HW4soln

HW4soln - ECE215A/Materials206A Winter 2008 Prof. Brown/ECE...

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ECE215A/Materials206A Winter 2008 Prof. Brown/ECE Dept/UCSB 1 Homework 4 Solutions (first four problems only, for Quiz#1) (1) Energy of lattice wave a) Monotomic linear lattice, mass m, spacing a, nearest neighbor interaction (spring constant) C. Consider longitudinal wave: () cos ru u t s k a ss ω ∆≡ = . The total energy of the wave is the sum over all atoms of the energy of each atom. From mechanics, we know that the instantaneous energy of a harmonic oscillator has a kinetic energy term and a potential energy term. The potential term depends only on the force constant and the displacement of the oscillator from equilibrium. Thus the linear lattice can be repeated as a series of springs with masses (atoms) attached. The potential energy associated with the sth spring is 11 22 PE C u u C u u s s s →− = ++ The kinetic energy associated with the sth mass is 2 2 du s KE M M dt υ  →=   By summing over all atoms we get the total energy: 2 2 1 du s Ut M C u u dt =+ + ∑∑ (instantaneous) b) For longitudinal wave cos uu t s k a s =− , we have sin du s ut s k a dt ωω and ( ) {} 2 22 2 2 sin cos cos S U t M u t ska Cu t ska t ska ka + The last term has the form 2 cos cos , with t ska ka αβ α α ω β  −− = − =  . So we can use the trigonometric identity ( ) cos cos cos sin sin βα −= + to write 2 2 sin cos cos 1 sin sin S M u C u ωα + 
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ECE215A/Materials206A Winter 2008 Prof. Brown/ECE Dept/UCSB 2 () ( ) 1 1 22 2 2 2 2 sin cos cos 1 2cos cos 1 sin sin sin sin 2 2 s Mu C u ω
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This note was uploaded on 01/17/2012 for the course ECE 215A taught by Professor Brown during the Winter '08 term at UCSB.

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HW4soln - ECE215A/Materials206A Winter 2008 Prof. Brown/ECE...

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