Sound Waves - AP Physics 1
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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?
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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Alice measures the wavelength and frequency of a sound wave as
At what speed is the sound traveling?
Alice measures the wavelength and frequency of a sound wave as
At what speed is the sound traveling?
We know from the question that
Frequency is the inverse of period:
The velocity of a wave is its wavelength multiplied by its frequency:
We know from the question that
Frequency is the inverse of period:
The velocity of a wave is its wavelength multiplied by its frequency:
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You are standing on the sidewalk when a police car approaches you at 
with its sirens on. Its sirens seem to have a frequency of 500 Hertz. After the police car passes you and is driving away, what will be the new frequency you hear?
You are standing on the sidewalk when a police car approaches you at
The doppler effect follows this formula:
In this equation, 
is the new frequency you will hear, 
is the speed of sound, 
is the velocity of the moving sound-emitting thing, and 
is the initial frequency of the sound.
Plugging the given values in, we can describe the initial situation as:
Note that the velocity is negative 
because the car is driving towards you.
Therefore,
When the police car is driving away, the situation is described with a positive velocity:
Therefore,
The doppler effect follows this formula:
In this equation,
Plugging the given values in, we can describe the initial situation as:
Note that the velocity is negative
Therefore,
When the police car is driving away, the situation is described with a positive velocity:
Therefore,
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The ukulele is a short instrument, relative to a guitar. How does this affect the frequencies of sounds that these two instruments produce? Assume the two instruments use the same strings.
The ukulele is a short instrument, relative to a guitar. How does this affect the frequencies of sounds that these two instruments produce? Assume the two instruments use the same strings.
The speed of sound in air is constant, assuming that the temperature of the air is constant. When the length of the string is shortened, by the principles of standing waves, this creates a higher frequencies. Assuming that the two instruments use the same strings is equivalent to stating that the two instruments have strings of equal linear mass density. This situation represents a standing wave, thus we can relate the following equation for the first harmonic:
Where, 
is the length of the string and 
is the wavelength. Then we can use the following equation to relate wavelength and speed (which is known) to frequency:
Since the velocity of sound in a fixed medium is constant, we see that a shorter length, 
corresponds to a shorter wavelength, 
. Thus when 
decreases, frequency, 
must increase, to keep velocity constant.
The speed of sound in air is constant, assuming that the temperature of the air is constant. When the length of the string is shortened, by the principles of standing waves, this creates a higher frequencies. Assuming that the two instruments use the same strings is equivalent to stating that the two instruments have strings of equal linear mass density. This situation represents a standing wave, thus we can relate the following equation for the first harmonic:
Where,
Since the velocity of sound in a fixed medium is constant, we see that a shorter length,
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A sound played from a speaker is heard at an intensity of 100W from a distance of 5m. When the distance from the speaker is doubled, what intensity sound will be heard?
A sound played from a speaker is heard at an intensity of 100W from a distance of 5m. When the distance from the speaker is doubled, what intensity sound will be heard?
Intensity is related to radius by the inverse square law:
This equation is derived from the concept that the energy from the sound waves is conserved and spread out over an area, producing the 
term. Applying this concept, when the radius doubles, the intensity decreases by a factor of 4. The correct answer is 
.
Intensity is related to radius by the inverse square law:
This equation is derived from the concept that the energy from the sound waves is conserved and spread out over an area, producing the
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A stopped pipe (closed at both ends) sounds a frequency of 500Hz at its fundamental frequency. What is the length of the pipe?
A stopped pipe (closed at both ends) sounds a frequency of 500Hz at its fundamental frequency. What is the length of the pipe?
A stopped pipe can be modeled with the following equation:
Rearrange the equation to solve for L, then plug in given values and solve.
Rearrange the equation to solve for L, then plug in given values and solve.
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What is the beat frequency between a 305Hz and a 307Hz sound?
What is the beat frequency between a 305Hz and a 307Hz sound?
Frequency of beats is determined by the absolute value of the difference between two different frequencies. Thus, the beats frequency is 2Hz. Note that beat frequency is always a positive number.
Frequency of beats is determined by the absolute value of the difference between two different frequencies. Thus, the beats frequency is 2Hz. Note that beat frequency is always a positive number.
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An open pipe (open at both ends) has a fundamental frequency of 600Hz. How long is the pipe?
An open pipe (open at both ends) has a fundamental frequency of 600Hz. How long is the pipe?
An open pipe can be modeled by the following equation:
Rearrange the equation to solve for 
then plug in given values and solve.
An open pipe can be modeled by the following equation:
Rearrange the equation to solve for
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Consider a 37cm long harp string with a fundamental frequency of 440Hz.
Calculate the speed of the standing wave created by plucking this string.
Consider a 37cm long harp string with a fundamental frequency of 440Hz.
Calculate the speed of the standing wave created by plucking this string.
Use the following equation to find the velocity of the wave, using its fundamental frequency and the length:
Use the following equation to find the velocity of the wave, using its fundamental frequency and the length:
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Consider a 37cm long harp string with a fundamental frequency of 440Hz.
Suppose the string is pressed down in such a way that only a 10cm length of string vibrates. What is the speed of the wave produced when the string is plucked in terms of 
the speed of the wave when all 37cm of the string vibrate?
Consider a 37cm long harp string with a fundamental frequency of 440Hz.
Suppose the string is pressed down in such a way that only a 10cm length of string vibrates. What is the speed of the wave produced when the string is plucked in terms of
The speed of a wave is a property of the medium, it is not affected by the length of the string. The frequency may change, but the speed remains constant. There is no change in the speed.
The speed of a wave is a property of the medium, it is not affected by the length of the string. The frequency may change, but the speed remains constant. There is no change in the speed.
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Consider a 37cm long harp string with a fundamental frequency of 440Hz.
If only half of the string is allowed to vibrate, what frequency will be heard?
Consider a 37cm long harp string with a fundamental frequency of 440Hz.
If only half of the string is allowed to vibrate, what frequency will be heard?
Since the speed of the wave does not change based on the length of the string and we know it has a fundamental frequency of 440Hz, a string of half the length will vibrate at twice the frequency, 880Hz. This makes sense as it will sound higher in pitch. You can try this with a rubber band on a shoebox. Plucking it while placing your finger halfway along the band will result in a higher pitched sound.
Since the speed of the wave does not change based on the length of the string and we know it has a fundamental frequency of 440Hz, a string of half the length will vibrate at twice the frequency, 880Hz. This makes sense as it will sound higher in pitch. You can try this with a rubber band on a shoebox. Plucking it while placing your finger halfway along the band will result in a higher pitched sound.
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Consider a 37cm long harp string with a fundamental frequency of 440Hz.
What is the wavelength of the second harmonic of the string?
Consider a 37cm long harp string with a fundamental frequency of 440Hz.
What is the wavelength of the second harmonic of the string?
The wavelength of the second harmonic of a standing wave on a string is just the length of the string. For the second harmonic, an entire cycle occurs on the length of the string. Therefore, the wavelength of the second harmonic for this string is 37cm or 0.37m. The wavelength for the first harmonic, or fundamental, is twice the length of the string, as this is when one half a cycle occurs over the length of the string.
The wavelength of the second harmonic of a standing wave on a string is just the length of the string. For the second harmonic, an entire cycle occurs on the length of the string. Therefore, the wavelength of the second harmonic for this string is 37cm or 0.37m. The wavelength for the first harmonic, or fundamental, is twice the length of the string, as this is when one half a cycle occurs over the length of the string.
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If the first harmonic of a string has frequency 
, what is the frequency of the 
harmonic of that string in terms of 
?
If the first harmonic of a string has frequency
Harmonics describe the relationship between wavelengths and frequency on a string. A given string will have a wave speed associated with it. Given this wave speed, the harmonics are the frequencies at which half-wavelengths occur along the length of the string. For instance, the first harmonic is the frequency at which a half-wavelength occurs over the string. The second harmonic completes one wave over the string. The string is 1.5 times the wavelength of the third harmonic and so on. Therefore, the 
harmonic has a frequency of 
.
Harmonics describe the relationship between wavelengths and frequency on a string. A given string will have a wave speed associated with it. Given this wave speed, the harmonics are the frequencies at which half-wavelengths occur along the length of the string. For instance, the first harmonic is the frequency at which a half-wavelength occurs over the string. The second harmonic completes one wave over the string. The string is 1.5 times the wavelength of the third harmonic and so on. Therefore, the
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A student at a concert notices that a balloon near the large speakers moving slightly towards, then away from the speaker during the low-frequency passages. The student explains this phenomenon by noting that the waves of sound in air are __________ waves.
A student at a concert notices that a balloon near the large speakers moving slightly towards, then away from the speaker during the low-frequency passages. The student explains this phenomenon by noting that the waves of sound in air are __________ waves.
Sound is a longitudinal, or compression wave. A region of slightly more compressed air is followed by a region of slightly less compressed air (called a rarefaction). When the compressed air is behind the balloon, it pushes it forward, and when it is in front of the balloon, it pushes it back. This only works if the frequency is low, because the waves are long enough so that the balloon can react to them.
Sound is a longitudinal, or compression wave. A region of slightly more compressed air is followed by a region of slightly less compressed air (called a rarefaction). When the compressed air is behind the balloon, it pushes it forward, and when it is in front of the balloon, it pushes it back. This only works if the frequency is low, because the waves are long enough so that the balloon can react to them.
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A guitar player hits a wrong note. The note he's supposed to hit is 
, but he's sharp and the note turns out to be 
. Starting from rest, how fast (and in which direction) do you have to run until the note sounds correct to you?
A guitar player hits a wrong note. The note he's supposed to hit is
In order to find the velocity needed, we use the Doppler equation:
Where 
is the speed of sound in air, 
is the velocity of the observer, and 
is the velocity of the source, which is zero in this case since the guitar player is not moving. Since the guitarist is sharp, your adjustment must decrease the perceived frequency by 
, which means we use the negative sign in the numerator of the Doppler equation. Plug in the given values into the equation and solve for 
.
Because we need the note to be the final note to be lower in frequency than the original frequency, we need to run away from it.
In order to find the velocity needed, we use the Doppler equation:
Where
Because we need the note to be the final note to be lower in frequency than the original frequency, we need to run away from it.
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Given that 20kJ of energy are hitting a window pane over a period of 5s with dimensions 2.5m by 4m, what is the sound level in decibels?
Given that 20kJ of energy are hitting a window pane over a period of 5s with dimensions 2.5m by 4m, what is the sound level in decibels?
The formula for intensity, 
is:
Where 
is power in watts and 
is the surface area. The surface area 
in our case is:
The power 
can be given as 
or in our case:
Solve for intensity.
The formula for sound level 
is:
, where 
is the intensity and 
is the threshold of hearing, which is 
The formula for intensity,
Where
The power
Solve for intensity.
The formula for sound level
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At a distance of 
from a fan exerting 
of mechanical energy, estimate the sound level if the threshold of hearing is 
At a distance of
First we need to solve for intensity 
, given by:
, where 
is power, 
is the distance from the source of the sound.
In our case, 
and 
, therefore
To solve for sound level 
, we do
, where
is the intensity and 
is the threshold of hearing.
In this problem,
, and 
First we need to solve for intensity
In our case,
To solve for sound level
In this problem,
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By what factor will the sound level in decibels change if the intensity is increased by a factor of 
?
By what factor will the sound level in decibels change if the intensity is increased by a factor of
Recall that the formula for sound level 
given in decibels is given by:
, where 
is the intensity and 
is the threshold of hearing.
Sound level is proportional to intensity by:
If the intensity is increased by a factor of 
, sound level would increase by a factor of 
Recall that the formula for sound level
Sound level is proportional to intensity by:
If the intensity is increased by a factor of
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The speed of sounds is the fastest in which of the following media?
The speed of sounds is the fastest in which of the following media?
The speed of sound is fastest in the least compressible media of the lowest density. Sound does not propagate in a vacuum. Air and water are compressible media, so sound does not travel as fast in these as it does in glass, an incompressible medium. In general, the speed of sound is greatest in solids, and within each phase, faster as density decreases.
The speed of sound is fastest in the least compressible media of the lowest density. Sound does not propagate in a vacuum. Air and water are compressible media, so sound does not travel as fast in these as it does in glass, an incompressible medium. In general, the speed of sound is greatest in solids, and within each phase, faster as density decreases.
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