Formation of Standing Waves

Transcript

0:00Introduction

Hello everyone! Let's start with a small puzzle. Tie one end of a long string to a wall, and keep shaking the other end steadily. Each wave runs to the wall, bounces back, and meets the waves still coming in. Shake it at just the right rate, and the string stops looking like a travelling wave. It swings up and down in loops, and some points do not move at all! So how can a point stay perfectly still, when two moving waves pass through it? Keep that puzzle in mind; we will settle it today. Standing waves come up often in JEE, so let's begin.

0:38Two waves running in opposite directions

So, what makes this pattern? Two identical waves, running in opposite directions. The first one is y sub one equals capital A times sine of, k x minus omega t. The small one at the bottom just labels the wave, x is the position along the string, t is the time, capital A is the amplitude, and omega, the Greek letter like a curly w, is the angular frequency. And k, the number multiplying x, fixes the wavelength; we will use it in a moment. The opposite signs of the x and t terms mean it runs towards plus x. Okay? Now the second wave is the same wave, running back towards minus x. Let me ask you this: what changes in its equation? Only the sign in the middle, that is correct! So y sub two equals capital A times sine of, k x plus omega t; same signs, so it runs towards minus x. Right? Now we add them, using the formula for sine C plus sine D, which turns a sum of two sines into a product. We get y equals y sub one plus y sub two, which is two capital A, sine k x, cos omega t, where cos is the cosine. Now look carefully: is there any k x minus omega t left inside it? So, is it still a travelling wave? No, it is not, and that is the key point! Sine k x depends only on where the point is, and cos omega t only on the time. So the shape does not slide along the string; it stands in place and swings. That is why we call it a standing wave. Got it? So every point oscillates as cos omega t, with the same frequency, but with its own amplitude: two capital A times the modulus of sine k x. The two straight bars mean modulus, so the amplitude is never negative. Clear?

2:28Nodes and antinodes

Now, back to our puzzle. Where is this amplitude zero, so that a point never moves at all? Wherever sine k x is zero, yes, exactly! Such a point never moves, and it is called a node. At a node, the two waves cancel each other at every instant, and that solves our puzzle. Okay? Sine k x is zero when k x is zero, pi, two pi, and so on; the three dots mean and so on. And k is two pi over lambda, where lambda, the Greek letter like an upside-down y, is the wavelength. So x equals zero, lambda over two, lambda, and so on. Right? Midway between two nodes, the modulus of sine k x is one, so those points swing the most, with amplitude two capital A, and they are the antinodes. Now tell me, how far apart are two neighbouring nodes? They are lambda over two apart, and so are two neighbouring antinodes. Many students say lambda here, so be careful. Got it? And since an antinode sits midway, a node to the next antinode is half of that, lambda over four. With me?

3:36How the loops move, and what happens at the ends

Now, how does the string move? The part between two adjacent nodes is called a loop, and all its points move in phase: they go up together, and come down together. So while one loop goes up, what does the neighbouring loop do? It comes down, that is right! Neighbouring loops move in opposite phase, because sine k x changes its sign as we cross a node. Clear? And the two waves carry equal energy in opposite directions, so the pattern goes nowhere, and no energy is carried along the string. That is a big difference from a travelling wave. Okay? Now, back to our string tied to the wall. The wave reflects from that fixed end. So at the wall, do we get a node or an antinode? We get a node, of course, because a fixed end cannot move! And when a wave reflects from a free end, an end that can move up and down, we get an antinode there. Got it?

4:33Example 1: wavelength, frequency, speed, amplitude, nodes

Now, an example. Example one: y equals zero point zero four, sine five pi x, cos one hundred pi t, in S I units, so metres and seconds. Find the wavelength lambda, the frequency f, the speed of the component waves, their amplitude, and the first three nodes from x equals zero. Pause the video and try it first. Done? Okay, let's check. Comparing with two capital A, sine k x, cos omega t, we get k equals five pi, and omega equals one hundred pi. Okay? So lambda equals two pi over k, which is two pi over five pi, and that gives zero point four metre; the small m means metre. Right? Next, f equals omega over two pi, which is fifty hertz, written H z. And each component wave runs at v equals omega over k, one hundred pi over five pi, which gives twenty metres per second, written m s to the power minus one. With me? Now the amplitude of the component waves. The number in front is zero point zero four, so is that our answer? No, no, careful here, because that is exactly the trap! That number is two capital A, the amplitude at an antinode. So capital A, the amplitude of each component wave, is half of it, which is zero point zero two metre, not zero point zero four. Clear? Finally, sine five pi x is zero when five pi x is zero, pi or two pi, so the nodes are at x equals zero, zero point two and zero point four metre. They are zero point two metre apart, which is lambda over two, just as we learnt. Got it?

6:14Example 2: component waves of a standing wave

Now a slightly harder one, and this time we go backwards. Example two: y equals six sine of pi x over three, cos forty pi t, with y in centimetres and t in seconds. Write down the two component waves. Pause the video and try it first. Done? Okay, let's check. We must break a product into a sum of two sines. Which formula from trigonometry does that? Two sine a cos b equals sine of, a plus b, plus sine of, a minus b, and that is the one! Here the first angle, a, is pi x over three, and the second angle, b, is forty pi t. Okay? So we write six as three times two, and the two sine a cos b sits ready inside. Right? That gives y equals three sine of, pi x over three plus forty pi t, plus three sine of, pi x over three minus forty pi t. The tall round brackets just hold each angle together. Clear? So the component waves are three sine of, pi x over three, plus or minus forty pi t, in centimetres, c m on screen. The plus-minus sign gives both waves at once, one running each way. And each has amplitude three, half of six, just like in example one. Got it?

7:33Quick revision: 2A sin kx cos wt, nodes, fixed and free ends

So, quick revision. Two equal waves running in opposite directions add up to two capital A, sine k x, cos omega t, a standing wave. Nodes are lambda over two apart, and from a node to the next antinode is lambda over four. And a fixed end gives a node, while a free end gives an antinode. Now go and try the practice problem on this page. All the best!