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