In a linear chain all the atoms have equal mass but are connected alternatively by springs of fo

step 1 This is problem number 5. This problem stated, consider the normal modes of chain in which the f

In a linear chain all the atoms have equal mass but are...

Key Concepts

Force Constant Variation

The concept of force constant variation introduces different strengths in the interatomic bonds within the crystal lattice. When springs with alternating force constants connect the atoms, it results in non-uniform restoring forces, which in turn modify the vibrational behavior of the system. This variation is essential in understanding the appearance of two branches in the dispersion relation, as it breaks the uniformity seen in simple monoatomic chains.

Acoustic and Optical Branches

In lattice dynamics, particularly for systems with a basis or alternating interactions, the solutions of the dispersion relation divide into two branches. The acoustic branch is characterized by lower frequencies where neighboring atoms oscillate in phase, while the optical branch corresponds to higher frequencies with out-of-phase oscillations. Even when the masses are equal, the presence of alternating force constants can lead to the splitting of the vibrational spectrum into these two distinct branches.

Dispersion Relation

The dispersion relation describes the relationship between the frequency of vibrations (?) and the wavevector (k) in a periodic structure. It is a fundamental concept in lattice dynamics that determines how vibrational waves propagate through a material, accounting for the periodic arrangement of atoms and the interactions between them.

Normal Modes

Normal modes refer to the independent patterns of vibration that occur in a system such as a linear chain of atoms. In these modes, every atom oscillates with the same frequency, and the collective motion can be described using a plane wave ansatz. This concept is crucial in deriving the frequency-wavevector relationship by simplifying the many-body problem into solvable equations.

Transcript

00:01 This is problem number 5.

00:03 This problem stated, consider the normal modes of chain in which the force constant between nearest neighbor atoms are alternatively c and the 10c.

00:18 So here that means there are atoms and here the suppose this is the central atom.

00:39 This is the force constant between these two atom is c and in these two atom is 10.

00:45 10 c and here this is 10 c and this is c this is c this is 10 c and c alternately the force constant in different and let the mass of this atom is m and suppose this is atom number one and the displacement of this is vs or suppose this is the s atom then it will be the s plus 1 and it will be the s minus 1 okay let's write and the nearest neighbor distance between the atoms is a by 2 which is given the question nearest neighbor distance is given and which is a by 2 and the force constant are different between the these two atom on this side this fourth constant 10 c and on this side it is it is c it is different force constant and here find omega k at k is equal to 0 and k is equal to pi by a so we have to find out this frequency that is omega k at k is equal to 0 and k is equal to pi by a okay let's write down the newton's law which is the equation 16a that is m d2 s the equation 16a is m d t u s by d t square where u s is the displacement of the s that is equal to 10 vs minus 1 minus u s plus c v s minus u s plus c vs minus minus us so let's represent this in diagram let us make again this diagram slightly large diagram suppose this is one atom this is another atom this is another atom and this is another and the mass of each atom is same that is they are same atom and mass is m and the nearest distance is the a by two so let us assume this is the s atom so the displacement of this atom is u s and the displacement of this is vs vs and the force constant between these is is c and here it is 10c similarly it will be 10c and it is c so here that it will be it is vs it is u s then it will be vs minus 1 so this is vs minus 1 this is vs minus 1 let's start this vs minus 1 and this will be us minus 1 us minus 1 us minus 1 when you write the newton's law then if you are assuming this then it will be vs minus this vs minus 1 minus u .s plus c this that is vs minus us plus c this that is vs minus us so first we write for this we have assumed this atom for this atom or this lattice we have read the again that is vs minus us us or the fourth constant is c and for this vs minus 1 minus u .s that is the fourth constitution is 10 c okay now solve this equations it is 10 it will be 10 vs minus 1 we have missing spring constant it is 10 c let's write again this this is 10c 10 c vs minus 1 and this 10 c minus u s and minus 10 c u .s plus c vs minus c u .s and again write this it will be 10 c vs minus 1 and this is u .s and this is u .s that is it will be minus 11 u .s and this is c vs minus 11 u .s let's suppose this is the equation number 1 d2 u s by d t square and this is equation number 1 similarly we can write for the vs m d2 vs by d t square is equal to so here for this for this we have to write that is us minus vs and the force constant that is c and here it will be us plus one and us plus one minus p s so here this will be c u s minus vs plus the plus the force constant between the ann of the atom that is elize u s plus one minus v s now rearrange this one it will be cus minus c vs plus 10 cus plus 1 minus 10 c vs from here we can write this 10 cus plus 1 and this is 10 c vc and this is c vc and this is c vs so it will be minus 11 vs and it is 11 that is cus minus 11 is and this is second equation that is second atom which for which the displacement is v .a.

09:16 This is the equation number 2.

09:19 Now let us assume the solution of these two equation.

09:25 Let the solution us is equal to u.

09:34 Not exponential i s k a n exponential minus i omega t similar in the solution for the second equation v s is equal to v0 exponential i s ka minus i omega t where k is the wave vector this is equation number 3 now substitute the equation 3 in equation 1 and 2 then you will get two equation that is minus m omega square u .0 is equal to 10 c v0 exponential minus i k a plus c v0 minus 11 c u not and this can be taken this side it will be 11c minus m omega square and u not is common is equal to v0 c is common and it will be 1 plus exponential minus i k a and let us assume this is the equation number 4 similarly we can get another equation and which will be 11c minus m omega square v0 is equal to u .0 c 1 plus exponential plus i k a and this equation number 5 now we have to find out the omega k at k is equal to 0 and the k is equal to pi by a so let's the case number 1 at k is equal to 0 at this value is e power i k a plus minus i k a that e power 0 that it will be 1 now equation 4 and 5 becomes this is 11 c now equation 4 and 5 becomes 11c minus m omega square u...

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