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### Course: Physics archive > Unit 11

Lesson 2: AP Physics 1 free response questions 2015- Question 1a: 2015 AP Physics 1 free response
- Question 1b: 2015 AP Physics 1 free response
- Question 1c: 2015 AP Physics 1 free response
- Question 2ab: 2015 AP Physics 1 free response
- Question 2cd: 2015 AP Physics 1 free response
- Question 3a: 2015 AP Physics 1 free response
- Question 3b: 2015 AP Physics 1 free response
- Question 3c: 2015 AP Physics 1 free response
- Question 3d: 2015 AP Physics 1 free response
- Question 4: 2015 AP Physics 1 free response
- Question 5: 2015 AP Physics 1 free response

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# Question 5: 2015 AP Physics 1 free response

Fundamental frequencies (first harmonics) of strings.

## Want to join the conversation?

- Just curious, why would the end that is moving on the oscillator not be an antinode of the string? Where the string is attached would have to be moving up and down to generate the wave, so then wouldn't it be a standing wave with string length as a fourth of the wavelength?(3 votes)
- This wouldn't happen because the string is fixed to the block on the end (and the oscillator could act as a "hand"), so the wave acts as a fixed end reflection. Because of this, the reflection of the wave, when it gets to the edge of the table, will be on the opposite side than what it originally was on. If the wavelength was four times the string length, the standing wave would be open where the oscillator is (which is what I assume you were getting at with the oscillator being an antinode). The issue with this is that the reflection of the wave then goes back to the oscillator and will hit it when the oscillator is in it's equilibrium (middle) position (due to the relationship between period and wavelength), so it will act as another fixed end that will reflect the wave back on the other side of the string. A situation where the end of the object
**is**an antinode can be seen in closed pipes such as trumpets!(5 votes)

- At10:43, why can't the next frequency be a fraction higher ?(3 votes)
- Can anyone help derive the given equation of velocity of the standing wave, thank you so much!(3 votes)
- Didn't the question establish the fact that the string is massless? Therefore, It should not be able to oscillate as a wave because its linear mass density is 0. Can someone explain this to me?(2 votes)
- There is a mass at the end of the string, which is means μ (m/l) is not 0(2 votes)

## Video transcript

- [Voiceover] The figure
above shows a string with one end attached to an oscillator and the other end
attached to a block there. There's our block. The string passes over a
massless pulley that turns with negligible friction. There's our massless pulley that turns with negligible friction. Four such strings, A, B, C,
and D are set up side-by-side as shown in the figure
in the diagram below. So this is a top view,
you can see oscillator. This is a top view of the
oscillator string pulley mass system, and we have four of them. Each oscillator is adjusted
to vibrate the string at its fundamental frequency f. So let's think about what
it, I'll keep reading then we'll think about what
fundamental frequency means. The distance between each
oscillator and the pulley L is the same. So the length between the
oscillator and the pulley is the same, and the mass
of each block is the same. So the mass is what's providing
the tension in the string. However, the fundamental
frequency of each string is different. So let's just first of all think about what the fundamental frequency is, and then let's think about
what makes them different. So the fundamental frequency, one way you could think about it is it's the lowest frequency
that is going to produce a standing wave in your string. So it's the frequency that
produces a standing wave that looks like this. It's a standing wave where the
string is half a wavelength. I guess there's two
ways to think about it. It's the lowest frequency we could produce the standing wave, or it's
the frequency at which you're producing the
wave, the standing wave, with the longest wavelength. So the string at the fundamental frequency is just going to go to, is gonna vibrate between those two positions. And you see here that the wavelength here is twice the length of the string. And if you want to see
that a little bit clearer, if I were to continue with this wave, I would have to go another
length of the string in order to complete one wavelength. Or another way to think about it, you're going up, down, and
then you're going down back. It's reflecting back off of this end here. And as we mentioned, essentially
what the mass is doing is providing the tension, the
force of gravity on this mass is providing the tension in this string. So the oscillators is vibrating
at the right frequency to produce this, the lowest
frequency where you can produce this standing wave. So let's answer the questions
now, and we have four of this set up and they all
have different frequencies. The equation for the velocity
v of a wave on a string is v is equal to the
square root of the tension of the string divided by
the mass of the string divided by the length of the string, where F sub T is tension of the string and m divided by L is
the mass per unit length, linear mass density of the string. And hopefully this makes sense. It makes sense that it's
going to be that if, if the tension increases, that
your velocity will increase, the tension, you could think
of the atoms of the string of how much they're pulling on each other. And so if there's higher tension, well, they're going to be able
to move each other better, or I guess you could say
accelerate each other better as the wave goes through the string. And also make sense that
the larger the mass, if you hold everything else equal, that you're gonna have a slower velocity. Mass, you could view it as a measure of inertia,
it's how hard it is to accelerate something. So if the string itself,
especially the mass per unit length, if there's a
lot of mass per unit length, actually, let me circle
that because that's actually the more interesting thing. If there's a lot of mass per unit length, it makes sense that for a
given amount of the string, it's gonna be harder to
accelerate it back and forth as you vibrate it. And so this part right over
here would be inversely related to the velocity. It's not proportional though. You have this square root here, but they're definitely, if the linear mass density increases, then you're gonna have a slower velocity. And if your tension increases,
you're going to have a higher velocity. So hopefully this makes
some intuitive sense. And they ask, what is the
difference about the four string shown above that would
result in having different fundamental frequencies? Explain how you arrive at your answer. And then part b is
student graphs frequency is a function of the inverse
of the linear mass density. Will the graph be linear? Explain how you arrived at your answer. All right, let's answer each of these. So a, part a, what is different
about the four strings because they all have different
fundamental frequencies? So the fundamental frequency, let's just go back to
what we know about waves, that the velocity of a wave
is equal to the frequency times the wavelength of the wave. Or you could say, you could
divide both sides by lambda, you could say that the frequency of a wave is equal to the velocity
over the velocity over the wavelength. So if we're talking about
the fundamental frequency, if we're talking about
the fundamental frequency, funda, let me just, let me write frequency and
then let me write fundamental real small here. Fundamental, the fundamental frequency, is going to be the velocity of our waves divided by the wavelength is
going to be twice the length of our string, divided by two L. And if you look at the
expression that they gave us for the velocity of
the wave on the string, well, this is going to be equal to the square root of the
tension in the string divided by, divided by
the linear mass density. Divided by the linear mass density. And all of that is going to be over two L. Now all of them have different
fundamental frequencies, but let's think about
what's different over here. They all have the same tension. They all have the same tension. How do I know that? Well, what's causing the
tension is the masses hanging over the pulleys. So the weight of those masses. So that's all going to be
the same for all of them, and they all have the same length. They all have the same length. So these are all the same. So the only way that you're
going to have different fundamental frequencies is
if you have different masses. So different masses. So that has to be different. Different. So to answer their question, the strings must have different mass, well, they would have different
linear mass densities, but since they're all the same length, they would also have different masses. So let me write this down. Strings, strings must have different, different masses and mass densities, and mass densities, densities, since all other variables driving, driving fundamental frequency are the same, fundamental frequency are the same, are the same. All right, let's tackle part b now. A student graphs frequency as a function of the inverse of linear mass density. Will the graph be linear? Explain how you arrived at your answer. So student graphs frequency. Let me underline this. They're graphing frequency as a function of linear mass density. So we actually can write this down. If we wanted to write frequency as a function of linear mass densities, so we could write as a
function of m over L, well, this is going to be equal to, is going to be equal to, we could rewrite this expression, actually, we could just
leave it like this. This is the same thing as one over two L times the square root of the tension divided by the linear mass density. Or you can do this as being equal to, you could say this is a square root of the tension over two L, and I'm putting all of
this separate because we're doing it as a function
of our linear mass density. So we can assume that all of
this is going to be a constant. And then times the square root of one over the linear mass density. So if you're plotting, if
you're plotting frequency as a function of this, it's not going to be,
the graph is definitely not going to be linear. You see here, one, I have
the reciprocal of it, and then I take the square root of it. So let me write this down. This f or graph of f of two L, definitely not linear. Let me write it that way. F of m over L graph, definitely, definitely not linear. And you could point out
that it has a reciprocal and it's a square root. Involves, involves square root and inverse, or I could say and
reciprocal of the variable. And reciprocal, reciprocal of the linear mass density of m over L. So definitely not linear. All right, let's tackle 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 string D below, mark and label the points
on the string that have the greatest average vertical speed. So one way to think about
the fundamental frequency, that's our first harmonic. So if they're talking
about the first harmonic and the string is what I showed before, that's the lowest frequency
that produces a standing wave or it's the frequency
that produces the largest standing wave, I guess
you could say the one with the largest wavelength. And if it was a first harmonic, the part of the string that moves the most is going to be that center. But our second harmonic is
the next highest frequency that produces a standing wave, and that's going to be
a situation in which the wave, instead of having
a wavelength twice the length of the string, it's
gonna have a wavelength equal to the length of the string. So now, it's going to look like this. So it's going to vibrate between this, actually, let me draw it a
little bit neater than that. It's going to vibrate between, between this. And so it's gonna vibrate between that and this. And this right over here. So when you see this, this version of it, where now that we have
half the wavelength, the wavelength is exactly
the wavelength of, it's actually L this time, the part that moves the most is here. So that's going to move the most. And here. I didn't draw it perfectly. But one way to think
about it, it's exactly 1/4 and 3/4 of the way. Exactly halfway is not
going to move as much, is not gonna move because
it's a standing wave now or it's gonna move very imperceptibly. These two places are where
you're gonna move the most at the first, or sorry,
at the second harmonic. The second harmonic being
the next highest frequency that produces a standing wave again.