Organ Pipes and the Resonance Tube

Transcript

0:00Introduction

Hello everyone! Take a pipe open at both ends, and blow across one end. You hear a note. Now close the far end with your palm, and blow again. Does the note go up or down, and by how much? We will answer it soon, and then check it with real numbers in our first example. Organ pipes and the resonance tube come up often in JEE, so let's begin.

0:23Open ends and closed ends

So, on the left are pipes closed at one end; the thick black line is the closed end. On the right, pipes open at both ends. The solid and dashed curves show how far the air moves, at two opposite moments. The formula beside each picture, we will build one by one in a moment. Now, can the air move to and fro at a closed end? It cannot, that is correct! So a closed end is a displacement node, and that is a pressure antinode, where the pressure changes the most. On the left, the two curves meet at the closed end. Okay? Now the open end. The air there is free, so it moves the most: an open end is a displacement antinode and a pressure node. On the right, the curves are farthest apart at the open ends. Right?

1:12Open pipe: all harmonics

Now, the open pipe. We need an antinode at both ends. So what is the simplest pattern that fits? One node in the middle, yes, exactly, the top right picture! So capital L, the length, is half a wavelength, lambda over two; lambda is the Greek letter like an upside down y. Then f sub one, the fundamental, equals v over two capital L, with v the speed of sound. With me? One more node gives the bottom right picture: capital L equals lambda, and f sub two equals two times f sub one. In general, f sub n equals n times v, over two capital L, for n equal to one, two, three, and so on, as the dots say: all harmonics. Clear?

1:59Closed pipe: odd harmonics only

Now close one end. We need a node at the closed end, and an antinode at the open end. What is the simplest pattern now? Just a quarter of a wave, very good, the top left picture! Node to antinode is lambda over four, so capital L equals lambda over four, and f sub one equals v over four capital L, okay? Now the JEE trap. An open pipe also gives two times f sub one. Can a closed pipe give two times f sub one too? It cannot, and many students slip here! The next fit is three quarters of a wave, the bottom left picture: f sub three equals three times f sub one. In general, f equals two n minus one, in brackets, times v, over four capital L, and two n minus one is always odd, so odd harmonics only. Got it? Careful with overtones, the notes above the fundamental. In a closed pipe, overtone number m is harmonic number two m plus one. The first overtone is the third harmonic, not the second. Right? And now our opening puzzle. For the same length, the open fundamental, v over two capital L, is two times the closed one, which is v over four capital L, so closing one end halves the frequency. With me?

3:24End correction

Now, a small correction for real pipes. The antinode lies a little beyond an open end, by about e equals zero point six times r, where r stands for the pipe's radius. This e is the end correction. Clear? So a closed pipe, with one open end, acts a bit longer: capital L plus e equals lambda over four. Okay? Now the open pipe. How many times do we add the end correction there? Twice, that is correct, because it has two open ends! So capital L plus two e equals lambda over two. Right?

4:00The resonance tube

Now, the resonance tube. The water surface closes it, so the air column is a closed pipe whose length we can change. We sound a tuning fork of frequency f at the top, and when the column fits, the sound becomes loud: that is resonance. With me? The first resonance is at l sub one, the small l with a one below it: l sub one plus e equals lambda over four. The next, l sub two, plus e, equals three lambda over four. Got it? But we do not know the end correction. So how can we get rid of it? Subtract the two, yes, exactly! The end correction cancels: l sub two minus l sub one equals lambda over two. The double arrow means therefore, so v, which is f times lambda, equals two f times, l sub two minus l sub one, free of the end correction. Clear? And taking three times the first equation away from the second gives e equals, l sub two minus three l sub one, over two. Okay?

5:05Example 1: first two notes of a 0.5 m open and closed pipe

Our first example. Find the first two resonant frequencies of a pipe zero point five metres long, open, and then closed at one end. The speed of sound is three hundred and forty metres per second; the small minus one on the s means per second. Pause the video and try it first. Done? Okay, let's check. Open pipe first. Two times capital L is one metre, so v over two capital L gives three hundred and forty hertz, the H z on screen. Then two v over two capital L gives six hundred and eighty hertz. Right? Now close one end. Four times capital L is two metres, so v over four capital L is three hundred and forty over two, which gives one hundred and seventy hertz. Exactly half: the note goes down, just as we found! Okay? Now the second resonance of the closed pipe. Two times one hundred and seventy gives three hundred and forty hertz, right? Oh no, careful, that is the same trap! Even harmonics are missing in a closed pipe, so the next is three v over four capital L, three times one hundred and seventy, which is five hundred and ten hertz, not three hundred and forty. Clear?

6:24Example 2: speed of sound and end correction, 512 Hz fork

Now a harder one. Here, a fork of five hundred and twelve hertz gives resonances at air columns of sixteen centimetres and fifty centimetres. Find the speed of sound and the end correction. Pause the video and try it first. Done? Okay, let's check. The first column is a quarter wave, so is lambda just four times sixteen centimetres? Oh no, careful, that ignores the end correction! Sixteen centimetres plus e is the quarter wave. So use the difference, l sub two minus l sub one: in metres, zero point five zero minus zero point one six gives zero point three four metres, which is lambda over two. With me? So v equals two f times the difference: two times five hundred and twelve, times zero point three four. That gives about three hundred and forty eight metres per second; the wavy equals sign means approximately. Okay? Now e equals, l sub two minus three l sub one, over two. Three times zero point one six is zero point four eight, and zero point five zero minus zero point four eight gives zero point zero two. Half of that is zero point zero one metre, one centimetre.

7:45Quick revision: harmonics, end correction, resonance tube

Right? So, quick revision. First, an open pipe gives all harmonics, and a closed pipe gives odd harmonics only. Second, the end correction, about zero point six times r, goes in once for a closed pipe, capital L plus e equals lambda over four, and twice for an open pipe, capital L plus two e equals lambda over two. Third, in the resonance tube, v equals two f times, l sub two minus l sub one, and e equals, l sub two minus three l sub one, over two. Now try the practice problems on this page. All the best!