The Resonance Tube

Transcript

0:00Introduction

Hello everyone, here is a small puzzle before we start. Sound covers more than three hundred metres in just one second. Now measure its speed with only a tuning fork, a tube, some water and a scale, and no stopwatch. That sounds impossible, right? But by the end of this video, you will do it yourself. This experiment belongs to experimental skills in JEE, and questions on it do come up. So, let's begin.

0:28The set-up: a fork over a column of air

Look at the figure on screen: the same tube twice, at the first resonance and at the second, with the water in grey. We hold a vibrating tuning fork of frequency small f over the open end, though the fork is not drawn. The air column is closed below by the water. Ignore the orange curves and the formulas for now, okay? Now we lower the water level slowly, and the sound becomes loud at certain lengths only. Let me ask you this: why is it loud only at those lengths? It is loud because only then does a natural frequency of the air column equal the frequency of the fork. The column is driven at resonance, so it vibrates with a large amplitude.

1:11Inside the column: node, antinode, end correction

Clear? Inside the column, a stationary longitudinal wave is set up, so the air moves up and down along the tube. The orange curves only show how large that movement is at each height. At the bottom, those curves close to a single point at the water. So tell me, is the water surface a node or an antinode? Yes, exactly, it is a node, a displacement node, which is why the curves close to a point there. Right? At the top, the curves open out at the upper dashed line, not at the mouth of the tube. So the antinode lies a little above the open end, okay? That extra length is the end correction, small e, the gap between the two dashed lines. It is approximately, that is the wavy equals sign, zero point three times d, the same as zero point six times r. Here d is the internal diameter of the tube, and r is its radius. Clear?

2:05The first two resonances

Take the first resonance, on the left. The arrow marks l one, the length from the open end down to the water, and the small one below the l only labels this resonance. From a node to the next antinode is a quarter of a wavelength, written lambda, this Greek letter. Here that distance is l one plus e, which equals lambda over four, okay? Now the second resonance, on the right, with length l two. From the water upwards we count node, antinode, node, antinode, which makes three quarters of a wavelength. The sketch is not to scale, so go by this count. So l two plus e equals three times lambda, over four. Right? We want lambda, but small e sits in both lines. So what would you do to get rid of it? Yes, exactly, you subtract the first line from the second, and small e cancels. That leaves l two minus l one equals lambda over two, so successive resonances are lambda over two apart. With me? So lambda equals two times the bracket, l two minus l one. Now small v, the speed of sound, is f times lambda, so v equals two times f, times the same bracket. This is the heart of the experiment, and the figure shows it too. Clear?

3:24Why two resonances, and the three-times trap

Now, one resonance position would already give the wavelength. So why do we use two? We use two because the first position alone gives lambda over four only if the end correction is known. Taking the difference removes it. Got it? Now here is a trap. One length goes with a quarter of a wavelength, and the other with three quarters. So is l two exactly three times l one? No, it is not, so be careful. It is l two plus e that equals three times the bracket, l one plus e, so the end correction spoils the neat three times. Right? But that line gives us a bonus. Rearrange it, and small e equals, l two minus three times l one, all over two. So the same two readings also give the end correction, as the figure shows. With me?

4:17Doing it properly in the lab

Now some care in the lab. Strike the fork on a rubber pad, and hold it horizontally, okay? Its prongs should vibrate in a vertical plane, just above the mouth of the tube, without touching it. And find each resonance position both while lowering the water level and while raising it, then take the mean.

4:37The speed you find is at room temperature

Clear? Your result is the speed at room temperature only, because small v is proportional to the root of capital T, the absolute temperature. The curly sign on screen means proportional to, okay? Near room temperature, v with a small t below it, the speed at small t degrees Celsius, is approximately v nought, plus zero point six one times small t. Here v nought, with the small zero, is the speed at zero degrees Celsius. On screen, m s to the power minus one means metres per second, and the little circle with capital C means degrees Celsius. With me? And one small fact to keep in mind: moist air carries sound slightly faster than dry air.

5:17Example 1: 512 Hz fork, resonances at 16.0 cm and 49.0 cm

Now the examples, the easier one first. With a fork of five hundred and twelve hertz, the H z on screen, the first two resonances are found at sixteen point zero centimetres and forty nine point zero centimetres. Find the speed of sound. Pause the video and try it first. Done? Okay, let's check. So l two minus l one is forty nine point zero minus sixteen point zero, which gives thirty three point zero centimetres, okay? So is the wavelength simply thirty three point zero centimetres? No, it is not, and this is the trap. Successive resonances are only lambda over two apart, so lambda is two times thirty three point zero, which gives sixty six point zero centimetres. Right? Now use the formula from the figure, v equals two times f, times the bracket, with thirty three point zero centimetres written as zero point three three zero metres. Two times five hundred and twelve times zero point three three zero gives about three hundred and thirty eight metres per second. And that is our puzzle solved, with no stopwatch anywhere. Clear?

6:26Example 2: 480 Hz fork, one resonance, tube diameter 3.0 cm

Now the harder one. Here, a fork of four hundred and eighty hertz gives the first resonance at seventeen point zero centimetres, in a tube of internal diameter three point zero centimetres. Find the speed of sound. Pause the video and try it first. Done? Okay, let's check. There is only one resonance this time, so nothing cancels. So is lambda over four simply seventeen point zero centimetres? No, it is not, because the antinode lies above the open end, and that is why the question gives the diameter. Small e is approximately zero point three times d, so zero point three times three point zero gives zero point nine centimetres. With me? Now lambda over four is l one plus e, and seventeen point zero plus zero point nine gives seventeen point nine centimetres. Right? Four times that gives lambda, seventy one point six centimetres, which is zero point seven one six metres. Clear? Finally, v equals f times lambda. Four hundred and eighty times zero point seven one six gives about three hundred and forty four metres per second. If you got this on your own, very well done!

7:40Quick revision: v = 2f(l2 - l1) and the end correction

So, quick revision. First, l one plus e equals lambda over four, and l two plus e equals three times lambda, over four. Two resonances remove small e, and v equals two times f, times the bracket, l two minus l one. And the end correction is about zero point three times d, or from the readings, l two minus three times l one, all over two. Now go and try the practice problems on this page. All the best!