Superposition of SHMs

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Give one particle three simple harmonic motions at once, with the same amplitude and the same frequency. Three pushes, so it should swing three times as far, right? No, not necessarily, because with the right phases, it does not move at all! This comes up often in JEE, and the secret is to add the motions like vectors. So, let's begin.

0:27Same line, same frequency: add them like vectors

First case: two motions on one line, with the same angular frequency omega, the Greek letter like a rounded w. x sub one equals capital A sub one times sine omega t, and x sub two equals capital A sub two times sine of, omega t plus delta, in brackets. The amplitudes are capital A sub one and capital A sub two, with the small subscripts as labels, and delta, the Greek d, is the phase difference. With me? Here is the key idea. Think of each motion as a rotating vector, called a phasor, of length capital A sub i, at the angle delta sub i; the small i just says which motion. And we add the phasors like vectors. Okay? Both phasors rotate together, so the angle between them stays fixed and the sum is again an SHM, a simple harmonic motion, with the same omega. x equals x sub one plus x sub two, equals capital A times sine of, omega t plus phi; phi, the Greek letter like a circle with a line through it, is the new phase. Right? Here is a trap question: is the new amplitude simply capital A sub one plus capital A sub two? No, no, careful, that holds only in one special case! We add vectors at an angle delta, so 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. Clear? And the phase, measured from the first motion: tan phi equals capital A sub two times sine delta, over, capital A sub one plus capital A sub two times cos delta. Got it? So when does simple addition work? For delta equal to zero, cos delta is one, and capital A is capital A sub one plus capital A sub two. For delta equal to pi, cos delta is minus one, and capital A is the modulus of, capital A sub one minus capital A sub two; the vertical bars mean modulus. Okay?

2:25Sine plus cosine: phasors at right angles

Now a common special case, a sine plus a cosine. Cos omega t equals sine of, omega t plus pi over two, so the two phasors are at right angles. With me? So cos delta is zero, and the middle term drops out. Small a times sine omega t, plus small b times cos omega t, equals the root of, small a squared plus small b squared, times sine of, omega t plus phi. Right? With delta equal to pi over two, the phase formula gives tan phi equals small b over small a. And the amplitude, the root of small a squared plus small b squared, is just Pythagoras. Clear?

3:01Different frequencies: never SHM, not always periodic

Next, let the two frequencies be different. Now the phasors rotate at different speeds, so the angle between them keeps changing, and so does their sum. So the sum is never an SHM, okay? But it can still repeat itself. It is periodic only if the ratio of the two frequencies is rational, and then its period is the least common multiple, the LCM, of the two periods. Got it? Let me ask you this. Is x equals sine omega t plus sine of pi omega t periodic? Very good if you said no, it is not periodic. The frequency ratio is pi, which is irrational, so no common period exists. Each term repeats on its own, but the sum never does, so be careful. With me?

3:46Perpendicular SHMs: the path of the particle

Last case: two motions at right angles, one along x and one along y. x equals capital A times sine omega t, and y equals capital B times sine of, omega t plus delta. This time, we want the path that the particle traces. Okay? If delta is zero, y over x is always capital B over capital A, a straight line through the origin. If delta is pi, a minus sign appears and the line tilts the other way; the plus minus sign covers both. Right? For delta equal to pi over two, y equals capital B times cos omega t, and since sine squared plus cos squared is one, x squared over capital A squared, plus y squared over capital B squared, equals one. That is an ellipse. Clear? Now, what does this ellipse become if the two amplitudes are equal? It becomes a circle, exactly! The ellipse has its axes along x and y, and with capital A equal to capital B, they are equal. Got it? For any other delta, the path is a tilted ellipse; but delta equal to three pi over two again gives an upright ellipse, traced the other way. With frequencies in the ratio one to two, the path depends on the phase; for x equals sine omega t and y equals sine two omega t, it is a figure of eight. These patterns are called Lissajous figures. Okay?

5:12Example 1: add 4 sin wt and 4 sin(wt + pi/3)

Example one, an easy one. Add x sub one equals four times sine omega t, and x sub two equals four times sine of, omega t plus pi over three. Pause the video, and try it first. Done? Okay, let's check. Both amplitudes are four, but the answer is not eight, since they are not in phase. Delta is pi over three, or sixty degrees, where the little circle means degrees, and cos sixty degrees is one half. With me? Now use the formula: sixteen plus sixteen, plus two times four times four, which is thirty two, times cos sixty degrees. That last term is sixteen, so we get the root of forty eight, which is four root three. Clear? Now let us find the phase. Two equal phasors form a rhombus, and its diagonal bisects the angle. Half of pi over three is pi over six. Right? So x equals four root three times sine of, omega t plus pi over six. One SHM again, nice and simple!

6:16Example 2: three equal SHMs 120 degrees apart cancel to zero

Now, remember our puzzle? Let's solve it. Three S H Ms of equal amplitude, small a, and the same frequency, differ in phase by two pi over three from one another. Find the resultant. Pause the video, and try it first. Done? Okay, let's check. Take the phases as zero, two pi over three and four pi over three. In degrees, that is zero degrees, one hundred and twenty degrees and two hundred and forty degrees. Okay? Add the first two with our formula. Cos one hundred and twenty degrees is minus one half, so under the root, small a squared plus small a squared minus small a squared leaves small a squared, and its square root is small a. The phasors are equal, so the sum bisects the angle, at sixty degrees. Clear? Now, the third phasor is at two hundred and forty degrees. Where does it point, compared with this resultant? It points exactly the opposite way, yes! Two hundred and forty degrees is sixty degrees plus one hundred and eighty degrees. So small a minus small a is zero. Got it? And here is the topper's view. Place the three phasors head to tail; each turns by one hundred and twenty degrees, so they close into a triangle. With me? So the resultant is zero, and the particle stays at rest. Three pushes and no motion, so our puzzle is solved! Right?

7:40Quick revision: phasor sum, rational ratio, line or ellipse

So, quick revision. Same line, same omega: add the phasors, with the cos delta formula. Different frequencies: never an SHM, and periodic only for a rational ratio. At right angles with the same omega: a line, an ellipse or a circle, decided by delta and the amplitudes. Now try the practice problems on this page. All the best!