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Unformatted text preview: a period, except that the period is shifted by an amount of kx , and therefore the integral is not changed because of it. Therefore, dk dt = 1 2 v 2 y 2 m cos 2 ( kx t ) = 1 4 v 2 y 2 m We wont show this here, but dP dt = dK dt so that dE dt = 1 2 v 2 y 2 m Exercise : Find the above average by averaging over a wavelength (instead of averaging over a period). Change of linear mass density Usually the frequency f is determined by the external force. As the speed depends only on the hardware, not on the frequency, then = v f implies that increasing f results in a smaller . Consider a string with two parts: one with linear mass density 1 and the other with 2 . How is the wave changing? Assume that the wave propagates initially in the medium 1 . It is the local physics that determines the speed. Therefore, v 1 = 1 , v 2 = 2 . What about the interface? Continuity of the string requires that the string moves the same way on either side of the interface, i.e., the amplitude is unchanged ( f is also the same from the requirement of continuity at the interface; also, it is required by conservation of energy).requirement of continuity at the interface; also, it is required by conservation of energy)....
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This note was uploaded on 07/19/2011 for the course PH 113 taught by Professor Liorburko during the Spring '09 term at University of Alabama  Huntsville.
 Spring '09
 LIORBURKO
 Physics

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