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finalsoln_07 - Physics 9B-C Final Exam Solutions Cole UC...

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A A net ƒ = π /2 Phy 9B-C, Final Solutions Page 1 of 3 12/15/05 A A net = Physics 9B-C Final Exam Solutions Cole, UC Davis 20 points per problem/200 points total [1] (a) False (b) blue (c) right (d) minimum (e) 1/ r (f) dimmer (g) increase (h) least (i) black (j) Î S ≥
 0. [2] a) (8 pts) Plug the solution into the wave equation. First calculate the second derivatives: & 2 y t 2 = ω 2 A sin kx ω t ( ) Also, 2 y x 2 = k 2 A sin kx ω t ( ) . Plug into the wave equation: - 2 A sin( kx - t) + u 2 k 2 A sin( kx - t ) + å A sin( kx - t ) = 0 Divide by the A sin and solve for : ∑
 = u 2 k 2 + α which is the dispersion relation. b) (4 pts) v p = ω k = u 2 + α k 2 c) (4 pts) v g = d ω dk = d dk u 2 k 2 + α = 1 2 u 2 2 k ( ) u 2 k 2 + α õ v g = u 2 k u 2 k 2 + α d) (4 pts) Write the phase velocity in terms of ¬ using k = (2 π )/¬: v p = u 2 + α 4 π 2 λ 2 Thus, longer wavelengths will have a greater phase speed. [3] A) (5 pts) Match the boundary conditions. The free ends must have antinodes, and there must be a node at each support. The fundamental obeying the boundary conditions is shown to the right where, ¬ = L õ k = 2 π λ = 2 π L = 2 π 1.2 m õ k = 5.23/m. b) (5 pts) ∑
 = vk = (3000 m/s)(5.23/m) = 1.57 ª 10 4 /s c) (10 pts) For a standing wavefunction the possibilities are sin-sin, sin-cos, cos-sin, cos-cos.
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