Waves - AP Physics 1
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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?
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
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.
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
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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?
One of your family members has convinced you to participate in a 10k race with them. You are running at a steady pace of
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.
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
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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?
You are commuting to work one morning on your bike. You travel at an average rate of
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:
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:
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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?
A group of storm chasers has found a tornado that is traveling north at a rate of
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:
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:
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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 
.
You are driving your car by a very loud concert and moving at
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
Remember the equation for the doppler effect for a moving observer:
Now, identify the given information:
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
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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?
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
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:
To solve this problem we can examine the doppler effect equation:
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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?
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?
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.
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.
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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?
A person showing off their new sound system is driving towards a group of people at a speed of
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.
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
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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.
What is the apparent frequency
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:
The formula for the doppler effect is:
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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?
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
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.
The first thing we need to do is convert the wavelengths given to a usable frequency using:
Where
Plug in known values into the Doppler equation. We may assume the Earth is stationary relative to the star.
Simplify.
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What is the wavelength of a 
wave if its velocity is 
?
What is the wavelength of a
The relationship between wavelength and velocity is given by the equation;
The question gives us the velocity of the wave and its frequency. Using these values, we can solve for the wavelength.
The relationship between wavelength and velocity is given by the equation;
The question gives us the velocity of the wave and its frequency. Using these values, we can solve for the wavelength.
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A pendulum of length 
with a ball of mass 
is released at an angle 
away from the equilibrium point. Which of the following adjustments will result in an increase in frequency of oscillation?
A pendulum of length
For a pendulum on a string, the period at which it oscillates is:
Period is the reciprocal of frequency.
Calculate the frequency of oscillation.
The only parameter that the frequency depends on is the length 
. Decreasing length increases frequency.
For a pendulum on a string, the period at which it oscillates is:
Period is the reciprocal of frequency.
Calculate the frequency of oscillation.
The only parameter that the frequency depends on is the length
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The wavelength of a ray of light travelling through a vaccuum is 
. What is the period of this wave?
The wavelength of a ray of light travelling through a vaccuum is
In this question, we're given the wavelength of light in a vaccuum. Since we know that the speed of light in a vaccuum is equal to 
, we can first calculate the frequency of the light, and then use this value to calculate its period.
To begin with, we'll need to make use of the following equation to solve for frequency.
Then, using this value, we can calculate the period by noting that the period is the inverse of frequency.
And for completion's sake, it is worth noting that the period refers to the amount of time it takes for one wavelength to pass, which in this case is a very, very small fraction of a second!
In this question, we're given the wavelength of light in a vaccuum. Since we know that the speed of light in a vaccuum is equal to
To begin with, we'll need to make use of the following equation to solve for frequency.
Then, using this value, we can calculate the period by noting that the period is the inverse of frequency.
And for completion's sake, it is worth noting that the period refers to the amount of time it takes for one wavelength to pass, which in this case is a very, very small fraction of a second!
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How does the period of a pendulum react to its length quadrupling?
How does the period of a pendulum react to its length quadrupling?
The equation for the period of a pendulum is as follows: 
When quadrupling the length, the square root of four can be taken out which doubles the period. 
The equation for the period of a pendulum is as follows:
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A buoy is rocking in the ocean when a storm hits. From the lowest point to the highest point, the buoy goes up a total of 
. Over the last 20 minutes, the buoy has moved up and down a total of 
.
What is the frequency of the buoy's oscillation?
A buoy is rocking in the ocean when a storm hits. From the lowest point to the highest point, the buoy goes up a total of
What is the frequency of the buoy's oscillation?
The buoy moves a total of 
. From the problem, you know the buoy moves a total of 
with every wavelength (10 meters up and 10 meters coming back down). This means that in 20 minutes, the buoy went through 
cycles. In order to find the frequency, you divide the total amount of cycles over the total amount of time in seconds that passed:
The buoy moves a total of
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Waves are hitting the side of a pier in a regular fashion. During 1 minute, 
waves that hit the side of the pier. Calculate the period 
of the waves during this time in seconds.
Waves are hitting the side of a pier in a regular fashion. During 1 minute,
In order to calculate the period 
, we need to find the frequency, 
, at which the waves hit the pier. We are told that 
waves hit within a 
window. Therefore,
. We can then calculate the period 
from this information.
In order to calculate the period
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A string produces the following frequencies: 
, 
, 
, and 
.
What is the fundamental frequency of this string?
A string produces the following frequencies:
What is the fundamental frequency of this string?
Since the frequencies produced, differ by 
we know that the fundamental frequency must be the difference. This is because the fundamental is the very first frequency and the lowest one that can be produced.
Therefore the correct answer is 
.
Since the frequencies produced, differ by
Therefore the correct answer is
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The frequency of a sound wave traveling through water at 
is record to be 
, what is the period of the wave?
The frequency of a sound wave traveling through water at
Frequency is simply represent as the inverse of Period as represented below
where f is frequency and T is period. To solve the equation for period we just reverse the inverse relation
Frequency is simply represent as the inverse of Period as represented below
where f is frequency and T is period. To solve the equation for period we just reverse the inverse relation
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If the wavelength of a light wave is 
, what is the period of this wave?
If the wavelength of a light wave is
To answer this question, it's important to have an understanding of the interrelationship between wavelength and period. Also, because we're told that this is a light wave, it's also essential that we know the speed at which light travels.
First, we can show an equation that relates wavelength and wave speed.
Moreover, we can relate the frequency of a wave to its period, both of which are inversely related to each other.
Using these two expressions, we can combine them into the following expression.
Next, we can rearrange the equation in order to isolate for the variable representing the period of the wave.
Lastly, we just need to plug in the value for wavelength given to us in the question stem, along with the speed of light, to obtain our answer.
What this answer tells us is that it will take 
in order for a single wavelength of light to cross a given point - a very, very short amount of time!
To answer this question, it's important to have an understanding of the interrelationship between wavelength and period. Also, because we're told that this is a light wave, it's also essential that we know the speed at which light travels.
First, we can show an equation that relates wavelength and wave speed.
Moreover, we can relate the frequency of a wave to its period, both of which are inversely related to each other.
Using these two expressions, we can combine them into the following expression.
Next, we can rearrange the equation in order to isolate for the variable representing the period of the wave.
Lastly, we just need to plug in the value for wavelength given to us in the question stem, along with the speed of light, to obtain our answer.
What this answer tells us is that it will take
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Which of the following is a standing wave?
Which of the following is a standing wave?
A simple definition of a standing wave is a wave that is self-reinforcing, which is to say that reflection of the wave through the medium results in some areas of amplification (anti-nodes) of the wave and some areas of nullification (nodes). In other words, resonance must occur, and that usually suggests confinement of the wave in some fashion.
A fan and a bus make noise and vibration, but the sound does not resonate. It is transmitted, but not confined. Light with a specific wavelength has no "resonant" character, and neither do waves striking a pier. If the waves were confined in a harbor so that they could amplify, it might be possible to produce a standing wave. Microwaves trapped inside a microwave oven have this feature, producing antinodes of intense heating and nodes where no energy is transmitted into the food; this is the reason that microwave ovens have rotating platforms to make heating of the food item more uniform.
A violin string will be seen to have discrete, stable regions of motion and lack of motion, the requirements of the standing wave phenomenon. The points of reflection on the string are the two ends. The vibration of the wave is confined within the string, amplifying the sound as the nodes overlap.
A simple definition of a standing wave is a wave that is self-reinforcing, which is to say that reflection of the wave through the medium results in some areas of amplification (anti-nodes) of the wave and some areas of nullification (nodes). In other words, resonance must occur, and that usually suggests confinement of the wave in some fashion.
A fan and a bus make noise and vibration, but the sound does not resonate. It is transmitted, but not confined. Light with a specific wavelength has no "resonant" character, and neither do waves striking a pier. If the waves were confined in a harbor so that they could amplify, it might be possible to produce a standing wave. Microwaves trapped inside a microwave oven have this feature, producing antinodes of intense heating and nodes where no energy is transmitted into the food; this is the reason that microwave ovens have rotating platforms to make heating of the food item more uniform.
A violin string will be seen to have discrete, stable regions of motion and lack of motion, the requirements of the standing wave phenomenon. The points of reflection on the string are the two ends. The vibration of the wave is confined within the string, amplifying the sound as the nodes overlap.
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