The document discusses the Doppler effect, which occurs when the observer and source of waves are moving relative to each other, causing the observed frequency to differ from the emitted frequency. It provides four examples of different relative motions between observer and source. When the observer moves toward the source, the observed frequency increases, and when they move apart it decreases. The same applies if the source moves toward or away from a stationary observer. Complex scenarios with multiple motions are also considered. Two example problems are worked through to demonstrate applying the Doppler effect concepts.

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’=f(v/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

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’=f(v/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

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…

10. Hint
 Which person will observe a larger frequency?
 Try visualizing the question!
Person A Person B

11. Answer
 That’s right!
 Person A will observe a lower frequency that 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. 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

16. Thank you for
watching!
Thanksfor helping us!