Homework04_solution - Homework # 4 Solution Problem #1 (a)...

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Homework # 4 – Solution Problem #1 (a) The total energy of the wave is the sum of the kinetic energy E kin and the potential energy E pot . The kinetic energy is the sum of the kinetic energies of all atoms, i.e. 2 1 2 n kin n du EM dt ⎛⎞ = ⎜⎟ ⎝⎠ , where M is the mass of atoms and n du dt is the velocity of n -th atom. The potential energy is the potential energy of all the “springs” connecting atoms. For two atoms this energy is the same as that for a harmonic oscillator, i.e. 2 1 2 Cx , where C is interatomic force constant and x is the change in distance between the atoms from the equilibrium distance. Therefore, for atoms n and n +1 having displacements and n u 1 n u + respectively this energy is ( 2 1 1 2 nn Cu u + ) . The total potential energy is given by () 2 1 1 2 pot n n n EC u u + =− . The total energy of the wave is therefore 2 2 1 11 22 n kin pot n n du E EE M Cuu dt + =+= + ∑∑ . (b) Time-averaged kinetic energy per atom is 2 2 sin n kin du E M M A qna t MA Nd tN 1 4 ω ωω == = ⎡⎤ ⎣⎦ , where we took into account that 2 1 sin 2 qna t = . Time-averaged potential energy per atom is ( ) ( ) { } {} 2 2 2 1 2 2 cos ( 1) cos 1 cos ( cos 2 cos ( cos 2 1 1 cos (2 2 cos 1 cos sin 2 2 pot n n n n n E C u u CA q n a t qna t NN CA q n a t qna t q n a t qna t N qa CA q n a t qa CA qa CA + = + = +− + − − +− = ⎧⎫ + − = = ⎨⎬ ⎩⎭ 2 2
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This note was uploaded on 09/21/2011 for the course PHYSICS 101 taught by Professor Wormer during the Spring '08 term at Aarhus Universitet.

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Homework04_solution - Homework # 4 Solution Problem #1 (a)...

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