410ahw12soln - HW#12Solution Phys410Fall 2011 Prof. Ted...

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Unformatted text preview: HW#12Solution Phys410Fall 2011 Prof. Ted Jacobson Room 4115, (301)405-6020 www.physics.umd.edu/grt/taj/410a/ jacobson@umd.edu 16.31 (Earthquake distance) t p = d/c p and t s = d/c s , so t s- t p = d ( c p- c s ) /c p c s , hence d = ( t s- t p ) c p c s / ( c p- c s ) = (12min)(3 . 0)(5 . 25) / 2 . 25)(1km / s) = 5040km. S12.1 Fourier modes of string in Lagrangian and Hamiltonian L = Z 1 2 ( y 2- Ty 2 ) dx where the dot and prime denote /t and /x respectively. (a) y ( x,t ) = y n ( t )sin k n x . The boundary condition sin ka = 0 = k = n/a , with n an integer. It suffices to include n 1. Now y = y n sin k n x and y = k n y n cos k n x , so the Lagrangian involves the integrals R a (sin k n x )(sin k m x ) dx = R a (cos k n x )(cos k m x ) dx = ( a/ 2) nm . Thus Z y 2 dx = X n,m y n y m ( a/ 2) nm = ( a/ 2) X n y 2 n Z y 2 dx = X n,m k n k m y n y m ( a/ 2) nm = ( a/ 2) X n k 2 n y 2 n so L = X n ( 1 2 m y 2 n- 1 2 m 2 n y 2 n ) with m = a/ 2, n = vk n , and v = p T/ . Thus n = n 1 , with 1 = v/a . (b) The oscillators are uncoupled. Each y n has its own Lagrange eqn which is the har- monic oscillator eqn y n =- 2 n y n , with solutions y n = A n cos( n t + n ). (c) p n = m y n , H = n ( 1 2 m p 2 n + 1 2 m 2 n y 2 n ). (d) The period of a guitar string vibration (e.g. 1/440 second for A above middle C) is much shorter than the time scale over which the note is bent, so E/ is approximately invariant. Thus for each mode E 2 /E 1 = 2 / 1 = 14 / 12 1 . 17, so the energy in each mode increases by about 17%, and the total energy increases by the same percentage.increases by about 17%, and the total energy increases by the same percentage....
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This note was uploaded on 12/29/2011 for the course PHYSICS 410 taught by Professor Jacobson during the Fall '11 term at Maryland.

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410ahw12soln - HW#12Solution Phys410Fall 2011 Prof. Ted...

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