This course is designed for high school students preparing to take the AP* Physics 1 Exam. * AP Physics 1 is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product.

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From the course by University of Houston System

Preparing for the AP Physics 1 Exam

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This course is designed for high school students preparing to take the AP* Physics 1 Exam. * AP Physics 1 is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product.

From the lesson

Waves and Sound

Topics include mechanical waves, sound, simple harmonic motion and applications of topics. You will watch 3 videos, complete 3 sets of practice problems and take 2 quizzes. Please click on the resources tab for each video to view the corresponding student handout.

- Dr. Paige K. EvansClinical Associate Professor

Math - Mariam ManuelInstructional Assistant Professor

University of Houston

Sound is a mechanical wave that is longitudinal.

Â To create a sound, a vibrating object moves air molecules back and

Â forth in the direction of the wave's velocity.

Â If a source or medium does not exist, there can be no sound wave.

Â Additionally, the speed of sound is not constant.

Â LIke all mechanical waves,

Â its speed depends on the medium through which it travels.

Â For instance, the speed of sound increases in air with increasing temperature,

Â as seen in the equation here.

Â Note that in this equation, T is the temperature of the air in Celsius.

Â When temperature is not given in a problem,

Â you can typically assume a room temperature of about 20 degrees Celsius,

Â so that the speed of sound is about 343 meters per second.

Â Okay, so

Â the speed of a wave depends entirely upon its medium, not the frequency of the wave.

Â And so here I've got a 440 hertz tuning fork, and it's struck on a hot day that's

Â 38 degrees Celsius or so, and it wants us to calculate the wavelength of sound.

Â Well, what immediately occurs to me is our wave equation.

Â That the velocity equals frequency times wavelength.

Â And if I can find out the speed, which we can probably do because we're given

Â temperature, and we know the frequency, we can solve for the wavelength.

Â So let's go ahead down here and solve for the velocity first.

Â We were told earlier the velocity of sound in air wave, would follow this function.

Â And that the higher the temperature goes, the faster those air molecules move,

Â and the faster sound vibrates.

Â So let's go ahead and sub in the numbers they gave us then.

Â It has to be in Celsius, which they already did give us, so

Â we don't have to worry about any conversions.

Â 38 degrees Celsius yields for me 353.8 meters per second.

Â That's the speed of sound on a day like that.

Â The frequency was 440 Hz.

Â We have that, we can sub both values in.

Â Notice that the speed of sound we got is faster than what

Â a typical day would be, like 343.

Â So that also seems to fit our expectations, and that's always a good way

Â to check to make sure you're not going off in the wrong direction.

Â So the velocity, 353.8, the frequency, 440.

Â Solve for the wavelength.

Â The wavelength of this sound will be 0.804 meters.

Â Notice that the wavelength of sound is about a meter long.

Â These aren't incredibly small or incredibly large, and that's also a good

Â way to kind of get a good feel to check your answers when it comes to sound.

Â >> Imagine a police car is driving by you at high speed with its siren on.

Â You may have noticed that the pitch it produces changes as it passes you by.

Â This occurs because the waves produced by the moving cars are no longer symmetrical.

Â Suppose that a police siren is a note at 800 hertz.

Â This means 800 new wave crests are produced every second.

Â Since the police car is moving forward, the waves in front of the car will

Â get closer together as they are created than the waves behind the car.

Â An observer would hear a higher pitch when listening to the waves

Â as the car approaches, and a lower pitch as the car drives away.

Â The equation to determine the frequency heard by an observer can be seen here.

Â Note that the observer can also move relative to the car, and so

Â we must take into account.

Â Signs in this equation can be very tricky, and rules for these are summarized here.

Â To check your work, try to visualize whether a specific motion would increase

Â or decrease a frequency overall.

Â Here is an example in which we can apply the Doppler effect equation.

Â So in this Doppler effect question, they're telling me that the police car has

Â a siren with a frequency of 400 hertz.

Â This means that 400 hertz represents the actual frequency.

Â Because remember,

Â the d\Doppler effect allows us to solve for the apparent frequency.

Â So 400's the actual.

Â They tell me that this police car is moving with a velocity

Â of 5 meters per second, or a speed of 5 meters per second.

Â So here I've got my cop car moving at 5 meters per secondd,

Â and for some reason, this criminal seems to be moving towards the police

Â at a speed of 2 meters per second.

Â Okay, maybe they want to get caught.

Â So I've got a speed of 2 meters per second.

Â They want me to calculate the frequency as heard by the criminal.

Â So the criminal is my detector or observer.

Â So when I go ahead and write out my equation.

Â Which is apparent frequency equals original frequency

Â multiplied by the speed of sound in the air.

Â Plus or minus the speed of the observer.

Â Divided by the speed of sound plus or

Â minus the speed of the source.

Â Which our source in this case is the police car that has the siren that's

Â producing this sound.

Â So let's figure out whether my speeds over here are plus or minuses.

Â Okay, so let's go ahead and look at our speed for the observer.

Â My observer is moving with a speed of 2 meters per second, and

Â this criminal is moving towards the source.

Â So that is positive.

Â On the other hand, my source is moving 5 meters per second,

Â and is moving towards the observer or the detector, so that is negative.

Â So now I can go ahead and plug these values in.

Â So I have an original frequency of 400.

Â The speed for sound, unless they give you a temperature or another value

Â to work with, you use 343.

Â So I have plus 2.

Â And then over here I have 343- 5.

Â And when I solve for this, I

Â end up with an apparent frequency of 408 hertz.

Â That's greater than the original frequency, and

Â keep in mind that both the source and detector are moving towards each other.

Â So now let's look at the next part.

Â The next part says, suppose instead the criminal is running away from the police

Â at 2 meters per second.

Â Calculate the frequency heard by the criminal in this scenario.

Â So now I have the police that's still moving

Â towards our observer with 5 meters per second.

Â And our observer, the criminal,

Â is running away from the police rather than towards it at 2 meters per second.

Â So let's go ahead and consider these.

Â The speed of the observer is the 2 meters per second, but because this observer is

Â running away from the police, well running away from the police is bad and negative.

Â So I've got negative over here.

Â My observer is moving away from the source.

Â The source, the police is moving towards this criminal, chasing after him or her.

Â And so that is also negative.

Â The source is moving towards the observer.

Â So we go ahead and plug this into our equation.

Â Still have the same original frequency.

Â Still using the speed of sound as 343 meters per second.

Â The really tricky part with Doppler effect questions is just being very

Â careful about the signs, being careful about what goes into your numerator,

Â what goes into your denominator.

Â The observer's speed is what goes into the numerator,

Â the source's speed is what goes into the denominator.

Â So, when I go ahead and solve for this over here,

Â I end up with an impaired frequency of 403 hertz.

Â >> A vibrating string with a standing wave is the foundation of

Â every stringed instrument.

Â The lowest frequency at which a standing wave can exist in a string

Â is called the fundamental frequency, or first harmonic.

Â This occurs when there are two nodes

Â at either end at the points where the string is attached.

Â Notice that in this scenario,

Â the standing wave has a wavelength that is twice the length of the string.

Â If you increase the frequency of the wave, until another standing wave forms,

Â the next harmonic observed would have three nodes.

Â In this scenario, the wavelength is equal to the length of the string.

Â Note that the speed of the wave has not changed,

Â because the medium has not changed.

Â The frequency, however, has increased, and the wavelength has decreased.

Â This frequency is called the first overtone, or

Â the second harmonic, and is twice that of the fundamental frequency.

Â The equation seen here describes the resonant frequencies for

Â strings at any harmonic.

Â In this equation, n can equal 1, 2, 3, 4, and so on.

Â N represents the harmonic number, and L represents the length of the strength.

Â This equation can also represent standing waves inside of pipes,

Â which are open in both ends.

Â In an open pipe, however, it is now the air itself that is vibrating.

Â In this scenario, anti-nodes are present at the open ends of the tube,

Â because here, the air is free to vibrate.

Â The v in this equation represents the speed of sound in the medium.

Â In this case, either the speed of the sound in the string or

Â in the ear inside of a pipe.

Â A closed pipe is one in which one side of the pipe is closed.

Â This end of the pipe is a node, while the open end of the pipe is an antinode.

Â Notice the fundamental in this scenario is different from the previous case.

Â Only one-fourth of the wave is present in the tube.

Â The wavelength is four times the length of the tube.

Â The second harmonic in which twice the length of the wave would fit in the tube

Â is not possible, because a node would not exist at the open end of the pipe.

Â Instead, the next standing wave occurs when three-fourths of the wave

Â is present in the closed pipe.

Â This is the third harmonic.

Â To describe the natural frequencies of a closed pipe,

Â we use the equation seen here.

Â F = (nv) / (4L), where n = 1, 3, 5, 7.

Â Even harmonics do not exist, because the open end must be an antinode.

Â Let's look at an example.

Â So before I solve for any frequencies in this question, I first need to solve for

Â what the speed of sound is.

Â The way that I'm going to do that is,

Â I'm going to use my equation, Vsound = 331 + .6T and

Â in this case, the temperature is 24 degrees.

Â So my speed of sound is 331 + .6(24),

Â and it has to be Celsius, which this is.

Â So my speed of sound in this case is

Â 345.4 meters per second.

Â So now I'm going to solve for this question.

Â They want me to figure out what the fundamental frequency in the first three

Â overtones are, given that the length is 20 centimeters, or 0.2 meters.

Â And so I'm going to start with this one that's open, both ends open.

Â So, the length of this, I've got half of a wavelength as shown over here.

Â So I can go ahead and solve for

Â this by using my fundamental equals V over 2L.

Â Which would be 345.4

Â / 2 times 0.2.

Â Remember, I'm only solving for the first harmonic, or the fundamental.

Â And this gives me 863.5 hertz.

Â Now if I want to get the first overtone, well the first overtone.

Â Is our second harmonic, which is 2

Â times the fundamental, 2f1.

Â So if I do 2 times the 863.5,

Â that gives me 1727 hertz.

Â Then the second overtone is the third harmonic.

Â So that would be 3 times the fundamental.

Â That gives me 2590.5 hertz.

Â Finally, my third overtone is the fourth harmonic,

Â so it's 4 times the fundamental.

Â Notice how it's just 2 times the fundamental, 3 times, then 4 times.

Â And it would continue on this way.

Â And here is the frequency, 3454 hertz.

Â So, that's when I'm solving for an open pipe.

Â Now let's look at a closed one.

Â So it's closed at one end over here.

Â And so I've got half of a wavelength that's being represented over here for

Â this entire length.

Â So I can go ahead and solve for

Â it by using my fundamental equals

Â V over 4l, since l is represented

Â by one fourth of a wavelength.

Â So I have 345.4 divided

Â by 4 multiplied by .2, and

Â this gives me 431.75 hertz.

Â Now recall that with closed pipes, I'm going to use odd numbers

Â as I continue to solve for my overtones or my following harmonics.

Â So my first overtone, my second harmonic rather,

Â it's going to be 3 times the fundamental.

Â Which gives me 1295.25 hertz.

Â My second.

Â Overtone.

Â They're harmonic, so I used three.

Â Now I'm going to use five, the next odd number.

Â Gives me 2158.75 hertz,

Â and I'll solve for the third overtone.

Â I'm going to use 7, that's my next odd number.

Â And 3022.25 hertz is that answer.

Â And those are all of the answers that we needed to solve for.

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