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# hw2sol - HW 2 Physics 1B Spring 2010 C M Cooper Teaching...

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HW 2: Physics 1B Spring 2010 C. M. Cooper, Teaching Assitant W. Gekelman, Professor April 12, 2010 Abstract This hw covers all of chapter 15. Topics include Traveling and Standing waves. Make sure you can do the two wave addition problems with phasors. When doing the rest of these problems, start getting used to finding the equations that link what you need to what you got. Remember, the summary of these equations is what you will need and use on the exams. Please note that the graph for problem 6 appears halfway through problem 5. Latex does what it wants sometimes. 1 Question 15.22: Investigating a UFO This is a one step problem that uses the intensity relation. Remember, this equation has to do with the definition of intensity of a point source spreading out over a sphere. I = P A = P 4 πr 2 (1) Since the sum of all the power over a sphere must be a constant at all distances, we get the relation P = 4 πI 1 r 2 1 = 4 πI 2 r 2 2 . Solving for r 2 gives r 2 = r 1 r I 1 I 2 = 7 . 5 m s 0 . 11 W/m 2 1 . 0 W/m 2 = 2 . 5 m (2) Remember, this is the new radius, so the amount you can move forward is r 1 - r 2 = 7 . 5 m - 2 . 5 m = 5 . 0 m (3) 2 Question 15.28: Interference of Triangular Pulses This is literally just adding the segments together. If you can’t see this visually, write out the segments of the pulses and add them piecewise. 1

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Figure 1: Try to visualize the pulses sitting on top of each other and add the values at the individual points 3 Question 15.36: Standing wave derivation What an excellent opportunity to use PHASORS . Remember, the advantage is that you have a better intuition for manipulating exponentials than you do sin’s and cos’s. It might be a little more work but you don’t need to look anything up. This is useful on an exam. Here is the derivation using them We will make use of the fact that e iωt + e - iωt = cos( ωt ) + i sin( ωt ) + cos( - ωt ) + i sin( - ωt ) = = cos( ωt ) + i sin( ωt ) + cos(+ ωt ) - i sin(+ ωt ) = 2 cos( ωt ) (4) and similarly e iωt - e - iωt = cos( ωt ) + i sin( ωt ) - cos( - ωt ) + i sin( - ωt ) = = cos( ωt ) + i sin( ωt ) + - cos(+ ωt ) + i sin(+ ωt ) = 2 i sin( ωt ) (5) therefore we can write - cos( kx + ωt ) + cos ( kx - ωt ) = 1 2 ( - e i ( kx + ωt ) - e - i ( kx + ωt ) ) + ( e i ( kx - ωt ) + e - i ( kx - ωt ) ) = 1 2 - e ikx e iωt - e - ikx e - iωt + e ikx e - iωt + e - ikx e iωt = 1 2 e ikx ( e - iωt - e iωt ) + e - ikx ( e iωt - e - iωt ) = 1 2 ( e ikx - e - ikx )( e - iωt - e iωt ) = 1 2 2 i sin( kx ) - 2 i sin( ωt ) = 2 sin( kx ) sin( ωt ) (6) the trigonometric solution is trivial and is only for punks who aren’t cool enough to use phasors.
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