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# AP Physics 1 review of Waves and Harmonic motion

## Video transcript

- [Lecturer] Hooke's Law
tells you how to find the force exerted by an
ideal or linear spring. And it's a simple law. It tells you the amount
of force that spring is gonna exert will be
proportional to the amount that spring has been
stretched or compressed from its equilibrium or natural length. Which, in equation form
just says that the magnitude of the spring force, is gonna
equal the spring constant multiplied by the amount the
spring has been stretched or compressed. Note this x is not the
length of the spring. The x is how much that spring
has been stretched from or compressed from the
equilibrium position or the unstretched position. So what's an example problem involving Hooke's law look like? Let's say an ideal spring is hanging from the ceiling at rest, and it has an unstretched length L1. And then you hang a mass
M from the spring at rest and it stretches the
spring to a length L2. What is an expression for the
spring constant of the spring? So the force of gravity has to be balanced by the spring force. That means the magnitude
of the spring force is equal to the magnitude
of the force of gravity. The spring force is always K times X. What is X gonna represent? It's not gonna be L1 or L2. X is how much the spring
has been stretched from its equilibrium position
which would be L2 minus L1. And if we solve for K we
get mg over L2 minus L1. What's a simple harmonic oscillator? A simple harmonic oscillator's
any variable who's change can be described by a
sine or cosine function. What does that function look like? It looks like this. So the variable that's
changing is a function of time, which could be the vertical
position of a mass on a spring. The angle of a pendulum or any other simple harmonic oscillator
is gonna be equal to the amplitude of the motion, which is the maximum
displacement from equilibrium, times either sine or cosine of two pie times the frequency of the motion times the variable t. Which, since frequency
is one over the period, you could write as two pie over the period times the time variable t. How do you know whether
you use sine or cosine? Well sine starts at zero and goes up. And cosine starts at a
maximum and goes down. So if you know the behavior of your oscillator at t equals zero. You can decide whether
to use cosine or sine. Something that's important
to know is how to find the period of an oscillator. The period of a mass
on a spring is gonna be two pie times the square
root, the mass connected to the spring divided by
the spring constant k. Note that this does not
depend on the amplitude. If you stretch that mass
farther, it'll go faster and it has farther to go which cancels out and the period remains the same. And the formula for the
period of a pendulum, which is a mass swinging on a string, is gonna be two pie times the square root the length of the string,
divided by the magnitude of the acceleration due to gravity. Which also does not
depend on the amplitude, as long as the angles are small. And it doesn't depend on the mass either. How do you find this period on a graph? Well if you're given
the graph of the motion of a simple harmonic oscillator
as a function of time, the interval between peaks is
gonna represent the period. Or the time is takes for
this oscillator to reset. So what's an example problem involving simple harmonic motion look like? Let's say in a lab a mass M on Earth can either be hung on a string of length L and allowed to swing back and forward with a period T pendulum, or
hung on a spring of spring constant k and allowed
to oscillate up and down with a period T spring. If a 2M mass were used
instead of the 1M mass, what would happen to the
period of the two motions? Well the period of a pendulum
doesn't depend on the mass. And so the period of the
pendulum would not change. The answer would have to be D. What are waves? Waves are disturbances that
travel through a medium and transfer energy in momentum
over significant distances without transmitting any mass
itself over those distances. What does medium mean? This is a fancy word for
the material through which the wave can travel. So you can classify a wave
by the medium it's in. But you could also
classify waves by the type of disturbance you've created. For transverse waves, the
disturbance of the medium is perpendicular to the wave velocity. By wave velocity, we mean
the direction in which the disturbance travels. And by oscillation of the medium, we mean the direction in which
the particles of the medium actually move. For a wave on a string, the
particles move up and down but the disturbance travels to the right. So this would be a transverse wave. For longitudinal waves, the
oscillation or disturbance of the medium is parallel
to the wave velocity. The classic longitudinal wave is sound. If a sound wave were traveling
rightward through the air, it would look like a compressed region, and the air itself would
move back and forth, right and left, parallel to the direction the wave disturbance is traveling. Which makes sound waves longitudinal. For every type of wave, the
speed of that wave disturbance is gonna be equal to the
wavelength of the wave divided by the period. In other words, if you
watched a wave crest, that wave crest would move
one wavelength every period. And since the speed is distance per time, the speed of the wave crest
would be one wavelength per period. You could find the wavelength
on a graph of y versus x by finding the distance between crests. And if you're wondering
why this doesn't represent the period? It's because this is a
graph of the wave versus x, versus the horizontal
position, not the time. You could make a graph
of y versus the time. And what that would
represent is the motion of a single point on the
wave for all moments in time. And for this graph versus time, the interval between peaks is the period. So if you get a graph of a wave, you've gotta check whether
it's versus x or versus t. If it's versus x, peak to
peak is the wavelength. And if it's versus t, peak
to peak is the period. And since one over the period
is equal to the frequency, we can rewrite this speed formula as the speed of a wave equals
the wavelength of a wave times the frequency. And the way it's given
on the formula sheet on the AP exam is that
the wavelength of a wave is equal to the speed of the
wave divided by the frequency. Now this formula confuses
a lot of people though, because they think if you
increase the frequency, that'll increase the speed of the wave. But that's not true. Increasing the frequency
will cause the wavelength to decrease and the speed of
the wave will remain constant. The only way to change the speed of a wave is to change the properties
of the medium itself. In other words the only
way to change the speed of the waves on water would
be to change something about the water itself,
its density, its salinity, the temperature of that water. Changing frequency isn't gonna
change the speed of the wave. And neither will changing the amplitude. The only thing that changes
the speed of the wave is changes to the medium itself. And so what's an example problem
involving waves look like? Let's say a sound lab is
being conducted in a lab room with total cubic volume
V and temperature T. A speaker in the room is hooked
up to a function generator and it plays a note with
frequency f and amplitude A. Which of the following
would change the speed of the sound waves? Increasing the frequency would just make that sound a higher note. But it wouldn't change the
speed of the sound wave. Increasing the temperature of the room is a change to the medium
itself so this would change the speed of sound. Decreasing the amplitude's
just gonna make the sound seem softer and not appear as loud. And decreasing the total volume
of the space in the lab room doesn't actually effect the medium, it just gives you less of the medium. So the best answer here would be B. The doppler effect refers to
the change in the perceived frequency when a speaker or a wave source, moves relative to the observer. If a wave source and observer
are moving toward each other, the wavelength of the sound
wave is gonna decrease according to that observer. Which would make the
perceived frequency increase. This happens since if the
source is heading toward you, as that speaker emits wave pulses, the speaker moves toward the
wave pulse it just emitted and on this leading edge
the crests of the wave will be closer together. Since they're closer together,
the wavelength is smaller. And the rate at which
these crests are gonna hit the observer is gonna be higher. So this observer's gonna
hear a higher frequency than is actually being
played by the wave source when it is at rest. And for the observer on the trailing edge, since the wave source is
moving away from the pulses it just sent in this direction, these wave crests are gonna
be spaced further apart, which increases the
wavelength and decreases the rate at which these crests
are gonna hit the observer. So this observer's gonna hear a frequency that's less than the actual
frequency being played by the source when it's at rest. So what's an example problem involving the Doppler effect look like? Let's say the driver of a
car sees that they're heading straight toward a person
standing still in the crosswalk. So the driver continuously
honks their horn and emits a sound of frequency f horn. This is the frequency the horn plays when the car would be at rest. And the driver of the
car also simultaneously slams on the brakes and skids to a stop right in front of the person
standing in the crosswalk. What would that person in the crosswalk hear as the car's skidding to a stop? Well since that car is
heading toward the person, those wave crests are gonna
be spaced closer together, so this person will hear
a smaller wavelength and a higher frequency. But as the car slows down,
this effect becomes less and less dramatic. And once the car stops,
the wave crests will be spaced out their normal spacing, and the person will just
hear the regular horn frequency of the car. So first this person's gonna
hear a higher frequency, but that will eventually
just become the actual frequency of the horn once the car stops and there's no more
relative motion between the person and the car. When two waves overlap in the same medium, we call it wave interference,
or wave superposition. While those waves are overlapping, they'll combine to form a wave shape that will be the sum of the two waves. In other words, while the
two waves are overlapping, to find the value of the total wave, you just add up the values
of the individual waves. So if you overlap two
waves that look identical, they would combine to form
a wave that's twice as big. And we call this
constructive interference. And if you overlap two
waves that are 180 degrees out of phase, they'll combine
to form no wave at all. We call this destructive interference. So even though while the
waves are overlapping, you add up their individual
values to get the total wave. After these waves are done overlapping, they'll pass right through each other. So if I sent a wave pulse down a string at another wave pulse, when these two pulses overlap, the string would be flat
but shortly after that, the wave pulses would continue
on their way unaffected. They don't bounce off of each other, or cause permanent damage. It's only while they're overlapping that you'll get the wave interference. So what's an example
problem involving wave interference look like? Let's say two wave pulses on a string head toward each other as seen
in this diagram to the right and we wanna know what would
be the shape of the wave when the wave pulses overlap? So to find the total wave
we'll add up the values of each individual wave. The blue wave's gonna
move to to the right, and the red wave's gonna move to the left. And we'll add up the individual values. Zero of the red wave
plus negative two units of the blue wave will add
up to negative two units for the total wave. And then positive two
units of the red wave plus negative two units of the blue wave is gonna equal zero
units for the total wave. Again positive two units of the red wave plus negative two units of the blue wave add up to zero units for the total wave. And then zero units for the red wave plus negative two units for the blue wave is gonna equal negative two
units for the total wave. So our total wave will look like this. Since this pyramid just
took a bite out of this blue rectangular wave. How do you deal with
standing waves on strings? Well to get a standing wave at all, you need waves overlapping that are going in opposite directions. But even if you have that,
you're not necessarily gonna get a standing wave. Only certain allowed
wavelengths will create a standing wave in that medium. And what determines the
allowed wave lengths is the length of the
medium and the boundaries of that medium. In other words the ends
of a string could be fixed or loose, if the end
of the string is fixed, it's gonna be a displacement node. Node is the word we use to refer to points that have no displacement. And if the end of a string is loose, that end would act as a
displacement anti node. Anti node being a point that
has maximum displacement. What would these standing waves look like? Standing waves no longer appear
to move along the medium, they just oscillate
back and forth in place. So this anti node would move
from the top to the bottom back to the top. But you wouldn't see this
crest move right or left, hence the name standing wave. However the nodes stay put. There's never any displacement at a node. Once you've determined
the boundary conditions and the length of that medium, the possible wavelengths are set. Because the only allowed standing waves have to start at a node and end at a node. The fundamental standing
wave refers to the largest possible wavelength standing wave. And in this case it would
be one half of a wavelength. So the length of the medium
would have to equal one half of the wavelength of the wave. Similarly for the second harmonic, we still have to start and end at a node, so the next possibility
would be an entire wavelength which means the length of
the medium would equal one wavelength of the wave. And if you get to the third harmonic, this is three halves of a wavelength. And there's no limit,
you can keep going here. To excite these higher standing waves, you gotta keep increasing the frequency, because you'll keep
decreasing the wavelength of the distance between peaks. So what would a standing wave look like if one of the ends were loose? In that case that end
would be an anti node, and the fundamental standing
wave would only take the shape of a fourth of a wavelength. Since it has to go from
a node to an anti node. That means the length of the
string would have to equal one fourth of a wavelength. The next possible standing wave would be three fourths of a wavelength. And the next possibility
would be five fourths of a wavelength. And again this progression keeps on going. So what's an example problem
involving standing waves on strings look like? Let's say one end of a string of length L is attached to the wall
and the other end is fixed to a vibrating rod. A student finds that the
string sets up a standing wave as seen here when the frequency of the rod is set to f nought. What are the speed of
the waves on the string? Well we know the string length is L. And we can figure out how
much of a wavelength this is. From here to here would be one wavelength. And there's another half. So the string length L
is equalling three halves of a wavelength or in other
words the wavelength here is 2L over three. And we know the speed
of the wave is always wavelength times frequency. So the speed of the wave here
is gonna be 2L over three, which is the wavelength of the wave, times the frequency and
so the best answer is D. How do you deal with
standing waves in tubes? So just like standing waves in strings, the wavelength of the
standing waves in a tube are determined by the length of the tube, and the boundary conditions of that tube. But this time instead
of the sting shaking, you're creating a standing
wave out of sound waves. Now for the boundary
conditions of this tube, an open end is gonna act like
a displacement anti node. Since the air at an open
end can oscillate wildly. And you get maximum air disturbance. But a closed end of a
tube is gonna act like a displacement node since
there'll be no air disturbance at a closed end. So what if both ends of the tube were open giving us anti node anti
node standing waves? Well the first possibility
of the largest wavelength standing wave would go from
anti node to anti node. And this is one half of a wavelength. So the length of this
tube would have to equal half of a wavelength. The next possibility would
still go from anti node to anti node. And this is equal to one whole wavelength. It might not look like is
but from valley to valley, is a whole wavelength. So the length of this tube
would equal one wavelength. And the next possibility
would equal three halves of a wavelength. And you should note this
is the same progression that we had for node node strings. So whether both ends are anti nodes, or both ends are nodes, if both boundary conditions are the same, you get this same progression that goes half of a wavelength,
one whole wavelength, three halves of a wavelength. Basically any integer or
half integer wavelength. And what if we closed one
of the ends of this tube? If we closed one end of the tube, that end would become a displacement node, since the air can't move at that position. Which would make it a node
and it would have to go to the open end which is an anti node. So the largest possibility this time would be one fourth of a wavelength. The next possibility would
still go from node to anti node and this would be three
fourths of a wavelength. And if you notice, this
is exactly the same as when we had node anti node strings. We had the same progression
of lambda over four, three lamba over four,
five lambda over four, any odd integer lambda over four, were the allowed standing wavelengths. So if one end has a
different boundary condition from the other end this is
gonna be the progression of allowed lengths of the medium. So what would an example
problem involving standing waves in tubes look like? Let's say you blow over the
top of a tube that's open at both ends and it resonates
with a frequency f nought. If the bottom of the tube is then covered and air is again blown
over the top of the tube, what frequency would be heard
relative to the frequency heard when both ends were open? Well when both ends are open, we know the standing wave's
gonna be an anti node to anti node. Which is one half of a wavelength. Which would equal the length of that tube. So the wavelength would
be two times the length of the tube. But when we close one of the ends, we turn that end from an
anti node into a node. So we'd have to go from
an anti node to a node. Which is only one fourth of a wavelength. So one fourth of a wavelength would equal the length of the tube. And that means lambda equals
4L, this wavelength doubled. So what would that do to the frequency? Well we know V equals lambda f, and we didn't change the medium here so the speed is gonna remain the same. So if we double the wavelength, we'd have to cut the frequency in half in order to maintain the
same speed of the wave. So when we close the bottom of this tube, we'd hear half the frequency we heard when both ends were open. Beat frequency refers to the phenomenon where two waves overlap
with different frequencies. When this occurs, the
interference of the waves at a point in space
turns from constructive, to destructive back to
constructive and so on. Which if this were a sound wave, you'd perceive as a wobble
in the loudness of the sound. And the reason this happens
is that if these waves started in phase and
they were constructive, since they have different frequencies, one wave would start to become
out of phase with the other. Eventually becoming destructive,
which would be soft. But if you wait longer, one of these peaks catches up to the next
peak in the progression, and the waves again become constructive which would be loud again. And the times this takes
to go from loud to soft to loud again is called the beat period. But more often you'll hear
about the beat frequency. Which is just one over the beat period. So the beat period is the
time it takes to go from loud to soft back to loud. And the beat frequency is the number of times it does that per second. How do you determine the beat
frequency or the beat period? Well the formula used to
find the beat frequency is actually really simple. You just take the difference
of the frequencies of the two waves that are overlapping. If there is no difference, if these waves have the same frequency, you'd have a beat frequency of zero which would mean you
hear no wobbles at all. The further apart these
two frequencies get, the more wobbles you
would hear per second. And then to find the beat period, you could take one over
the beat frequency. So what would an example problem involving beat frequency look like? Let's say these two waves were overlapping and we want to determine
the beat frequency. The period of the first
wave is four seconds. That means the frequency of the
first wave is one over four, or 0.25 hertz. And the period of the
second wave is two seconds, which means that the frequency
is one over two or 0.5 hertz. To get the beat frequency
you subtract one frequency from the other. 0.5 minus 0.25 would be 0.25 hertz.