Phys 101 learning object lo5 doppler effect vivian tsang

Doppler Effect by Vivian Tsang 14153143

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!

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 ’

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

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

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

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

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

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 personAand 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?

Hint Try visualizing the question!

Person A Person B

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!

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?

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

Let’s clear up the information Given information

f= 70x10^3 Hz

Vbaby=3m/s

Vwater=1484m/s

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!

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

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