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410ahw12soln

# 410ahw12soln - HW#12—Solution —Phys410—Fall 2011 Prof...

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Unformatted text preview: HW#12—Solution —Phys410—Fall 2011 Prof. Ted Jacobson Room 4115, (301)405-6020 www.physics.umd.edu/grt/taj/410a/ [email protected] 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 ( μ ˙ 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 = nπ/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) δ nm . Thus Z ˙ y 2 dx = X n,m ˙ y n ˙ y m ( a/ 2) δ 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) δ nm = ( a/ 2) X n k 2 n y 2 n so L = X n ( 1 2 m ˙ y 2 n- 1 2 mω 2 n y 2 n ) with m = μa/ 2, ω n = vk n , and v = p T/μ . Thus ω n = nω 1 , with ω 1 = πv/a . (b) The oscillators are uncoupled. Each y n has it’s own Lagrange eqn which is the har- monic oscillator eqn ¨ y n =- ω 2 n y n , with solutions y n = A n cos( ω n t + φ n ). (c) p n = m ˙ y n , H = ∑ n ( 1 2 m p 2 n + 1 2 mω 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/ω is approximately invariant. Thus for each mode E 2 /E 1 = ω 2 /ω 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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410ahw12soln - HW#12—Solution —Phys410—Fall 2011 Prof...

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