ph2a_final_soln

ph2a_final_soln - Plausible Final Questions Nate Bode...

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Unformatted text preview: Plausible Final Questions Nate Bode December 11, 2009 1 Tsunamis [15pts] (a) (3pts) We would expect the wavelength to be of the order as the scale length in the problem: L = 100 km. Moreover, the width of the gaussian will also be of the order of k , but the width of the gaussian is also of order the wavelength. Therefore, k k 2 2 L 6 . 28 10- 5 m- 1 . (1) Then ( kh ) 2 = 3 50 2 . 036. (b) (3pts) Using the appropriate values we note that gh 173 m / s. Then v ph ( k ) ( k ) k = q gh 1- 1 6 ( kh ) 2 172 m / s (2) (c) (4pts) Just using the definition of the group velocity we find: v g ( k ) k = k q gh k- h 2 6 k 3 ! = q gh 1- 1 2 ( hk ) 2 170 m / s (3) (d) (2pts) The time it will take to arrive at the beach, t arr , is just given by distance traveled velocity : t arr = D gh h 1- 1 2 ( hk ) 2 i D gh 1 + 1 2 ( hk ) 2 1 . 6 hr (4) (e) (3pts) The wave will spread by an amount given by t arr (difference in phase velocities for wavenumbers k and k + k ). x [ v ph ( k )- v ph ( k + k )] t D 2 ( hk ) 2 18 km . (5) 1 Comparing this to the wavelength of the wave we have that x L . 18 telling us that the wave has spread by about 18%. One should note that real tsunamis have wavelengths an order of mag- nitude larger and so have dispersions two orders of magnitude smaller (c.f. eqn 5). That is, real tsunamis do not experience significant dispersion and so their shape will remain constant as long as the depth remains constant. 2 Problem 2 by Jonathan Arnold December 11, 2009 Solution There is a string of length L and tension T stretched between two walls. The left third of the string has mass density 4 , while the right two-thirds has mass density . a. Write down the wave equation in each region in terms of ,T . We have: < x < L 3 : 2 y x 2- 4 T 2 y t 2 = 0 (1) L 3 < x < L : 2 y x 2- T 2 y t 2 = 0 (2) b. What are the boundary conditions at x = 0 , L 3 , and L ? There are 3 boundaries to consider. The conditions at each are x = 0 : y ( x = 0 ,t ) = 0 (3) x = L : y ( x = L,t ) = 0 (4) x = L 3 : y ( x = L 3 ,t ) and y ( x,t ) x x = L 3 continuous (5) for all t . c. Write down the general solution to the wave equation in each region. You can assume the initial string velocity is zero for our purposes, and make sure the boundary conditions at x = 0 and x = L are satisfied. (Hint: sin[ k ( L- x )] is a useful form of y ( x,t ) in the region L 3 < x < L .) 1 We can use < x < L 3 : y ( x,t ) = A 1 sin( k 1 x ) cos t (6) L 3 < x < L : y ( x,t ) = A 2 sin( k 2 ( L- x )) cos t (7) where k 1 = r 4 T = v 1 (8) k 2 = r T = 2 r 4 T = 2 v 1 . (9) Or, in terms of v 1 : y ( x,t ) = A 1 sin v 1 x cos t x , L 3 , A 2 sin 2 v 1 ( L- x ) cos t x L 3 ,L (10) d. By applying boundary conditions at x = L 3 , find the allowed normal mode frequencies,...
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ph2a_final_soln - Plausible Final Questions Nate Bode...

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