Graphs and Errors with the Resonance Tube

Transcript

0:00Introduction

Hello everyone, here is a small puzzle. One student uses a five hundred and twelve hertz fork, and carefully finds the first resonance at sixteen centimetres. Four times the frequency times this length gives three hundred and twenty eight metres per second, but the correct value is three hundred and thirty eight. This method comes out low every time. Reading errors can go either way, so why is this one always on the same side? You will answer that yourself. These questions do come up in JEE, so let's begin.

0:35The first resonance, with the end correction

Look at this figure from the previous lesson. Above the water, the curved lines show the air vibrating. The arrows mark l one and l two, small l with a small one or two below it, the lengths at the first and second resonance. The gap between the dashed lines is e, the end correction. Beside the tubes are that lesson's results, with lambda, the Greek letter, for wavelength, d for internal diameter, and the wavy sign for nearly equal. At the first resonance, the air column plus e makes a quarter wavelength, so l one plus e equals lambda over four. Since v, the speed of sound, equals f, the frequency, times lambda, this is v over, four times f, clear? Taking e across, we get l one equals v over, four times f, minus e, and the whole graph comes out of this. With me?

1:28Plotting l₁ to get a straight line

Now use the same tube with forks of different frequency, and measure l one each time. To get a straight line, should we plot l one against f, or against something else? If you said one over f, very good, and here is why. We write l one equals v over four, times one over f, minus e, which looks just like y equals m times x, plus c, so we plot l one against one over f, right? So what is the slope of our graph, and what is the intercept? The slope is whatever multiplies one over f, so it is v over four, and four times the slope gives the speed of sound. Notice that e does not affect the slope, okay? The intercept on the length axis is minus e, so the line cuts that axis below zero. Be careful, though, because e itself is a positive length, the size of that intercept. Clear?

2:24The error in the speed of sound

Now the speed itself comes from two resonances, because that removes the end correction. So v equals two times f, times the bracket, l two minus l one. Got it? Here is a trap: the lengths are subtracted in that bracket, so do their reading errors get subtracted too? No, no, careful, they do not subtract, because for a difference the absolute errors add. This little triangle, the Greek capital delta, means the error in that quantity. So delta v over v equals delta f over f, plus, delta l one plus delta l two, all over l two minus l one. With me? Now, if each length is read to the same one millimetre, which fork gives the smaller percentage error from the lengths, low frequency or high frequency? Yes, exactly, it is the low frequency fork, because l two minus l one is lambda over two, and a long wavelength makes that gap larger, so the length term is smaller. But the tube must be long enough for its second resonance. Right?

3:25Back to our puzzle: always too small

Now back to our puzzle: look at the left tube. The student measured only up to the open end. So why must the answer always come out too small? Very good if you said the end correction, because the true quarter wavelength reaches the top dashed line. So it is l one plus e, and the correct speed is four times f, times the bracket, l one plus e, okay? Ignoring e gives v equals four times f times l one. Since e is positive, the length used is always short, so this speed is always too small. That is a missing piece of length, not a reading slip. Clear? With our numbers, four times five hundred and twelve times zero point one six zero metres gives about three hundred and twenty eight metres per second, written m s to the power minus one. Right? The forgotten end correction was zero point five centimetres. With it, the speed is three hundred and thirty eight metres per second, so three hundred and twenty eight is about three per cent too low. Got it?

4:27Example 1: speed of sound with its error from two resonances

Now the examples, the easier one first. The fork is five hundred and twelve hertz, plus or minus one hertz, so it can be off by one hertz either way. The resonances are at sixteen point two and forty nine point eight centimetres, each read to zero point one centimetre. Find the speed of sound with its error. Pause the video and try it first. Done? Okay, let's check. The difference is thirty three point six centimetres, so v is two times five hundred and twelve times zero point three three six metres, about three hundred and forty four metres per second, okay? Now, is the error in l two minus l one just zero point one centimetre, like each reading? Oh no, careful, it is not, because each reading brings its own error, so zero point one plus zero point one gives zero point two centimetres. Clear? Now put this in our formula. The fork gives one over five hundred and twelve, about zero point two per cent. The lengths give zero point two over thirty three point six, about zero point six per cent. Together, that is zero point eight per cent. Right? Now zero point eight per cent of three hundred and forty four is nearly three. So v equals three hundred and forty four, plus or minus three, in metres per second. If you used zero point two centimetres for the length error, very good!

5:53Example 2: v and end correction from 256 Hz and 512 Hz forks

Now the harder one, back to our straight line. The first resonance occurs at thirty two point three centimetres with a two hundred and fifty six hertz fork, and at fifteen point seven centimetres with a five hundred and twelve hertz fork, in the same tube. Find the speed of sound and the end correction. Pause the video and try it first. Done? Okay, let's check. With v and e both unknown, what stays the same when only the fork changes? Yes, exactly, the tube and the air stay the same, so e is the same, and so is the speed of sound. So we write our line twice, in metres, once for each fork. With me? Both equations have the same e, so how do we get rid of it in one stroke? That is correct, we subtract, and e cancels, leaving the difference of the lengths on the left. On the right, v over four comes out, times a bracket that is just one over five hundred and twelve, so we get v over two thousand and forty eight, right? That left side is zero point one six six, so v is that times two thousand and forty eight, about three hundred and forty metres per second, okay? For e, use the five hundred and twelve hertz equation. Its first term is exactly the zero point one six six we just found. So e is zero point one six six minus zero point one five seven, which gives zero point zero zero nine metres, or zero point nine centimetres. If you got both, very well done!

7:25Quick revision: l1 against 1/f, error in v, why 4fl1 is low

So, quick revision. The graph of l one against one over f is a straight line of slope v over four, with intercept minus e on the length axis. In the error formula, delta f over f adds to the length term, where the two length errors add, over l two minus l one. And the first resonance alone, ignoring e, gives four times f times l one, which is always too small. Now go and try the practice problems on this page. All the best!