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Unformatted text preview: PHYS 302: Unit 10 Viewing Notes – Athabasca University PHYS 302: Vibrations and Waves – Unit 10 Viewing Notes
You may note that this is MIT lecture 11 and that the recording date is 12 days after the previous lecture included in
PHYS 302. Lecture 10 was the Exam Review provided as Supplementary Material in Unit 9, and presumably MIT
has a formal exam week during which there are no lectures.
The topic of this lecture is Fourier analysis. It turns out that our known ability to represent an arbitrary vibration in
terms of normal modes has already shown us the equivalent of Fourier analysis (2:10), but without calling it that.
We wish to examine how such a representation helps us to deal with initial conditions. In representing the vibratory
displacement as a superposition of normal modes we have y ( x, t ) B
n 1 n sin(kn x) cos nt , with kn n
L and n vkn . If we are working from an initial condition, then t=0 and y( x, 0) B1 sin( Lx ) B2 sin( 2 x ) B3 ...
L
(2:45). Fourier analysis can show us how each Bn corresponds to part of the initial condition, but first we need to
discuss some general considerations about Fourier series. For any singlevalued regular function a series can be
A0
2 formed f ( x) m 1 m 1 Am cos mx Bm sin mx , where the first, second, and third parts are singled out for discussion (4:45). We have to find the values for the Am and the Bm, given an initial value for the function at all
points. This will involve integral calculus again. If you do not have a deep acquaintance with integral calculus, you
should follow these viewing notes carefully. One important aspect of the definite integral is that it reflects the
average value of a function. The average value of a sequence of N numbers is their sum divided by N. Similarly, the
average value of a function is its definite integral over an interval, divided by the length of the interval. Consider the
interval π to π. Over this interval, the second term has cos functions, and cos(mπ) is 1 for any positive integer m.
These all end up at cos(mπ), which is 1, but they do it perfectly symmetrically, so the average value over this
interval is 0. You should sketch this. Similarly, in the third part the sin curves all begin and end at 0, with at least
one bump and valley in between, but the areas of the bump(s) and valley(s) are equal, so the average value is
again 0. This can also be generalized by saying that sin is an odd function, with sin(x)=sin(x). So integrating, or
taking the average value, over any interval centred on 0, must give 0. You should sketch this too. We are left with A
0 2 f ( x)dx dx 2 A0
2 A0 (6:30). This is the average value (constant A0
2 ) times the length of the interval (2π), so it is consistent with the average value’s relation to a definite integral mentioned above. We can thus f ( x)dx . To get the other A values, consider the integral f ( x) cos nxdx , with n a positive integer. The first part gives cos nxdx cos nxdx , which is 0, for the reason explained. The third part gives cos nx sin mxdx . You can convince yourself that this is 0 in several ways. First (and easiest), cos*sin is write A0 1 A0
2 A0
2 an odd function for any value of n or m (recall that both are positive integers), so right away that gives 0 on
integration. Second, you could make a sketch for a few cases and see how it works. Third, you could use the
complex exponentials, since exponentials are easy to integrate. Finally, you could use trig identities. The most
interesting case here is the second part, which has cos nx cos mxdx . It may not be obvious, but if m≠n, this is 0. The most likely way to show this is with integration of complex exponentials. However, if m=n, we have cos 2 nxdx . Obviously this cannot be zero, since the argument is always positive. Dr. Lewin suggests that you do this integral with complex exponentials, although if your memory is good, trig identities can help. You should get cos 2 nxdx (9:10). Thus, 1
Am f ( x) cos nxdx A m cos mx cos nxdx Am , or 1
f ( x) cos mxdx . By a similar procedure, we find that Bm 1 f ( x) sin mxdx . PHYS 302: Unit 10 Viewing Notes – Athabasca University The formal mechanism for Fourier analysis is found in the three equations A0 1
Am 1
f ( x) cos mxdx , and Bm 1 f ( x)dx , f ( x) sin mxdx . If we let m start at 0, we can forget the first of these (12:30). We can think of these as simple recipes, and use them as needed. The simplest case of a plucked
string is unusual and extreme. We let the whole string be lifted up to a value a over its whole length from 0 to L.
Since this action is not periodic, we imagine that the function extends from 0 to 2L and is of value –a between x=L
and x=2L. The period is 2L and the average value (and thus A0) is 0. We must convert linear measure to radians.
Since one period of 2L has to be 2π radians, the formulas must be modified to be y ( x) f ( x) Am 1
L 2L 0 A0
2 m 1 m 1 Am cos mx Bm sin mx , and the (two) recipes must be modified to
L
L f ( x) cos mx dx and Bm L 1
L 2L 0 f ( x) sin mx dx . For insight, the terms are estimated roughly and
L overplotted on the function. It is immediately clear that the cos terms cannot represent the function well, since cos is
an even function, while the function to be represented is odd (25:30). All the A values must be zero! The sin terms
(at least the first one) can represent the function well. However, the second term (B2) contributes in the wrong
places, as do all even Bn, so all of these terms must also be zero! Using the recipe Bm a
L L 2L a
sin mx dx L sin mx dx , where the integral was broken in two where the function changes value.
L
L
L 0 We can do this much like we can break a sum into two sums. We can do these integrals using the antiderivative
d
(indefinite integral). We recall that the derivative dx cos x sin x . This means that the antiderivative sin xdx cos x . Here, mL , so sin
1 mx
L difference of the antiderivative between the end points, so
is odd, cos mx
L cos mx
L 2L
L a
Bm L L
0 dx mL cos mx . The definite integral is the
L
a
Bm L L
m L a
cos mx 0 L
L L
m cos mx
L 2L
L (31:55). If m cos m L cos 0 cos m 1 1 1 2 . Also,
L cos mL2 L cos m L 1 (1) 2 . In this case (m is odd), in total,
L
L
m a
(2) L L
m 2 4a
m . However, if m is even, then the cos terms are zero in all cases. Thus, so is Bm. The French text gives a special case by Eq. 632; what is given in the lecture is more general. The final result is that
y( x, 0) 4a [sin Lx 1 sin 3 x 1 sin 5 x ...] (36:50). These terms can be drawn in, and the first two terms
3
L
5
L
look like they will start to reproduce the function, with B1≈1.27a, B3≈0.42a, etc. For a convincing demonstration, the
terms can be added in on an oscilloscope. The break and exam discussion continue (to 44:50).
Now, it is possible to record the time development of the string by multiplying in the time dependence of each
normal mode, giving (with y ( x, t ) 4a m vkm v mL and v T ) [sin Lx cos 1t 1 sin 3 x cos 3t 1 sin 5 x cos 5t ...] . The time development may be
3
L
5
L unexpected at first. If a triangular pulse is released, it travels out in both directions. Demos show how this works,
including all the harmonics oscillating, each with its own frequency.
(1:00:00) We can consider the energy in a signal made up of propagating waves. If we take A2+B2 at each frequency
f, we would get a power spectrum showing how much energy is present at each frequency. The Fast Fourier
Transform (FFT) can perform this spectral analysis with a digitized realtime signal on a computer. If enough data is
available, the FFT can extract power information even from a noisy signal such as that of Xrays from a pulsar.
From Fourier analysis of the data shown in the video lecture, one can deduce 401 Hz rotation of an object of about
the mass of the Sun (a neutron star). The FFT is a powerful and universallyused technique in signal analysis. 2 ...
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This note was uploaded on 02/18/2011 for the course PHYS 320 taught by Professor Martinconner during the Spring '10 term at Open Uni..
 Spring '10
 martinconner

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