02prob - Physics 9B-A Solution Set #2 Cole [Q15.13] The...

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Physics 9B-A Solution Set #2 Cole [Q15.13] The intensity of the wave is the average power per area perpendicular to the direction the wave is traveling. For the surface water wave the perpenidular area is a segment of a right circular cylinder centered at the point the rock hit the water. Area º = 2π rh. As wave expands, the radius grows, this area grows, and the energy in the wave front must be spread more thinly causing the amplitude to decrease as I á 1/r and I á A 2 õ A á 1/ r . [Q15.20] As you stretch a rubber band the frequency goes up! As a string (fixed both ends) is made longer the frequency of the fundamental goes down. f v L 1 2 = . For a rubberband, when it is stretched, the tension goes up. If we assume Hooke's law | F | = kx , the frequency should still decrease f kx x x 1 2 1 = / μ . The solution is to realize that the rubber band is not perfectly elastic but becomes stiffer as it is stretched: |F | = kx + cx 3 . The x 3 term is responsible for the frequency going up. [P15.26] In 15 ms the wave travels x = ( 40 cm/s)(0.015 s) = 6 mm and 2mm each .005 s after. a) The wave inverts at the reflection. In (b) the wave does not invert. [15.28] In 1/4 second each wave will travel 0.5 cm. After the first 1/4 s the two waves will be just touching. After 1/2 s they will be overlapping by 1/2, after 3/4 s they will be exactly on top of eachother which means the amplitude will be 2 cm, after 1 s they will only half overlap and after 1.25 s they will again be just touching at they move away from eachother. Phy 9B #2, 10/14/05 Page 1 of 4 UC Davis
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[15.43] Trick: Put all the things that change on one side of the equation and all the constants on the other. a) f v L 1 2 = Here, the tension in the string is constant so v is constant.
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02prob - Physics 9B-A Solution Set #2 Cole [Q15.13] The...

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