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## Class 11 Physics (India)

### Course: Class 11 Physics (India) > Unit 19

Lesson 5: Sound- Production of sound
- Sound Properties: Amplitude, period, frequency, wavelength
- Speed of Sound
- Relative speed of sound in solids, liquids, and gases
- Decibel Scale
- Why do sounds get softer?
- Ultrasound medical imaging
- Standing waves in open tubes
- Standing waves in closed tubes
- Wavelength and frequency for sound waves in tubes
- Pressure and displacement of sound waves in tubes
- Introduction to sound review

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# Standing waves in closed tubes

Find out why you can make music by blowing into empty bottles. Created by David SantoPietro.

## Want to join the conversation?

- Musicians sometimes talk about going up or down an octave. How do the concepts of the resonance region, anti-nodes, and nodes relate to the idea of an octave?(32 votes)
- every power of 2 * 1st harmonic is an octave above. For example:

880 = A in 2nd octave

440 = A in first octave

220 = A in small octave

110 = A in great octave(27 votes)

- If I have a cylindrical container open-open and another one open-close (that are identical), which one will produce lower notes? And why?(6 votes)
- If we consider that both of them are first harmonic, then "open open" will be higher.

As wavelength for "open open" (2L) is less than for "open close" (4L)..(4 votes)

- Not quite understanding why there can only be an odd number of nodes.(6 votes)
- it depends on the situation.

you can draw a standing wave (indeed create a standing wave using a spring) with odd and even number of nodes.

what makes you think there can only be odd numbers of nodes??

take a look online for examples

if you look here at 5.52, the standing wave has 10 nodes

https://www.youtube.com/watch?v=jz8IIk_bps0(0 votes)

- How does this apply to windpipe instruments? Like... I used to play the clarinet (and recorder in elementary school) but generally, the more holes you plug the lower the note. So if you don't plug any holes in the windpipe instrument then the L is shorter? Super random question, I know lol.(4 votes)
- Yes, by covering or opening various holes you are changing the place where the nodes are in the standing wave.(4 votes)

- Can we just count the number of antinodes and nodes and leave out the one at the open end and get n?

Like for the 3rd harmonic, we have 2 nodes and 1 antinode before the last antinode, so that makes a total of three, thus 4L / 3.

For the fifth harmonic, there are 3 nodes and 2 antinodes other than the last antinode, so that is a total of 5. Thus 4L / 5.

I mean I think this helps during exams. Just skip the antinode at the opening an count the other nodes and antinodes to get n. It works right?(4 votes)- i think your intuition heads toward a right direction

but to have a bit more concise equation

we better focus on nodes only

0. n = number of nodes in the middle (except for the two edges)

1. open-open tube

wavelength = 4L/2n

2. open-closed tube

wavelength = 4L/(2n+1)

3. closed-closed tube (e.g. a string with two standing edges)

wavelength = 4L/2n

one more thing, the difference between case1 and case3 is the type of starting (and ending) nodes. 1 starts and ends at the antinodes, while 3 at nodes(1 vote)

- I'm not quite sure on how to name the harmonics in this system. Do you call the harmonic where n=3 the 2nd or 3rd harmonic? Is n=5 the 3rd or 5th?(3 votes)
- If I blow air over a soda bottle, how do i know which harmonic I am playing? Is there a way to know how many nodes are in in there?

And what can I do to get the overtone of the current tone(like blow harder or softer..)?(3 votes) - what is meant by fundamental harmonic, second harmonic and so on. What is the relation between them? Do they occur at the same time or turn by turn? what effect is caused by them?(1 vote)
- The fundamental is the same thing as the first harmonic, and it is the mode of vibration where you have the fewest possible nodes in the standing wave. The second harmonic is the next highest frequency where you can get a standing wave. The third harmonic is the next, etc.(3 votes)

- 8:15why doesn't the number of nodes change as L becomes larger?(2 votes)
- Because the standing wave only exists at certain combinations of L and f, as given by the equation derived here.(0 votes)

- for the closed closed tube, why does the even number of harmonic waves do not exist?(1 vote)
- try drawing the patterns in closed tubes.

you will see that all 'ends' have a node and this restricts the number of harmonics available(2 votes)

## Video transcript

- [Voiceover] Okay, so last
time we saw that for an open open tube, or an open open pipe, a pipe where both ends were open, there only particular
wavelengths that were allowed, because you had to have
anti-nodes at both ends. So, you had that one. We had this one. We have this one, and we found the
wavelength of all of these. Then, we realized, wait a minute we can write down a formula. For any possible wavelength, in an open open tube, and it depends only on the
length of the tube L and N. N is which harmonic we're talking about. One is the fundamental,
two is the second harmonic, three is the third
harmonic, four and so on. This gave you ever possible wavelength. The questions is, can we do the same thing if we had an open closed end? What happens if one ends closed? What if we closed this end off, so that it's not open anymore. Well, let's do that right now. Here we go. Here's a closed end right here. This end is closed. This is more like a soda bottle, because this end would be open, the top that you drink out of
and the bottoms closed off. We've got air inside. If you blow over the top, what possible frequencies,
what possible wavelengths could you set up? Well, what do we know here? We know that this end, this air is open or this side is open, and so this air molecule
can oscillate wildly. What type of node is that going to be? That's going to be an anti-node, since it can oscillate a lot. This end over here, these
air molecules keeping bumping into the side. It's got to be a node over on this end, because there can't be any displacement. What do we do here? This ends a node. We know that. I'm going to
put this on the axis here. I know this end has no displacement. We're graphing displacement
again versus X. This is, in this case,
horizontal displacement. This end is going to oscillate wildly, so I know there's going
to be an anti-node. So, I'm going to put it right here. I'm going to draw the
simplest possible wave that can go from this anti-node to this node is going to look like this. Anti-node and it goes curve
right down to the node. That's it. What wavelength is this? Well, in terms of the length of this tube, so if this is L, what wavelength is this in terms of L? We've got to figure out how
much of a wavelength this is. That's always the trick. This
is what freaks students out. They don't like figuring out
how many wavelengths is this. It's not too hard. First thing I like to do is draw what I know one wavelength looks like. So, what does one wavelength look like? One wavelength looks like this. If we start up here, a wavelength is going all the way and then back to the same
point in the process. There we go. That's one entire wavelength, at least if this is a
versus X graph it is. So, how much of a
wavelength is this pink line for this first fundamental frequency? This first fundamental wavelength? It starts at the top, so
let's start at the top. It goes, and that thing
just goes till the first crossing of the axis, and that's the node. So, this thing crosses
the axis here. That's it. That's all we got. We got a big long ... what is this? This is
one-fourth of a wavelength. That's one-fourth, and it goes to here. That's another fourth. Here's another fourth,
and here's another fourth. So, this is only
one-fourth of a wavelength. That's hard for a lot of people to see. The whole things one wavelength. Half of it is a half,
so if I cut it in half, that's a half, and I cut
this half in half again, I get one-fourth. So, what I find out is that, okay, the length L equals one-fourth of a wavelength in this case. Our fundamental wavelength divided by four equals the length of this tube. So, if I want to know
what the wavelength is, that means my wavelength
for the fundamental open closed case is four L. I'm going to write that over here. (Lambda) equals four L, and this would be the
fundamental wavelength for this open closed tube. This is bigger. For open open is was
two L, so it's four L. How about the next one? All right, we got it
started in the anti-node, and we got it into the node. The next simplest case in ... well, we had no nodes in between. We only had one at the end. So, now I need a node in the middle. I'm going to create a node in the middle, and then I have to get all the way back to the node on this end. So, I've got one node here.
I've got a node at the end. I've got this anti-node here,
and an anti-node up here. How many wavelengths is this? Let's figure it out. Start at the top. All right, this goes all the way down past the node and back up. So, all the way down, past the node, back up, and ends at a node. That's how much of a
wavelength I've got here . So, how much is this? Well, it's one-fourth,
two-fourths, three-fourths. This time L is the length of my tube. Is my wavelengths fitting into L? In this case, it's
three-fourths of a wavelength. Again, if I solved that for wavelength, I'd get four L over three. So, the next possible wavelength is four L over three. If we solve for the next one, let's draw it, started an anti-node, end a node, the next possible one, well, the last one had one node, this time it's going to have two nodes. I'm going to come down here. One node, two node, and then it gets (back)
to the node at the end. So, a node at this end, there
has to be a node at this end, because it's closed. This time I have two nodes in the middle. I have a anti-node on this end, and two anti-nodes in the middle. So, how many wavelengths is this? All right, let's figure it out. Start at the top. All right, it comes down,
let's just trace it out. Down, hits that first node, down to the anti-node, back up to the node,
now, I've gotten to here. I keep going. I go up to
the top, I'm not done yet. Now, I'm back up to here. I have to keep going one
more fourth of a wavelength. This is more than a wavelength. This time L equals, well, this is one whole
wavelength just to this point right here. Then, I have to add one more fourth of a wavelength to that. So, this is one wavelength and a fourth, or another way to say that is that this is five-fourths of a wavelength. Five-fourths of a wavelength. Just to here is one whole wavelength, and I have to add one more fourth to that. So, if I solve for lambda, I get lambda is four L, ... we see a pattern. Patters are great. Four L over five. This next wavelength
possible is four L over five and shoot, we can do this now. Now, I see enough. If I want to write down
any possible wavelength, it's going to be four L. Look, four L, four L over
three, four L over five. The next one four L over seven. I get that the possible wavelengths for an open closed tube are four L over N. Except, instead of being
any possible integer, the only allowed integers
are the odd ones. So, I'm not allowed to put in two or four. It's got to be one,
three, five, and so on. Here's the formula. This is it. If I want to know what are
the possible wavelengths allowed in an open closed tube, this is it. It depends on
the length. It depends on N. Looks just like the case for open open, except it was two L for that case. This case is four L on top. Four times length of the tube, and on the bottom it's
only the odd integers. That means I only get the odd harmonics. This four L over three, I wouldn't call this the second harmonic, I'd call this the third harmonic. This five down here, I
wouldn't call this the third, I'd call this the fifth. We're missing all the even
harmonics on this case for an open closed tube. It's a little bit strange, but that's what happens when you have an anti-node at this end, and a node required at that end. You can try this. Look,
this is a function of L. If you want to test this out, next time you're drinking a soda. Look, if the L is large if the length of your tube is large, that means the wavelength should be large. If the wavelength is large, well, let's see. V equals lambda F. If you've got a big wavelength, that'll mean a small frequency, because the speed won't change. The speeds determined by the medium, and you're probably not
changing the temperature of the medium all that much. Maybe, you've got ice in
your drink or something, but if it melts it'll change a little bit. It's not going to change
the speed by that much. If you increase the length, you'll increase the wavelength, it'll decrease the frequency. Low frequency means a low note. You'll hear a lower bassy note. Try this, I want you to try this out the next time you're drinking a soda, or maybe it's a healthy beverage. Whatever it is. You got some soda here,
it's up to some level. Where ever the waters at acts
as the bottom of the tube. Right now, I only have a tube this length. The top is open, and the bottom is basically where ever there's water level is, because the air can't get
past that water level. This would only be a length of that. As you keep drinking, keep blowing over the top. As you keep blowing over the top, listen for what note you hear. You should keep hearing a lower note. The lower this gets,
once it's down to here, now your length ... the length of your tube is bigger. It's open at the top closed at the bottom. You'll hear even a lower note, once it gets down to here. When you finish, now it's even longer. You'll hear a much lower notes. The notes should get lower and lower. Not louder and louder,
but lower and lower, the frequency should sound lower and lower the lower your drink gets. The higher the length,
bigger the wavelength, smaller the frequency.