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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 r u u t ska s s 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 ( ) ( ) 1 1 2 2 1 1 2 2 PE C u u C u u s s s s s = + + The kinetic energy associated with the sth mass is 2 1 1 2 2 2 du s KE M M s s dt υ = By summing over all atoms we get the total energy: ( ) ( ) 2 1 1 2 1 2 2 s s du s U t M C u u s s dt = + + (instantaneous) b) For longitudinal wave ( ) cos u u t ska s ω = , we have ( ) sin du s u t ska dt ω ω = − and ( ) ( ) ( ) ( ) { } 2 1 1 2 2 2 2 sin cos cos 2 2 S U t M u t ska Cu t ska t ska ka ω ω ω ω = + The last term has the form ( ) 2 cos cos , with t ska ka α β α α ω β = = . So we can use the trigonometric identity ( ) cos cos cos sin sin α β α β α β = + to write ( ) ( ) ( ) 2 1 1 2 2 2 2 sin cos cos 1 sin sin 2 2 S U t M u Cu ω α α β α β = + +

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

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