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ps11soln

# ps11soln - 1 Physics 316 Cornell University Solution for...

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1 Physics 316 Solution for homework 11 Spring 2005 Cornell University I. EXERCISE 1 A. Consider the two dimensional wave equation for the vibration of a rectangular membrane, 2 z ∂x 2 + 2 z ∂y 2 = 1 v 2 2 z ∂t 2 . (1.1) The membrane has a boundary at x = 0 , y = 0 and x = L x , y = L y at which it does not oscillate, i.e. z = 0 . Show that z = A sin( k x x ) sin( k y y ) cos( ωt ) (1.2) can be a solution for specific k x , k y , and ω . To verify that the given form of z can be a solution, we substitute it into the given wave equation. After canceling common terms, we find that z is a solution if k 2 x + k 2 y = ω 2 v 2 . (1.3) The constraints from the fact that z = 0 on all the boundaries imply that k x = n x π L x and k y = n y π L y , with n x , n y = 1 , 2 , ... . This leads to a constraint on ω : ω n x ,n y = πv s n 2 x L 2 x + n 2 y L 2 y . (1.4) B. Show for L x = L y that there can be different solutions z ( x, y, t ) for the same frequency ω . This is called degeneracy . For L x = L y = L , we can rewrite ω n x ,n y as ω n x ,n y = πv L q n 2 x + n 2 y . (1.5) We can see that ω ab = ω ba , but the corresponding z ( x, y, t ) are different: z ab = A sin L x · sin L y cos( ωt ) z ba = A sin L x sin L y · cos( ωt ) (1.6) II. EXERCISE 2 Show that the spherical finite well potential, V ( r ) = V 0 H ( r - R ) with H ( x ) = if ( x> 0 , 1 , 0 ) , does not have a spherically symmetric bound state if V 0 < ¯ h 2 π 2 8 mR 2 . We consider the spherical finite potential well V ( r ) = V 0 H ( r - R ). In general, spherically symmetric states are found with the radial part of the Schr¨ odinger equation, - ¯ h 2 2 m u 00 ( r ) + V ( r ) u ( r ) = Eu ( r ) , (2.1) 2 the actual wavefunction being given by Φ( r ) = u ( r ) r . We can solve (2.1) in the two domains r < R and r > R independantly and then make sure that u ( r ) and its first derivative are continuous at r = R : - ¯ h 2 2 m u 00 1 ( r ) = Eu 1 ( r ) (2.2) - ¯ h 2 2 m u 00 2 ( r ) = - ( V 0 - E ) u 2 ( r ) . (2.3) Both solutions are already known: u 1 ( r ) = A 1 sin( k 1 r ) + B 1 cos( k 1 r ) (2.4) u 2 ( r ) = A 2 e k 2 r + B 2 e - k 2 r , (2.5) with k 1 = 2 mE/ ¯ h and k 2 = p 2 m ( V 0 - E ) / ¯ h . In order to have u 1 ( r ) r well defined at the origin, we must have u 1 (0) = 0, that is B 1 = 0. Since we want Φ to be normalizable, we also need to put A 2 = 0. We now apply the continuity conditions: u 1 ( R ) = u 2 ( R ) A 1 sin( k 1 R ) = B 2 e - k 2 R (2.6) u 0 1 ( R ) = u 0 2 ( R ) k 1 A 1 cos( k 1 R ) = - k 2 B 2 e - k 2 R . (2.7) Therefore, dividing (2.7) by (2.6), one gets k 1 cot( k 1 R ) = - k 2 . (2.8) 0 π 1 A B C k R 1 FIG. 1: Graphical solution of equation (2.10) for three different values of k V . A: k V R < π 2 (too small for any possible solution for k 1 R ). B: k V R = π 2 ; this is the limiting value in order to get a solution.

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