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Definition
| A wave with a direction of propagation that is perpendicular to its direction of oscillation |
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Definition
| A wave with a direction of propagation that is parallel to its direction of oscillation |
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Definition
| Any speed that is faster than the speed of sound in the substance of interest |
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| The sound produced as a result of an object traveling at or above Mach I |
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| An indication of how high or low a sound is, which is primarily determined by the frequency of the sound wave |
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Sound waves with low pitch have ______ frequency
Sound waves with high pitch have _______ frequency |
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Definition
Sound waves with low pitch have low frequency
Sound waves with high pitch have high frequency |
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Hint for test:
An instrument with a long tube will produce ____ pitch sounds because there is _____ wavelengths |
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Definition
| An instrument with a long tube will produce lower pitch sounds because there is longer wavelengths |
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| If an alarm clock is in a container with no air, could you hear it? ( do not count the container as a medium for the sound waves to travel) |
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Definition
| No, sound waves need air to oscillate |
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| Sound waves oscillate ________ to the direction in which the sound wave travels. |
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Definition
| Sound waves oscillate PARALLEL to the direction in which the sound wave travels. |
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Term
Know how to use this formula in a problem:
v= (331.5 + 0.6T)m/s
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Definition
V= speed of sound
T = temperature
m/s is the measurement of meters per second |
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Using this formula : v= (331.5 + 0.6T)m/s
Answer this problem
How fast is the speed of sound when the temperature is 23°C?
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Definition
v=(331.5 + 0.6 x 23)m/s
v= (331.5 + 13.8) m/s
V= 345.3m/s
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You will be asked a problem similar to this one so study it well.
A sound wave traveling through 20°C air has a wavelength of 4 meters. What is the frequency of the sound wave? |
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Definition
You need to formulas to answer this question
First use this formula: v= (331.5 + 0.6T)m/s
You need to determine the speed of sound first.
v= (331.5 + 0.6x 20)
v= 343.5
Now you need to use the second formula:
f = v/η
f = frequencyt
v= speed
η = wavelength
You would set it up like this:
f= 343.5 / 4
f= 85.875
Your final answer is 85.875 frequency |
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| Know which sound waves are longer or shorter than other sound waves |
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Definition
Infrasonic is the lowest
Sonic waves are in the middle
Infrasonic waves are the longest
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You will have a problem similar to this one on the test so study this well
During a thunderstorm, the temperature is 22°C. You see a strike of lightning and then hear the thunder 3 seconds later. How do figure how far away did the lightning strike? |
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Definition
You need 2 formulas to determine the answer.
The first formula is: v= (331.5 + 0.6T) m/s
This formula will determine the speed of sound.
v=(331.5 + 0.6 x 20)m/s
v=343.5
Now you need the second formula:
distance traveled = speed x time traveled
Now plug in your numbers:
distance traveled =343.5 x 4
distance traveled = 1374 meters
The final is 1374 meters away was the lightning strike |
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Sound waves travel fastest through _____
Sound waves travel faster through ________ than air
Sound waves travel through _____ the slowest. |
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Definition
Sound waves travel fastest through SOLIDS
Sound waves travel faster through LIQUIDS than air
Sound waves travel through AIR the slowest. |
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Term
A boy and girl are singing at church but the boy is not singing as loud as the girl. The boy is also singing the low notes and the girl is singing the higher notes
Are the frequencies of the boy's sound waves lower or higher than those of the girl?
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Definition
Since the boy is singing the lower notes, the boy's sound waves have lower frequencies and longer wavelengths.
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A boy and girl are singing at church but the boy is not singing as loud as the girl. The boy is also singing the low notes and the girl is singing the higher notes
Are the wavelength's of the boy's sound waves longer than or shorter than those of the girl? |
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Definition
Since the boy is singing the lower notes, the boy's sound waves have lower frequencies and longer wavelengths.
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A boy and girl are singing at church but the boy is not singing as loud as the girl. The boy is also singing the low notes and the girl is singing the higher notes
Are the boy's sound waves smaller or larger amplitude than those of the girl? |
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Definition
| It is smaller because the boy's voice is quieter than the girl's voice. |
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A boy and girl are singing at church but the boy is not singing as loud as the girl. The boy is also singing the low notes and the girl is singing the higher notes
Will the speed of sound of the man's voice be faster than that of the woman's sound waves. |
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Definition
Of course not, the speed of sound depends only the medium it travels through and the temperature of the air.
Both the boy and girl's voice travel through the same air so the speed of sound will not effect if their voice travels faster than the other one. |
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| A jet travels through the air at 15°C at Mach 2. What is the speed in meters per second? |
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Definition
You need to do 2 things to answer this question.
First, figure the speed of sound and put 15°C into the formula below:
v=(331.5 +0.6T)m/s
v=(331.5 +0.6 x 15)m/s
v= 340.5
Now you have the speed of sound but the jet is traveling at Mach 2 so you need to multiply it by 2 to get the final answer
2 x 340.5
The final is 681m/s
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Term
| You are driving in your car, and you hear an ice cream truck but the sound is stuck and it keeps on the same note. As you speed up, you notice the pitch of the stuck ice cream truck sound keeps getting higher. Are you driving toward or away from the ice cream truck? (Assume the true pitch stays constant) |
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Definition
| Since the pitch of the stuck ice cream truck sound is getting higher, you are moving so that the wavelength of the sound seems smaller. This means you are moving toward the ice cream truck, because that will cause you to encounter the crests of the sound waves faster than you would if were still. |
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There will be an extra credit question like this one
An amplifer takes a 20 decibel sound and turns it into a 50 decibel sound.
How many times larger is the intensity of the sound waves coming out of the amplifer as compared to the intensity of the sound waves going into the amplifer |
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Definition
One bel = 10 decibels
Convert the decibels into bels
20 decibels 1 bel
1 x 10 decibels = 2 bels
50 decibels 1 bel
1 x 10 decibels = 5 bels
Since the sound is 3 bels lounder after the amplifer, the increase in sound waves is 3 factors of ten higher.
Thus, the sound waves intenisty is:
10 x 10 x 10 = 1000 times after the amplifer as compared to before.
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Term
Know the parts of the sound wave:
crests
troughs
wavelengths
amplitude
frequency |
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Definition
Crests: highest point of the wave
Troughs: Lowest point of the wave
Wavelength: the distance between the crests
Amplitude: the height of the waves
Frequency: how many waves hit a certain point every second. |
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