A 170-cm-long open-closed tube has a standing sound wave at 250 Hz on a day when the speed of sound is 340 m/s . How many pressure antinodes are there, and how far is each from the open end of the tube?

Video transcript

Hey everyone. So this problem is dealing with standing sound waves. Let's see what it's asking us. A physics teacher conducting a lab experiment has a centimeter long pipe with one end open and the other closed to demonstrate standing sound waves. The teacher creates a standing sound wave with a frequency of 213 Hertz inside the pipe under conditions where the speed of sound is 348 m per second. She tasked the students with determining the following the number of pressure anti nodes inside the pipe and the distance of each anti node from the open end of the pipe. Our multiple choice answers here are a two pre two pressure anti nodes, 0.6 m and 1.2 m away from the end of the pipe B three pressure anti nodes at distances of 0.6 m 1.8 m and two m C three pressure anti nodes at 0. m 1.2 m and two m or D two pressure anti nodes at 0.4 m and 1. m. OK. So the first thing we can do with this problem is recall that our frequency when we have one closed end and one open end of a pipe is given by N V divided by four L where N is your number of harmonics. The number of harmonics will tell us the number of nodes and anti nodes. And so that is what we are going to solve for. So we'll rearrange the equation to isolate N which gives us four L F divided by B. And so we have four, multiplied by our length. It was given to us. And the problem is 200 centimeters. We're going to use standard units here. So we'll just uh rewrite that as two m and then multiplied by our frequency of 213 Hertz, all of that divided by our speed 340 m per second. And that gives us a harmonic value of 5.1. Now, because we can remember that harmonics are always integers. What that's telling us is that we are in the fifth in, we are in the fifth harmonic because we are in the fifth harmonic. We can recall that that means we have three anti nodes. Fifth harmonic is three nodes and three anti nodes. But we have one open and one closed and on the so when we look at our multiple choice answers that eliminates answer choice A and D kind of right off the bat. Now, we need to find the distance of each of those three anti nodes from the open end of the pipe. So the distances for the first or for the three anti nodes are given by D equals pi divided by four. Sorry D one equals, excuse me. Sorry. So D one is equal to Lambda divided by four, D two or the second anti node is three multiplied by lambda divided by four. And then D three is five multiplied by LAMBDA divided by four. We weren't given lambda our wavelength though, in the problem, we can recall that lambda is equal to speed divided by frequency. And so frequency, right, it was given as B divided by four L N V divided by four L. And so when we sub that in, we get four L V divided by N B speeds cancel. And our lambda can be rewritten as four L divided by. And so we are going to use that to finally solve for those distances. So we have T one equals four L divided by four N, the four is cancel. And then again, our length is just two m divided by N five. And that gives us 0.4 m. We're gonna, we are going to do that for those next two lengths. So we will have three multiplied by four L divided by four N or six meters divided by five, which gives us 1.2 m. And then four D three, we will have five multiplied by four L divided by four N or meters divided by five oops, sorry, 10, 10 m divided by five or two m. And so those are the distances those three anti notes that we have and we have that fifth harmonic. And so when we look at our multiple choice answers that aligns with answer choice C three pressure anti nodes 0.4 m 1.2 m and two m away from the open end of the pipe. All right. That's all we have for this one. We'll see you in the next video.