 ## Physics

Learn the toughest concepts covered in Physics with step-by-step video tutorials and practice problems by world-class tutors

18. Waves & Sound

# Standing Sound Waves

1
concept

## Standing Sound Waves 8m
Play a video:
Hey, guys, in this video, we want to talk about standing sound waves. Okay, so we're gonna apply what we learn about standing waves on strings to sound now. All right, let's get to it. Just like with waves on a string, when sound is reflected off of the surface, they interfere. Those sound waves interfere with whatever sound waves are coming at them because they're going forward, and then they're bouncing back. And whatever sound waves are also coming forward, there's gonna be interference. Okay. Standing sound waves, air produced by actual oscillations and air molecules, just like regular sound waves, are produced by actual oscillations and air molecules. So we have two things we need to worry about. We need to worry about that displacement of those actual air molecules and the pressure of the air. Remember the dent in the more densely packed those air molecules are the larger the pressure. It turns out that because of this ah, displacement node is always a pressure anti node, and vice versa. A displacement anti note is always a pressure node, and the reason for this is let's say that this point right here is a displacement note. What that means is those molecules they're not moving, but the molecules around them are oscillating back and forth. So we have these guys coming in, right? So they're gonna get really close, and then they're gonna move outwards, so they're going to get really far. So you can see that even though this line of air molecules right here is a displacement, know that they don't move. That area undergoes a very, very large pressure change. Okay, when the two surrounding air molecules come very close and when they move very far, there's low pressure, right When they move very closest high pressure. When they move very far, there's low pressure. So that displacement node undergoes a very, very large change in pressure. Now, what if this is a displacement anti node? Well, then it's gonna be moving back and forth, and so are the molecules near it. So when this guy is moving to the left right, the anti node, the molecules around it are moving with it, right? Like a wave. They're all moving with it. When the displacement anti note moves the other direction, all of the other ones move with it too. So those three lines sort of move as one. And there is no pressure difference because the density doesn't change, right. They move together, and because they're moving together, the density isn't changing. They're not getting any closer. They're not getting any further apart. So you can see that that density anti note is actually a pressure node. Right? Note is no displacement. That pressure is not changing at that density. Anti node. Okay, let me minimize myself here. Now, typically with standing sound waves, you are producing them inside of tubes. The open end of the tube is always a pressure node, which means what is the displacement? The opening is a displacement anti note, and the closed in is always a pressure anti node, which means the closed end is a displacement note. Okay, Your book and your professor probably show you images that compare the two. I'm gonna focus entirely on pressure. Okay? I want to talk about pressure nodes and pressure anti nodes. Alright. The two important scenarios to remember guys is a tube with both ends open. Has a pressure note of each end. So it is a node node scenario. A tube with one enclosed has a note at one end and an anti noted the other end. This is a node anti notes situation. Besides that, the equations are the same. I drew figures here for a tube of to open ends and the two with one open and one closed in. These are the pressure oscillations. Okay, so you see, the pressure is a node at every open end, and it's an anti node at every closed end. Okay, so the tubes on the left are node node. The tubes on the right are node Anti note. Okay. And remember the equations for node, node and node. Anti note for note, note standing waves. Regardless of whether they're sound waves or waves on a string, we have this equation for the wavelengths and this equation for the frequencies. Where n is any integer sighs still integer. There we go. Any integer now for note anti notes, standing waves regardless of whether they're waves on a string or sound waves. Right. This is your equation for wavelengths allowed. And this is your equation for frequencies allowed. Where in is on Lee Odd integers. Okay, once again, it's very, very, very important to remember that that end is Onley. Odd integers. All right, let me minimize myself for this problem. Ah, wind instrument is 1 m long and open at the mouthpiece, but plugged at the far end. What is the third highest harmonic frequency this instrument could make? Assume that the temperature is 20 degrees Celsius. Okay, if it's open at the mouthpiece One end but plugged at the far end It's open closed. So this is node anti node. If we want to find a harmonic frequency, we have to know then that the harmonic frequencies for note anti nodes are envy. Over four l Where in his Onley? Odd. What's the third highest? Remember, In can be etcetera. What's the third highest five? So the fifth harmonic is the third highest harmonic. That's 5/4 l. But before we can solve that, we need to know the speed of sound in this instrument. We're assuming that the temperatures 20 degrees so the speed is gonna be 3. 31 times one plus 20 over to 73 which is gonna be meters per second. Okay, so going back to filling the south, this is five times 3 43 over four times l. It's a 1 m long instruments and this comes out to 429 hurts. Okay, Exact same equation. Exact. Same process to solving for standing sound waves. Just recognized our you know, node or nodes. Anti node. Alright, guys, that wraps up our discussion on standing sound waves. Thanks for watching.
2
example

## Fundamental Frequency of Ear Canal 1m
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Hey, guys, let's do an example. The human ear can be modeled as a tube with one end open and one end closed the ear canal right outside the ear drum. If the length of the ear canal is roughly 2.5 centimeters, what would the fundamental frequency of standing waves in the ear be assuming the temperature inside the ear is out of a human body 37 degrees Celsius? Okay, if one end is open and one end is closed, this is a node anti node problem. If we're looking for the fundamental frequency, that's always a harmonic number of one, So remember any harmonic frequency for note. Anti Note is envy over for L, where in is odd, the fundamental frequency is just V over for L. Before we can solve this, we need to know the speed of sound in the ear canal. We're assuming that it's at body temperature, right? So it's the square root of one, plus the temperature in Celsius over to 73 times 331 m per second, and this equals meters per second. Okay, so the fundamental frequency is 53/4 times the depth of the ear canal, which is roughly 2.5 centimeters, or 30.20 to 5 m, which is 3500 and hurts. OK, just a straight up application off our equations for a node node standing waves. Alright, guys, that wraps up this problem. Thanks for watching.
3
example

## Third Harmonic for Waves in a Tube 2m
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