Waves - MCAT Chemical and Physical Foundations of Biological Systems
Card 1 of 490
A sound source with a frequency of 790Hz moves away from a stationary observer at a rate of 15m/s. What frequency does the observer hear?
The speed of sound is 340m/s.
A sound source with a frequency of 790Hz moves away from a stationary observer at a rate of 15m/s. What frequency does the observer hear?
The speed of sound is 340m/s.
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In this scenario the Doppler effect is described by the following equation.
Using the values from the problem, we know that vo is zero and vf is 15m/s. v is 340m/s and fs is 790Hz.
In this scenario the Doppler effect is described by the following equation.
Using the values from the problem, we know that vo is zero and vf is 15m/s. v is 340m/s and fs is 790Hz.
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A fire truck emits an 880Hz siren. As the truck approaches an obeserver on the sidewalk, he perceives the pitch to be 950Hz. Approximately what pitch does he hear after the truck passes and is moving away? Assume the truck's velocity remains constant, and that the velocity of sound in air is 340m/s.
A fire truck emits an 880Hz siren. As the truck approaches an obeserver on the sidewalk, he perceives the pitch to be 950Hz. Approximately what pitch does he hear after the truck passes and is moving away? Assume the truck's velocity remains constant, and that the velocity of sound in air is 340m/s.
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The equation for Doppler effect is 
, where the + sign applies when the source and observer are moving farther apart, and the - sign applies when they are moving closer together. In these equations, v is the speed of sound, 340m/s, 
is the frequency of sound emitted by the source, 
is the freqency perceived by the observer, and 
is the relative velocity between the source and observer.
We can apply this equation to the first part of the motion, as the truck moves closer to the observer, to solve for the velocity of the truck.
Now we can plug this velocity into the equation again for when the truck moves farther away from the observer and solve for 
.
The equation for Doppler effect is
We can apply this equation to the first part of the motion, as the truck moves closer to the observer, to solve for the velocity of the truck.
Now we can plug this velocity into the equation again for when the truck moves farther away from the observer and solve for
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Which of the following best describes the effect of the Doppler shift on the appearance of stars moving towards Earth?
Which of the following best describes the effect of the Doppler shift on the appearance of stars moving towards Earth?
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The Doppler shift equation for light is 
, where f is the source frequency, f' is the observed frequency, v is the relative velocity between source and observer, and c is the speed of light.
When the source and observer are moving closer together, v is positive, so the observed frequency is greater than the source frequency. Greater frequency also implies shorter wavelength, so visible light is shifted towards the blue end of the spectrum.
The Doppler shift equation for light is
When the source and observer are moving closer together, v is positive, so the observed frequency is greater than the source frequency. Greater frequency also implies shorter wavelength, so visible light is shifted towards the blue end of the spectrum.
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A fire truck emitting a siren at 
moves at 
towards a jogger. The jogger is moving at 
towards the fire truck. Take the speed of sound to be 
.
At what frequency does the jogger perceive the siren?
A fire truck emitting a siren at
At what frequency does the jogger perceive the siren?
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In order to solve this problem we must know how to utilize the Doppler formula.
is the velocity of the observer and 
is the velocity of the source. Notice that the frequency must increase as the observer and source move closer, and therefore the plus sign is used in the numerator and the minus sign is used in the denominator. Had the jogger been moving away from the fire truck, the subtraction function would be used in both the top and bottom.
In this case we see that the source is the fire truck, moving at 
, and the observer is the jogger, moving at 
. By plugging these numbers into the formula and 
for 
, we find the perceived frequency or 
to be 
.
In order to solve this problem we must know how to utilize the Doppler formula.
In this case we see that the source is the fire truck, moving at
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At a local concert, a speaker is set up to produce low-pitched base sounds with a frequency range of 20Hz to 200Hz, which can be modeled as sine waves. In a simplified model, the sound waves the speaker produces can be modeled as a cylindrical pipe with one end closed that travel through the air at a velocity of 
, where T is the temperature in °C.
As a person walks towards the speaker, the frequency he or she hears .
At a local concert, a speaker is set up to produce low-pitched base sounds with a frequency range of 20Hz to 200Hz, which can be modeled as sine waves. In a simplified model, the sound waves the speaker produces can be modeled as a cylindrical pipe with one end closed that travel through the air at a velocity of
As a person walks towards the speaker, the frequency he or she hears .
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This question is asking us how the frequency changes when one object moves directly towards another; thus, this is a Doppler effect problem. Remembering back to our Doppler effect formula, we know that
, where f is the frequency heard by the recipient (the person at the concert), vr is the velocity of the receiver, vs is the velocity of the source, and f0 is the original frequency.
In our case, the speaker is not moving, so vs is zero. vr is positive when the person is walking towards the speaker, so the frequency heard will be higher than the original frequency.
This question is asking us how the frequency changes when one object moves directly towards another; thus, this is a Doppler effect problem. Remembering back to our Doppler effect formula, we know that
In our case, the speaker is not moving, so vs is zero. vr is positive when the person is walking towards the speaker, so the frequency heard will be higher than the original frequency.
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At a local concert, a speaker is set up to produce low-pitched base sounds with a frequency range of 20Hz to 200Hz, which can be modeled as sine waves. In a simplified model, the sound waves the speaker produces can be modeled as a cylindrical pipe with one end closed that travel through the air at a velocity of 
, where T is the temperature in °C.
A person runs away from the speaker at 3m/s, while it creates a 200Hz sound wave. What frequency does he or she hear?
Use 340m/s for the speed of sound.
At a local concert, a speaker is set up to produce low-pitched base sounds with a frequency range of 20Hz to 200Hz, which can be modeled as sine waves. In a simplified model, the sound waves the speaker produces can be modeled as a cylindrical pipe with one end closed that travel through the air at a velocity of
A person runs away from the speaker at 3m/s, while it creates a 200Hz sound wave. What frequency does he or she hear?
Use 340m/s for the speed of sound.
Tap to reveal answer
This question is asking us how the frequency changes when one object moves directly away from another; thus, this is a Doppler effect problem. Remembering back to our Doppler effect formula, we know that
, where f is the frequency heard by the recipient (the person at the concert), vr is the velocity of the receiver, vs is the velocity of the source, and f0 is the original frequency.
In our case, the speaker is not moving, so vs is zero. vr is negative when the person is walking towards the speaker, so the frequency heard will be lower than the original frequency. We can calculate the heard frequency using our equation.
This question is asking us how the frequency changes when one object moves directly away from another; thus, this is a Doppler effect problem. Remembering back to our Doppler effect formula, we know that
In our case, the speaker is not moving, so vs is zero. vr is negative when the person is walking towards the speaker, so the frequency heard will be lower than the original frequency. We can calculate the heard frequency using our equation.
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The source of a sound moves away from the listener. The listener has the impression that the source is .
The source of a sound moves away from the listener. The listener has the impression that the source is .
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The formula for the Doppler effect is:
Only frequency of the sound is affected by the Doppler effect; velocity and amplitude remain unchanged. When the source is moving away from the observer the velocity of the source is added to the speed of light.
This increases the value of the denominator, decreasing the value of the observed frequency. Frequency corresponds to pitch or tone; a lower observed frequency will result in a lower observed pitch.
The formula for the Doppler effect is:
Only frequency of the sound is affected by the Doppler effect; velocity and amplitude remain unchanged. When the source is moving away from the observer the velocity of the source is added to the speed of light.
This increases the value of the denominator, decreasing the value of the observed frequency. Frequency corresponds to pitch or tone; a lower observed frequency will result in a lower observed pitch.
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You are jogging on the sidewalk at a rate of 
. A police car behind you is patrolling at a rate of 
when it turns on its siren. If the siren has a frequency of 
, is the frequency you perceive higher or lower than the frequency emitted?
You are jogging on the sidewalk at a rate of
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The doppler effect states that if two objects are moving closer together, perceived frequencies for emitted waves will be higher. If you are jogging away from the car at 
, but the car is traveling at 
, the overall distance between you and the siren is decreasing. You will hear a higher frequency than what the siren is emitting.
The numerator is subtracted when the observer moves away from the source. The denominator is subtracted when the source moves toward the observer. In this problem both terms would be subtracted, but the denominator would be decreased by more than the numerator. This would result in a fraction greater than one, and an overall increase in the final frequency.
The doppler effect states that if two objects are moving closer together, perceived frequencies for emitted waves will be higher. If you are jogging away from the car at
The numerator is subtracted when the observer moves away from the source. The denominator is subtracted when the source moves toward the observer. In this problem both terms would be subtracted, but the denominator would be decreased by more than the numerator. This would result in a fraction greater than one, and an overall increase in the final frequency.
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A star emits visible color at 
. If the star is moving away from a stationary observer, which of the following cannot be the wavelength observed?
A star emits visible color at
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The Doppler effect accounts for observed frequency versus actual frequency emitted by a sound or light source. The equation for the Doppler effect is:
The numerator terms are summed when the observer moves toward the source, and the denominator terms are summed when the source moves away from the observer. In this question, we know that the source moves away from the observer; therefore, the observed frequency will be less than the actual frequency because the denominator in the equation will increase.
The question, however, asks about wavelength. If the frequency decreases, then the wavelength must increase according to the equation:
Velocity will remain constant (the speed of light). Any change in frequency will cause a change in wavelength via their direct relationship in the given equation.
The Doppler effect accounts for observed frequency versus actual frequency emitted by a sound or light source. The equation for the Doppler effect is:
The numerator terms are summed when the observer moves toward the source, and the denominator terms are summed when the source moves away from the observer. In this question, we know that the source moves away from the observer; therefore, the observed frequency will be less than the actual frequency because the denominator in the equation will increase.
The question, however, asks about wavelength. If the frequency decreases, then the wavelength must increase according to the equation:
Velocity will remain constant (the speed of light). Any change in frequency will cause a change in wavelength via their direct relationship in the given equation.
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If the first overtone of a sound wave in a pipe with one end open and one end closed has a frequency of 300Hz, what is the frequency of the second overtone?
If the first overtone of a sound wave in a pipe with one end open and one end closed has a frequency of 300Hz, what is the frequency of the second overtone?
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In an open-closed pipe, we get only odd harmonics, since there must always be a displacement node at the closed end and a displacement antinode at the open end. So, the first overtone is the same as the third harmonic, which has 
. Similarly, the second overtone is the fifth harmonic, which has 
. Using the given value of 
in the first overtone to be 300Hz, we find 
and 
.
In an open-closed pipe, we get only odd harmonics, since there must always be a displacement node at the closed end and a displacement antinode at the open end. So, the first overtone is the same as the third harmonic, which has
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A string of length 2.5m is struck producing a harmonic. The wavelength produced is 1.25m. Which harmonic is the string vibrating at?
A string of length 2.5m is struck producing a harmonic. The wavelength produced is 1.25m. Which harmonic is the string vibrating at?
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A harmonic is a standing wave that has certain points (nodes) which do not go through any displacement, and certain points (antinodes) that are moving through maximum displacement during the wave's resonation.
Wavelength is related to harmonic on a string fixed at both ends through the formula 
, where 
is the length of the string and 
is the number harmonic. By plugging in the given length and wavelength, we can solve for 
.
The string must be vibrating at the fourth harmonic.
A harmonic is a standing wave that has certain points (nodes) which do not go through any displacement, and certain points (antinodes) that are moving through maximum displacement during the wave's resonation.
Wavelength is related to harmonic on a string fixed at both ends through the formula
The string must be vibrating at the fourth harmonic.
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At a local concert, a speaker is set up to produce low-pitched base sounds with a frequency range of 20Hz to 200Hz, which can be modeled as sine waves. In a simplified model, the sound waves the speaker produces can be modeled as a cylindrical pipe with one end closed that travel through the air at a velocity of 
, where T is the temperature in °C.
How are the first three harmonics of the base speaker designated?
At a local concert, a speaker is set up to produce low-pitched base sounds with a frequency range of 20Hz to 200Hz, which can be modeled as sine waves. In a simplified model, the sound waves the speaker produces can be modeled as a cylindrical pipe with one end closed that travel through the air at a velocity of
How are the first three harmonics of the base speaker designated?
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First, notice that the paragraph above tells us that the wave can be modeled as a pipe with one end closed. This is in contrast to the other possibility, where the wave is modeled as a pipe with two ends open. It is critical to recognize this difference, as the definition of sequential harmonics and the formula used to calculate them changes depending on whether both ends are open or not. In the situation where one end is closed, the harmonics are odd numbers, meaning that the first three harmonics are 1st, 3rd, and 5th. In the situation where both ends are open or both ends are closed, the harmonics are sequential, meaning that the first three harmonics are 1st, 2nd, and 3rd.
First, notice that the paragraph above tells us that the wave can be modeled as a pipe with one end closed. This is in contrast to the other possibility, where the wave is modeled as a pipe with two ends open. It is critical to recognize this difference, as the definition of sequential harmonics and the formula used to calculate them changes depending on whether both ends are open or not. In the situation where one end is closed, the harmonics are odd numbers, meaning that the first three harmonics are 1st, 3rd, and 5th. In the situation where both ends are open or both ends are closed, the harmonics are sequential, meaning that the first three harmonics are 1st, 2nd, and 3rd.
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At a local concert, a speaker is set up to produce low-pitched base sounds with a frequency range of 20Hz to 200Hz, which can be modeled as sine waves. In a simplified model, the sound waves the speaker produces can be modeled as a cylindrical pipe with one end closed that travel through the air at a velocity of 
, where T is the temperature in °C.
What is the closest distance a person could be standing away from the speaker in order to hear the loudest fundamental frequency of a 1.2m wavelength wave at the concert?
At a local concert, a speaker is set up to produce low-pitched base sounds with a frequency range of 20Hz to 200Hz, which can be modeled as sine waves. In a simplified model, the sound waves the speaker produces can be modeled as a cylindrical pipe with one end closed that travel through the air at a velocity of
What is the closest distance a person could be standing away from the speaker in order to hear the loudest fundamental frequency of a 1.2m wavelength wave at the concert?
Tap to reveal answer
This question asks us to incorporate information we learned in the pre-question text and new information in the question. First, we are told that we are looking to calculate a distance where maximal sound is heard, in other words, where the amplitude is at its maximum. This only occurs at an anti-node.
Additionally, we know that the speaker can be modeled as a one-end closed pipe, meaning that the wavelengths of the harmonics are calculated as 
.
In a one-end closed pipe model, each harmonic ends with maximal amplitude, so we can find L (the length of the pipe, i.e. where the person standing would hear maximal sound), as we know the fundamental harmonic has n = 1.
This question asks us to incorporate information we learned in the pre-question text and new information in the question. First, we are told that we are looking to calculate a distance where maximal sound is heard, in other words, where the amplitude is at its maximum. This only occurs at an anti-node.
Additionally, we know that the speaker can be modeled as a one-end closed pipe, meaning that the wavelengths of the harmonics are calculated as
In a one-end closed pipe model, each harmonic ends with maximal amplitude, so we can find L (the length of the pipe, i.e. where the person standing would hear maximal sound), as we know the fundamental harmonic has n = 1.
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At a local concert, a speaker is set up to produce low-pitched base sounds with a frequency range of 20Hz to 200Hz, which can be modeled as sine waves. In a simplified model, the sound waves the speaker produces can be modeled as a cylindrical pipe with one end closed that travel through the air at a velocity of 
, where T is the temperature in °C.
If the sound crew wanted to quadruple the intensity of the sound, they could .
At a local concert, a speaker is set up to produce low-pitched base sounds with a frequency range of 20Hz to 200Hz, which can be modeled as sine waves. In a simplified model, the sound waves the speaker produces can be modeled as a cylindrical pipe with one end closed that travel through the air at a velocity of
If the sound crew wanted to quadruple the intensity of the sound, they could .
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This question asks us to determine what would happen to the intensity of a sound wave if a variable could be changed. First, we need to remember the equation for intensity: 
, where A is the wave amplitude, rho is the density of the medium, f is the frequency of the wave, and v is the velocity of the wave.
This formula is important to understand qualitatively. For the MCAT, it is important to understand how the various factors impact the intensity of a wave. If we wanted to quadruple the intensity, we can see that we could double amplitude or double frequency. Seeing that we want the sound pitch to remain the same, we would want to double the amplitude and not the frequency. The relationship between intensity, frequency, and amplitude is important to understand.
This question asks us to determine what would happen to the intensity of a sound wave if a variable could be changed. First, we need to remember the equation for intensity:
This formula is important to understand qualitatively. For the MCAT, it is important to understand how the various factors impact the intensity of a wave. If we wanted to quadruple the intensity, we can see that we could double amplitude or double frequency. Seeing that we want the sound pitch to remain the same, we would want to double the amplitude and not the frequency. The relationship between intensity, frequency, and amplitude is important to understand.
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At a local concert, a speaker is set up to produce low-pitched base sounds with a frequency range of 20Hz to 200Hz, which can be modeled as sine waves. In a simplified model, the sound waves the speaker produces can be modeled as a cylindrical pipe with one end closed that travel through the air at a velocity of 
, where T is the temperature in °C.
If the sound crew wanted to halve the intensity of the sound, they could .
At a local concert, a speaker is set up to produce low-pitched base sounds with a frequency range of 20Hz to 200Hz, which can be modeled as sine waves. In a simplified model, the sound waves the speaker produces can be modeled as a cylindrical pipe with one end closed that travel through the air at a velocity of
If the sound crew wanted to halve the intensity of the sound, they could .
Tap to reveal answer
This question asks us to determine what would happen to the intensity of a sound wave if a variable could be changed. First, we need to remember the equation for intensity, 
, where P is the power of the wave, A is the area, 
is the wave amplitude, rho is the density of the medium, f is the frequency of the wave, and v is the velocity of the wave.
This formula is important to understand qualitatively. For the MCAT, it is important to understand how the various factors impact the intensity of a wave. If we wanted to halve the intensity, we can see that we could halve the power of the wave or double the area the wave is hitting; thus, we could double the size of the room, allowing the sound wave to have more room to spread out and increasing the area the wave interacts with.
This question asks us to determine what would happen to the intensity of a sound wave if a variable could be changed. First, we need to remember the equation for intensity,
This formula is important to understand qualitatively. For the MCAT, it is important to understand how the various factors impact the intensity of a wave. If we wanted to halve the intensity, we can see that we could halve the power of the wave or double the area the wave is hitting; thus, we could double the size of the room, allowing the sound wave to have more room to spread out and increasing the area the wave interacts with.
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At a local concert, a speaker is set up to produce low-pitched base sounds with a frequency range of 20Hz to 200Hz, which can be modeled as sine waves. In a simplified model, the sound waves the speaker produces can be modeled as a cylindrical pipe with one end closed that travel through the air at a velocity of 
, where T is the temperature in °C.
Find the intensity of a 35dB base wave produced by the speaker.
At a local concert, a speaker is set up to produce low-pitched base sounds with a frequency range of 20Hz to 200Hz, which can be modeled as sine waves. In a simplified model, the sound waves the speaker produces can be modeled as a cylindrical pipe with one end closed that travel through the air at a velocity of
Find the intensity of a 35dB base wave produced by the speaker.
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This question asks us to determine the intensity of a wave, given to us as a comparison of the wave intensity to the intensity of the limit of human hearing. In math terms, we know that the decibel scale is calculated as shown below.
I0 is the limit of human hearing (10-12 W/m2).
This question asks us to determine the intensity of a wave, given to us as a comparison of the wave intensity to the intensity of the limit of human hearing. In math terms, we know that the decibel scale is calculated as shown below.
I0 is the limit of human hearing (10-12 W/m2).
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A song comes onto your car radio and you lower the volume from a setting of 
to a setting of 
. How much has the intensity level decreased?
A song comes onto your car radio and you lower the volume from a setting of
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Regardless of the units measured, a ten-fold decrease in volume equals a 
decrease in intensity level.
Here, there is a twenty-fold decrease in sound intensity.
A ten-fold decrease would be a 
change, and a hundred-fold decrease would be a 
change. A 
decrease is closer to a ten-fold decrease than a hundred-fold decrease, hence the decrease in intensity level should be closer to 
.
Regardless of the units measured, a ten-fold decrease in volume equals a
Here, there is a twenty-fold decrease in sound intensity.
A ten-fold decrease would be a
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Which statement best explains why sound intensity is lessened when a wall is placed between the source and listener?
Which statement best explains why sound intensity is lessened when a wall is placed between the source and listener?
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When a sound wave contacts a surface, it transfers longitudinal vibration into the solid. Because of the increased density in the solid, these longitudinal compressions actually speed up and increase in wavelength. The frequency of the sound remains constant.
Only some of the sound energy is transferred to the solid, however. When the wave impacts the solid, some of the energy is bounced of the surface and reflected back to the source as an echo. This reduction of energy accounts for the lessened intensity experienced by a listener on the opposite side of the wall. Energy is also lost to internal reflection within the wall.
When a sound wave contacts a surface, it transfers longitudinal vibration into the solid. Because of the increased density in the solid, these longitudinal compressions actually speed up and increase in wavelength. The frequency of the sound remains constant.
Only some of the sound energy is transferred to the solid, however. When the wave impacts the solid, some of the energy is bounced of the surface and reflected back to the source as an echo. This reduction of energy accounts for the lessened intensity experienced by a listener on the opposite side of the wall. Energy is also lost to internal reflection within the wall.
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You are watching a launching space shuttle that produces 10MW of sound from a horizantal distance of 300 meters away. After the shuttle has risen 500 meters straight up from its launch point, what intensity of sound do you percieve?
You are watching a launching space shuttle that produces 10MW of sound from a horizantal distance of 300 meters away. After the shuttle has risen 500 meters straight up from its launch point, what intensity of sound do you percieve?
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As sound radiates from a point source, it does so in concentric circles. Therefore, we can calculate the intensity of sound using the following formula:
We simply need to calculate 
: your distance from the space shuttle.
The shuttle is 500 meters from the ground, and you are 300 meters from its launch point. We can calculate the distance from you to the current position of the shuttle by using the Pythagorean Theorem:
There is no need to take the square root, since it's already in the form we need to do the calculation. Plugging in this value to the original equation, we get:
As sound radiates from a point source, it does so in concentric circles. Therefore, we can calculate the intensity of sound using the following formula:
We simply need to calculate
The shuttle is 500 meters from the ground, and you are 300 meters from its launch point. We can calculate the distance from you to the current position of the shuttle by using the Pythagorean Theorem:
There is no need to take the square root, since it's already in the form we need to do the calculation. Plugging in this value to the original equation, we get:
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A guitar player uses beats to tune his instrument by playing two strings. If one vibrates at 550 Hz and the second at 555 Hz, how many beats will he hear per minute?
A guitar player uses beats to tune his instrument by playing two strings. If one vibrates at 550 Hz and the second at 555 Hz, how many beats will he hear per minute?
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Two waves will emit a beat with a fequency equal to the difference in frequency of the two waves. In this case the beat frequency is:
(beats per second)
Convert to beats per minute:
Two waves will emit a beat with a fequency equal to the difference in frequency of the two waves. In this case the beat frequency is:
Convert to beats per minute:
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