Superposition, Interference, Reflection and Transmission

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. On a long string, send a pulse bulging upwards from the left end, and an equal pulse bulging downwards from the right end. They run towards each other, and at one moment they overlap exactly. At that moment the string is perfectly straight, as if there were no wave at all! Yet a moment later, both pulses appear again and run on. So, at that straight moment, where is the energy hiding? Keep your answer in mind; we will settle it today. Questions on this come up often in JEE, so let's begin.

0:39Superposition and interference

So, what happens when two waves meet on a string? Each point takes both displacements at once, and they simply add. Small y, the displacement, equals y sub one plus y sub two; the small one and two at the bottom tell us which wave. This is the principle of superposition. Okay? Now take two waves of the same frequency, with amplitudes capital A sub one and capital A sub two. When they overlap like this, we call it interference. Let me ask you this: is the new amplitude simply capital A sub one plus capital A sub two? Not always; it depends on the phase difference delta, the Greek letter like a small d with a curl. Capital A equals the square root of, capital A sub one squared, plus capital A sub two squared, plus two times capital A sub one times capital A sub two times cos delta. The tick-shaped sign is the square root, and cos is the cosine. Right? So when the waves are in phase, delta is zero, and cos zero is one. Then capital A equals capital A sub one plus capital A sub two, and the waves fully help each other. Got it? And when delta is pi, the familiar Greek letter, the waves are in opposite phase. Cos pi is minus one, so capital A equals the modulus of capital A sub one minus capital A sub two; the two straight bars mean modulus, the size without the sign. Clear? For the intensity, capital I, the formula has the same shape. Capital I equals capital I sub one, plus capital I sub two, plus two times the square root of capital I sub one times capital I sub two, times cos delta. But careful: here the root sits only on the last term.

2:30Reflection at the end of a string

Okay? Next, send one upward pulse towards an end tied firmly to a wall. Does it come back the same way up, or upside down? It comes back upside down, yes, exactly! At a fixed end, or when the wave meets a denser string, the reflected pulse is inverted. That is a phase change of pi, which is half a cycle. Right? And at a free end, one that can move up and down freely, or when the wave meets a lighter string, there is no inversion. So remember it like this: fixed or denser, it flips; free or lighter, it does not. Got it?

3:09At a joint: part reflected, part transmitted

Now, the middle case: our string is joined to a second string. The wave arriving at the joint has amplitude capital A sub i, small i for incident. It runs at speed v sub one on the first string, and v sub two on the second. At the joint, part goes on, and part bounces back. With me? The part that goes on is the transmitted wave, and a small t marks it. Capital A sub t equals two times v sub two, over, v sub one plus v sub two, times capital A sub i; the long line is a fraction bar. Okay? The part that bounces back is the reflected wave, and a small r marks it. Capital A sub r equals v sub two minus v sub one, over, v sub one plus v sub two, times capital A sub i. Because of the minus sign on top, it can be negative, which means the reflected pulse is inverted. Right? So let me ask you this: when does capital A sub r come out negative? Yes, exactly, it is negative when v sub two is less than v sub one; the pointed sign means less than. And v equals the square root of capital T over mu, the tension over the mass per unit length; mu is the Greek letter like a u with a tail. So a heavier string under the same tension is slower, and it reflects inverted, just like a denser end. Clear?

4:34Back to our puzzle: equal and opposite pulses

Now, back to our puzzle. Two equal and opposite pulses overlap exactly, so at every point, y sub one plus y sub two is zero. Okay? So, for that one moment, the string is perfectly straight. But energy cannot just vanish, so where is it? It is all kinetic energy, exactly! The string looks straight, but its points are still moving, and that motion brings the two pulses back a moment later. So our puzzle is solved. With me?

5:07Example 1: 3 mm and 4 mm waves with phase difference pi/2

Example one: waves of amplitudes three and four millimetres, with a phase difference of pi over two, overlap. Find the resultant amplitude. On screen, m m means millimetres. Pause the video and try it first. Done? Okay, let's check. Many students write three plus four, seven millimetres, straight away. Is that right? No, careful, that is the trap! Here delta is pi over two, and cos of pi over two is zero, so the last term under the root drops out. Okay? So capital A is the square root of, three squared plus four squared, that is nine plus sixteen, which is twenty five. Its square root is five, so the answer is five millimetres, not seven; seven needs the waves in phase. Right?

5:59Example 2: reflected and transmitted amplitudes at a joint

Now a slightly harder one. Example two: a string is joined to a heavier one, with four times its mass per unit length, under the same tension. So a wave of amplitude capital A comes in from the light string. Find the reflected and transmitted amplitudes. Pause the video and try it first. Done? Okay, let's check. With four times the mass per unit length, is the new speed a quarter of the first? No, careful, mu sits under a square root! Mu sub two is four times mu sub one, so v sub two is the square root of capital T over four mu sub one. That is v sub one over two, just half. Clear? In both formulas, v sub one cancels top and bottom, so take v sub one as one and v sub two as one half; their sum is three halves. Then capital A sub t equals two times one half, over three halves, times capital A, which gives two thirds of capital A, okay? Now, before we work out capital A sub r, predict its sign. The wave is going into a heavier string. Will the reflected pulse be inverted? Yes, exactly, it will be inverted, because the wave enters a slower string, just like a denser end. Capital A sub r equals one half minus one, over three halves, times capital A, and minus one half over three halves gives minus one third. The minus sign confirms that the reflected pulse is inverted. Got it?

7:36Quick revision: resultant amplitude, reflection, junction

So, quick revision. The resultant amplitude is the square root of, capital A sub one squared, plus capital A sub two squared, plus two times capital A sub one times capital A sub two times cos delta. At a fixed or denser end the reflected pulse is inverted; at a free or lighter end it is not. And at a junction, both amplitudes have v sub one plus v sub two below; on top, two times v sub two for the transmitted wave, and v sub two minus v sub one for the reflected wave. Now go and try the practice problem on this page. All the best!