom’s motion can be ignored
compared to that of the frantic baby dolphin
 The baby is going towards its mom! It’s getting closer!
 The “echo” can be thought of as the waves reflected
from the mother dolphin

14. Let’s clear up the information
 Given information
 f= 70x10^3 Hz
 Vbaby=3m/s
 Vwater=1484m/s

15. Steps:
 Part I
 Let us consider the baby dolphin as the initial source of the
sound waves
 Use the doppler effect equation that we talked about in the
previous slides and find the frequency that the mom hears
her baby at
 Part II
 After the waves “hit” the mom, the mom now becomes the
source of the waves
 The baby then becomes the receiver of the sound waves
 Therefore, we perform the doppler effect equation a second
time!

16. Solution
Notice that the
mom has zero
speed throughout
our calculations
We first find the
frequency that
the mom hears
her baby at
Next, we find the
frequency that
the baby hears the
echo back at.
Note that fbaby
differs from f’baby

17. Thank you for
watching!
Thanksfor helping us!