Beats

Transcript

0:00Introduction

Hello everyone! Let's start with a small puzzle. Take a tuning fork of two hundred and fifty six hertz, and a second fork whose frequency we do not know. Strike them together, and the sound goes loud, soft, loud, soft, four times every second. With this throbbing and a little wax, can we find the second fork's frequency? We can, and by the end of this video, you will. This throbbing is called beats, and it comes up often in JEE, so let's begin.

0:31What beats are

So, take two sounds of nearly equal frequencies. We call them f sub one and f sub two: the f with a small one below it, and the f with a small two below it. And each sound has the same amplitude, capital A, okay? When the two sounds add up, here is y, the combined displacement, at time t. y equals two times capital A, times cos of, pi times f sub one minus f sub two, times t, times sine of, pi times f sub one plus f sub two, times t. Right? Now look at the two parts. The frequencies are nearly equal, so their difference is small, but their sum is large. So tell me, which part changes fast, and which part changes slowly? The sine part changes fast, yes, exactly, because it has the sum. That fast part is the tone we hear. The part with two times capital A times cos has the small difference, so it changes slowly. It acts as the amplitude, so the sound swells and fades. Clear? Now, careful. The slow cos part has a frequency of only one half of, mod f sub one minus f sub two. The two straight bars mean the modulus, the positive value of the difference. So the number of beats per second is also half the difference, right? No, no, careful, that is the trap! The sound is loudest when the cos is plus one, and also when it is minus one; the plus or minus sign on screen means both. So each cycle of the cos gives two loud moments. That is why the beats per second equal the modulus of f sub one minus f sub two, the full difference. Got it?

2:17Beats from two wavelengths

Sometimes a question gives wavelengths instead of frequencies. Lambda, the Greek letter like an upside down y, is the wavelength, and v is the speed of sound. If both sounds travel in one medium, they have the same speed v, and the frequency f equals v over lambda. Okay? So, now we have two wavelengths, lambda sub one and lambda sub two. How many beats per second do they give? Just find each frequency and subtract them. The beats per second equal v times the modulus of, one over lambda sub one, minus one over lambda sub two. And remember the condition: one medium, so one speed. With me?

3:00Making a frequency go up or down

Now, how do we change a frequency a little? We stick some wax on a prong of a tuning fork; a prong is one of its two arms. Does the frequency go up, or down? It goes down, yes, exactly! Loading a prong with wax lowers its frequency. Here is a small trick to remember it: loading lowers, and both words start with the same letter. Okay? And filing the prongs does the opposite: it raises the frequency. Clear? For a string, raising the tension raises its frequency. Tighten the string, and its frequency goes up.

3:37Finding an unknown frequency

Right? Now the main skill. An unknown sound gives n beats per second with a known frequency, f sub zero, the f with a small zero. Then the unknown frequency f equals f sub zero, plus or minus n. Okay? But is it plus, or minus? Can the number of beats alone tell us which one? It cannot, because the beats give only the size of the difference. So there are two candidates: f sub zero plus n, or f sub zero minus n. Right? To decide, the question changes one source a little, in one of the ways we just saw. Apply that same change to both candidates. Got it? Then keep the candidate that gives the new beat rate, and reject the other one. That is how the sign is decided. Clear?

4:26Example 1: beats from wavelengths 1.00 m and 1.01 m

Now our first example. Sounds of wavelengths one point zero zero metres and one point zero one metres travel at three hundred and thirty metres per second; the small minus one on the s means per second. How many beats are heard per second? Pause the video and try it first. Done? Okay, let's check. The wavelengths differ by zero point zero one metres. So the answer is three hundred and thirty over zero point zero one, right? No, no, wait, that is the trap! That gives thirty three thousand beats per second, which is nonsense. Beats come from the difference of the frequencies, not the difference of the wavelengths. With me? So, find each frequency. f sub one equals three hundred and thirty over one point zero zero, which gives three hundred and thirty hertz, the capital H and small z on screen. f sub two equals three hundred and thirty over one point zero one, which gives about three hundred and twenty six point seven hertz; the wavy equals sign means approximately. Okay? Now the beats: three hundred and thirty, times, one minus one over one point zero one. That gives about three point two seven, so about three point three beats per second. Always subtract the frequencies, never the wavelengths. Clear?

5:52Example 2: unknown fork frequency using beats and wax

And now, remember our puzzle from the start? Let's solve it. Fork capital A, of two hundred and fifty six hertz, gives four beats per second with fork capital B, the unknown one. When fork B is loaded with wax, the beat rate falls to two. Find the frequency of the unknown fork. Pause the video and try it first. Done? Okay, let's check. Four beats per second, so f sub capital B, the frequency of fork B, equals two hundred and fifty six, plus or minus four. That is two hundred and sixty hertz, or two hundred and fifty two hertz. With me? Now, wax lowers the frequency of fork B, the one we loaded. So it must be the lower one, two hundred and fifty two hertz, right? Oh no, careful, that is the trap! Let us test it. Wax lowers f sub B, so say it takes two hundred and fifty two down to two hundred and fifty; the arrow means changes to. The difference from two hundred and fifty six is now six beats, not two. Lowering it only moves it further from two hundred and fifty six, so the beats go up, not down. Okay? Now test two hundred and sixty. Say wax lowers it to two hundred and fifty eight. The difference from two hundred and fifty six is two, exactly the new beat rate. Got it? So the unknown fork has a frequency of two hundred and sixty hertz. See, the throbbing and a little wax gave us the whole answer, just as we promised at the start.

7:31Quick revision: beat formula, wax, test both candidates

So, quick revision. First, the beats per second equal the modulus of f sub one minus f sub two, not half of it. From two wavelengths in one medium, it is v times the modulus of, one over lambda sub one, minus one over lambda sub two. Second, wax lowers the frequency, while filing and more tension raise it. Third, an unknown frequency is f sub zero plus or minus n, so change one source and test both candidates. Now try the practice problem on this page. All the best!