Standing sound waves are produced in a pipe that is 1.20 m long. For the fundamental and first two overtones, determine the locations along the pipe (measured from the left end) of the displacement nodes and the pressure nodes if (a) the pipe is open at both ends

Video transcript

Hey everyone in this problem. We have a woodwind player that generates standing waves in a 32 cm flute. The resonating air column is open at both ends. And were asked to draw a diagram showing the locations of the displacement notes and anti nodes along the flute for the 1st and 2nd harmonic. Now, the first thing we want to note is that we're told this is an open pipe. If this is an open pipe, that means it's open at both ends. And so both ends are going to be displacement anti note. And when we have an open end it's a displacement anti note. So we have two open ends. That means both of them are going to be displacement anti notes. Okay, now we're talking about an open pipe. So we think about the possible wavelengths we can have recall that those will be given by lambda. N Is equal to two L over N where L is the length of the pipe. And our possible end values here are 123 dot. Okay, the natural number. So the positive integers. So let's start with our first harmonic for our first harmonic. We're going to have any equals one. And when when n equals one, we have that lambda one is equal to two L. Okay, so we have two times 32 centimeters divided by our end value of one, which gives us a wavelength of 64 centimeters. So we know the length of our pipe, we know the wavelength of the first harmonic. Let's figure out where we have nodes and anti nodes. Well we know that X equals zero cm. We have a displacement anti note. Okay, we know that because the end is open. So this is a displacement anti note. Now recall that if we go from X equals zero to X equals a quarter of the wavelength. We will then be at a displacement note. Okay, so each quarter of a wavelength we go from anti node to node to anti node to node. So at x equals one quarter of lambda one, which is equal to 1/4 times centimeters, Which is equal to 16 cm. We will be added displacement note. All right, so we have an anti node. We have a displacement note at 16 centimeters. Our pipe is 32 centimeters. So let's move on to the next displacement anti node and see if that's still within the length of our pride. So we started at X equals zero. Then we looked at a quarter of the wavelength. Now a half of the wavelength we will be back to another anti note. So at 1/2 times 64 cm. This gives us 32 cm which is going to be the end of that flute. We have a displacement anti note which makes sense because we were told both ends were open. And if both ends were open, then both And should be displacement anti notes. And now we found by finding the wavelength that the second end is indeed an anti note as we expected. And now we're at 32 cm. We've looked at the entire length of that flute. Any other displacement nodes and anti nodes would happen further with a longer length. Okay, so let's go ahead and draw what we have for our first harmonic. We have our flute with open ends. Okay, so we have Anti nodes at both ends. This is x equals zero cm. We have 16 cm halfway through and then 32 cm at the end. And we know that we get a note at 16 cm. And so are flute is going to look something like this. Okay, those are gonna be the displacement anti nodes at each end and a displacement node in the middle. Alright, so we've got our diagram for the first harmonic. Now let's do the second harmonic and it's gonna be a very similar process. So for the second harmonic We're gonna have an equals two. And so our wavelength lambda two is gonna be two times L 32 cm over two, Which gives us a wavelength of 32 cm. Now again at X equals zero. We're going to have a displacement anti note because it is an open end. We know that at a quarter of the wavelength. So a quarter of 32 cm which is going to give us eight cm. We're going to have a displacement note at half of the wavelength. We will be back to a displacement anti node that will give us 16 centimeters. Half of 32 centimeters is 16 centimeters. We will be back to a displacement anti note. Now our displacement and you know this is where we stopped last time Or the first harmonic. But here we're only at 16 cm. We have a length of 32 cm so we can actually keep going and see what happens in the second part of this flu. So we're looking every quarter. Okay, every quarter of a wavelength we change from a node to an anti node. Back to a node back to an anti node. So we have gone 01 quarter one half, which is two quarters. Now we go to three quarters. So at X equals three quarters of the wavelength. So three quarters times 32 centimeters. This gives us an X value of 24 centimeters. We're gonna be back to a displacement node. Alright well we have a displacement note at 24 cm. Again, our length is 32 cm. So let's keep going. X equals, we have zero a quarter half three quarters we're back to one. So at X equals 32 centimeters which is the length of our flute. We're going to have a displacement anti node which is what we expected because we have an open it. Now, we're out of room here to draw it, let me go back up to the top and we have a bit of space here. So I'm going to draw the second harmonic in here and I'm just going to put it in a box. So we don't get confused. Second harmonic is going to look like this. We have our flute, it is open at both ends. So we know we have anti nodes there. Now we're gonna break this into four pieces. So we have 8cm 16 cm and 24 cm. Those were all values of importance when we looked where we have anti notes and notes. And we found that we get a node a displacement note at 8cm A displacement anti note at 16 cm A displacement node at 24 cm and a displacement anti note at cm. And so our second harmonic is going to look something like this for our flu open up both ends. So now we have our first harmonic and our second harmonic. Let's look at our answer choices and see which one matches. So we have an open flute. So we know that we have anti nodes at both ends. We found for the first harmonic that we would have anti nodes at both ends and a node in the middle At 16cm. So we're looking at either option a or option B, option C&D have more nodes and anti nodes than we found. And then for the second harmonic we found that we have anti nodes at zero centimeters at 16 centimeters and at centimeters and we found that we had a displacement node at eight centimeters and at 24 centimeters. And that matches with the diagram in A in B. We have more anti nodes and nodes for the second harmonica than we found. And so the correct answer here is going to be A. That matches the diagrams we drew for the 1st and 2nd harmonic of that flute. Thanks everyone for watching. I hope this video helped see you in the next one.