- Good morning.
Today we are going to solve
free response question number five
from the 2015 AP Physics 1 exam.
♫ Flipping Physics ♫
We have a string.
On the left to the string
is attached to a frequency oscillator.
On the right it is attached
to a block with a mass,
Capital M.
The block and string go over
a massless and frictionless pulley
and the oscillator is set to oscillate
at a frequency which will set up
a standing wave in the string.
Specifically the fundamental frequency.
There are actually four different similar lab set ups,
each with a different string and each oscillating
at a different fundamental frequency.
We are given an equation for the velocity
of a wave on a string.
The velocity of a wave on a string
equals the square root of
F sub T: The force of tension in the string divided by
the mass per unit length of the string
or what is also called
the linear mass density of the string.
And please do not confuse this lowercase m
in the mass per unit length of the string
which represents the mass of the string
with this capital M which represents the mass of the block
hanging from the right end of each string.
And this capital M, the mass of the block
is the same for each string.
Billy, could you please read part A.
- Sure. Part A...
- Hold up. I don't remember that equation
for the velocity of a wave on a string.
Have we learned that before?
- Bo, you are correct.
We have actually not seen this equation before
but it's not all that unusual for the AP test writers
to put in a topic that you haven't seen before
or an equation you haven't seen before
and expect you to be able to use
your knowledge of the AP physics 1 topics
to go beyond the curriculum a little bit.
Billy, please?
- Part A: What is different about
the four strings shown above that would result
in their having different fundamental frequencies?
Explain how you arrived at your answer.
- Bobby, could you please answer the question?
- Okay. We know the velocity of a wave equals
its frequency times its wavelength
and each of the four strings has the same length Capital L
and is vibrating at its fundamental frequency
therefore the wavelength for each string is the same.
Each string is vibrating at a different frequency
therefore the velocity of the wave
on each string must be different.
Because the mass attached to the end
of each string, capital M, is the same
the force of tension will also be the same
therefore using the equation they gave us
velocity equals the square root of the quantity
force of tension divided by linear mass density
in order for the velocity to be different for each string
the mass per unit length must be different for each string.
- Very nice, Bobby.
Billy, could you please read part B.
- A student graphs frequency as a function
of the inverse of the linear mass density.
Will the graph be linear?
Explain how you arrived at your answer.
- Go ahead, Billy.
- Okay, frequency is on the y axis
and one over mass per unit length is on the x axis.
We can combine equations from part a
and solve for the relationship
between the x and y axis variables.
Velocity equals frequency times wavelength
and velocity also equals the square root
of the force of tension divided by the linear mass density.
Let's square the whole equation
and then divide by wavelength squared.
We now have frequency squared equals the quanity
force of tension divided by wavelength squared
times the inverse of linear mass density.
The slope intercept form of a line is y equals m, or slope
times x plus b, the y intercept, or b is zero
and in order to get a linear relationship
we would need a frequency squared on the y axis
not frequency.
So no, we should not expect
a linear relationship when graphing frequency
as a function of the inverse of the linear mass density.
- And Billy, please read part C.
- Part C: The frequency of the oscillator
connected to string D is changed
so that the string vibrates in its second harmonic.
On the side view of the string below
mark and label the points on the string
that have the greatest average vertical speed.
- Bo, part C is yours.
- Okay, well, because antinodes are the location
on a standing wave with the greatest maximum amplitude
and every point on the string takes the same amount of time
to go through one full cycle, antinodes are locations
of greatest average vertical speed.
At its first harmonic, or the fundamental frequency,
the string has two nodes, one on either end,
and one antinode in the middle
halfway between the two nodes.
The second harmonic increases both of those numbers by two
so we now have three nodes.
One node on either end and one node in the middle.
And two antinodes, each antinode located
halfway between nodes.
So the two locations of greatest average vertical speed
are at the two antinodes.
One fourth of L towards the middle from both ends.
- That is it.
I hope you enjoyed my solutions
to the 2015 AP physics 1 exam
You could also enjoy my review videos.
I have review videos which cover every AP Physics 1 topic.
That would be in the link which just popped up.
Thank you very much for learning with me today.
I enjoyed learning with you.