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# viewnote10 - PHYS 302 Unit 10 Viewing Notes Athabasca...

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PHYS 302 : Unit 10 Viewing Notes – Athabasca University 1 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 1 ( , ) sin( )cos n n n n y x t B k x t , with n n L k and n n vk . If we are working from an initial condition, then t =0 and 2 1 2 3 ( ,0) sin( ) sin( ) ... x x L L y x B B B (2:45). Fourier analysis can show us how each B n corresponds to part of the initial condition, but first we need to discuss some general considerations about Fourier series. For any single-valued regular function a series can be formed 0 2 1 1 ( ) cos sin A m m m m f x A mx B mx , where the first, second, and third parts are singled out for discussion (4:45). We have to find the values for the A m and the B m , 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(- m π ) is -1 for any positive integer m .

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viewnote10 - PHYS 302 Unit 10 Viewing Notes Athabasca...

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