AP Physics 1 : Doppler Effect

Study concepts, example questions & explanations for AP Physics 1

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Example Questions

Example Question #1 : Doppler Effect

A police car is at a red light. A car behind him foolishly tests his luck and runs the red light. The officer immediately flicks on his siren and begins to chase the car, accelerating at a rate of  . The frequency of the siren is 500Hz. If you are standing at the red light, what frequency do you percieve 6 seconds after the police officer began chasing the car?

Possible Answers:

Correct answer:

Explanation:

We simply need to know the Doppler effect equation to solve this problem:

Since the observer (you) is not moving, we can rewrite the equation as:

The reason for using addition in the denominator is explained below.

We know all of our values, so we can simply solve for the perceived frequency:

For the Doppler effect equation, don't waste your time trying to memorize whether the signs should be . We have purposely made them the same to emphasize the following idea. Think about the situation practically. The police car is moving away from you. Therefore, we know that the percieved frequency is going to be less than the source. How do we make the frequency lower? We either lower the numerator or increase the denominator. Since we can only manipulate the denominator, we will use the + sign. Knowing this will also help you immediately eliminate all answers in which the frequency is increased.

Example Question #2 : Doppler Effect

One of your family members has convinced you to participate in a 10k race with them. You are running at a steady pace of . A man who throughouly enjoys running 10k races is behind you with a megaphone, telling jokes, providing encouragement for other racers, and simply enjoying himself a bit too much. If he is traveling at a pace of  and his voice has an average frequency of 800Hz, what is the percieved wavelength of sound that you hear?

Possible Answers:

Correct answer:

Explanation:

This is a two-step problem. First, we need to calculate the percieved frequency that you here, and then convert that frequency into a wavelength.

For calculating percieved frequency, we need the Doppler equation:

Plugging in our values we get:

The reasoning for the signs used is explained below.

Now we need to conver that frequency to a wavelength using the equation:

If you don't know this equation, use your units to reason it out!

Plugging in our values, we get:

 

For the Doppler effect equation, don't waste your time trying to memorize whether the signs should be . We have purposely made them the same to emphasize the following idea. Think about the situation practically. You are moving away from the source of the sound. This will lower the percieved frequency. How do we lower the frequency? Make the numerator smaller. Therefore, we subtract your (the observer's) velocity. The source is moving toward you. This will increase the percieved frequency. How do we increase the frequency? Decrease the denominator. Therefore, we also subtract his (the source's) velocity.

Example Question #3 : Doppler Effect

You are commuting to work one morning on your bike. You travel at an average rate of . A police car is traveling down the street towards you at a rate of . If the police siren emits a maximum frequency of , what is the maximum frequecny that you here?

Possible Answers:

Correct answer:

Explanation:

We need to know the equation for the Doppler effect to solve this problem:

Instead of memorizing the sign conventions, we can think about the situation pratically. You are traveling towards the source, which will increase your perceived frequency; therefore, we'll add your velocity. The source is traveling toward you, which will further increase the perceived frequency. Since the source's velocity is in the denominator, we will subtract it.

Plugging in values for each variable:

Example Question #4 : Doppler Effect

A group of storm chasers has found a tornado that is traveling north at a rate of . The storm chasers are south of the storm, trying to catch up. If they are traveling at a rate of  and percieve an average frequency of  coming from the storm, what is the actualy frequency of sound that the storm is emitting?

Possible Answers:

Correct answer:

Explanation:

We need the Doppler effect equation to solve this problem:

Instead of memorizing sign conventions, we can think about the situation practically. The chasers are traveling toward the storm, increasing their percieved frequency. Since their velocity is in the numerator, we'll add it to the speed of sound. The storm is traveling away from them, decreasing their percieved frequency. That velocity is in the denominator, so we will also add it.

We are solving for actual frequency (not the perceived frequency), so we need to rearrange the equation:

We have values for each variable, allowing us to solve:

Example Question #24 : Waves

You are driving your car by a very loud concert and moving at . If the frequency of a particular long note is 740 Hz, what is the frequency of the note you hear as you approach the concert? What is the frequency of the note you hear as you move away from the concert? The speed of sound in air is .

Possible Answers:

Moving toward the concert: 

Moving away from the concert: 

Moving toward the concert: 

Moving away from the concert: 

Moving toward the concert: 

Moving away from the concert: 

Moving toward the concert: 

Moving away from the concert: 

Correct answer:

Moving toward the concert: 

Moving away from the concert: 

Explanation:

Remember the equation for the doppler effect for a moving observer: 

Now, identify the given information:

 (This is the speed of sound in air.)

When you are moving towards the concert, the plus sign is used. Therefore, the apparent frequency is

When you are moving away from the concert, the negative sign is used. Therefore, the apparent frequency is 

Example Question #25 : Waves

An astronomer observing a distant star finds that the frequency of light she is observing is less than she expected. She concludes that there is motion between her telescope and the star. What type of motion would explain the astronomer's observation?

Possible Answers:

The star must be moving away from the Earth

The Earth must be moving toward the star

The Earth is moving away from star or the star is moving away from the Earth.

The Earth is moving toward the star or the star is moving toward the Earth

The Earth must be moving away from the star

Correct answer:

The Earth is moving away from star or the star is moving away from the Earth.

Explanation:

The doppler shift of a wave can be caused by the motion of either the observer (the astronomer) or the source (the star). Since the frequency is decreased, the motion must be away, either by the star or the Earth or both.

Example Question #5 : Doppler Effect

A person showing off their new sound system is driving towards a group of people at a speed of . The group of people is stationary. What is the observed frequency of the sound if the sound emitted by the car has a frequency of 421Hz? 

Possible Answers:

Correct answer:

Explanation:

The difference between the observed frequency and the frequency of the source due to relative velocity is called the Doppler effect, and is given by the following formula:

where  is the observed frequency,  is the source frequency,  is the speed of sound,  is the speed of the observer, and  is the speed of the source. The speed of the observer is equal to zero, because the group of people is stationary. The speed of sound was given to us as . The frequency of the source was given as 421Hz. We must convert the speed of the source into , which when converted is equal to . Finally, we must determine whether the sign in the denominator term should be a plus or minus; because we are approaching the group of people, the observed frequency should be higher, therefore a minus sign is appropriate (makes a smaller denominator which makes a larger result). Our final answer after plugging everything in is 442Hz.

Example Question #4 : Doppler Effect

What is the apparent frequency  of a wave traveling at a speed  and frequency  that is moving away from you at a speed , and that you are moving towards at a speed ? Note:  in this context is not referring to the speed of light.

Possible Answers:

Correct answer:

Explanation:

The formula for the doppler effect is:

, where  is the apparent frequency of the wave,  is the actual frequency,  is the wave velocity,  is the velocity of the detector,  is the velocity of the source. In our case, the detector (which is you) is moving towards the source at speed , making the numerator a plus sign. The source is moving away from us at speed , making the denominator also a plus sign. Knowing this, we can now plug into our equation:

Example Question #5 : Doppler Effect

Samira is standing on the corner of Market St. and 20th Ave. A rock band on the back of a flatbed truck drives past her. At the moment she is even with Samira, the lead singer yells at a frequency of . If the truck drives away at a constant velocity of , what is the apparent frequency of the singer to Samira?

Possible Answers:

Correct answer:

Explanation:

To solve this problem we can examine the doppler effect equation:

 where  is the velocity of the observer,  is the velocity of sound through air, vs is the velocity of the source of the sound,  is the original frequency of the source, and is the source's apparent frequency to the observer. Because we are dealing with a stationary observer and a source in motion, we will add the velocity of the source to the velocity of sound. We can then plug in the values given by the problem into the above equation:

Example Question #8 : Doppler Effect

An astronomer on Earth looks at a distant star and analyzes the spectral lines for hydrogen from that star.  They find the wavelength of hydrogen from the star to be , but the astronomer knows that the wavelength of hydrogen on Earth is .  How fast is the distant star moving away from the Earth?

Possible Answers:

Correct answer:

Explanation:

The first thing we need to do is convert the wavelengths given to a usable frequency using:

Where  is frequency,  is the speed of light , and  is the wavelength.  Once we have the frequencies, we can find the velocity of the star in regards to Earth:

Plug in known values into the Doppler equation. We may assume the Earth is stationary relative to the star.

Simplify.

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