Problems 1. Monatomic linear lattice. Consider a...

Problems 1. Monatomic linear lattice. Consider a longitudinal wave $u_s = u \cos(\omega t - ska)$ which propagates in a monatomic linear lattice of atoms of mass $M$, spacing $a$, and nearest-neighbor interaction $C$. (a) Show that the total energy of the wave is $E = \frac{1}{2} \sum_s M \left(\frac{du_s}{dt}\right)^2 + \frac{1}{2} C \sum_s (u_s - u_{s+1})^2$, where $s$ runs over all atoms. 4 Phonons I. Crystal Vibrations 103 (b) By substitution of $u_s$ in this expression, show that the time-average total energy per atom is $\frac{1}{4} M\omega^2 \overline{u}^2 + \frac{1}{2} C (1 - \cos Ka)\overline{u}^2 = \frac{1}{2} M\omega^2 \overline{u}^2$, where in the last step we have used the dispersion relation (9) for this problem.

Transcript

00:01 This is the equation based on the elastic wave and here this problem is built the mono -atomic linear lattice and consider a longitudinal wave and defined by this us is equal to u cos omega t minus s k a that is for this is the equation number one which propagate in in a monotomic linear lattice of atoms or mass m and the spacing a and the nearest neighbor and neighbor distance is c and in the part a question we have to show that part a so that the total energy of the wave is given by a is equal to half m s d u s by d t square plus half c summation as u s minus u s plus one square where s runs over all the atom so we have to prove this you have to solution we know that the total energy total energy is the sum of that is e t is equal to kinetic energy of all the atoms that is kinetic energy of atom plus elastic potential energy which is store in the bonds in the bond of atoms and this is equation number one that means we have to find out the kinetic energy of the atoms and the elastic potential energy now we know that kinetic energy is half mb square where m is the mass of the atom and the v is the velocity and this v can be presented as the m and v the d s by d t here d u s by d t poly square and this is the velocity and here u s this is the displacement this is s displacement for the s atom now this kinetic energy will be total kinetic energy for all atom total kinetic energy for all atoms that is k e is equal to half m and this velocity here this is for the one single atom sorry this is for one single atom and for all atom this d u by d t we replaced by that d u s by d t square and there there will be a summation here this is and this u is replaced by the u s and d t square so this is the kinetic energy for all atoms now from hoax law we know that hoax law the elastic potential energy elastic potential energy is equal to half k x square but in case of the atom this case this formula is because of this f is equal to minus x but here we are using this f is equal to minus c and this us minus us plus one we have this is the displacement and c is the force constant hence replace this potential energy formula by this in case of the this monotron linear chain that k is replaced by c and this x is replaced by u s minus us plus one then this total kinetic elastic potential energy for all atoms total elastic potential energy for all atoms is and is p .e is equal to half and k is replaced by c and this x is replaced by this us minus us plus one whole square thus the e total and because we are using for all atom then it will be a synest it is c now et is equal to k e plus p e that may e is equal to that is e in the question is equal to half and summation as d u .s by d t equal square plus half c summation as u s minus u s plus one square this is the solution of the this proof of the part a now in the part b by substitution of the u s in this expression so that the time average total energy per atom is given by 1 by 4m omega square u square plus half c 1 minus k u square is equal to half m omega square u square okay then that means this is uh v s velocity so the this e can…

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