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Solutions for Computational Materials Science, An Introductions, 2nd Edition by Lee (All Chapters included)

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Complete Solutions Manual for Computational Materials Science, An Introductions, 2nd Edition by June Gunn Lee ; ISBN13: 9781498749732.....(Full Chapters included)...Chapter 1 Introduction Chapter 2 Molecular Dynamics (MD) Chapter 3 MD Exercises with XMD and LAMMPS Chapter 4 First-Principles Meth...

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  • October 26, 2024
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  • Computational Materials Science 2e Lee
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SOLUTIONS MANUAL FOR
COMPUTATIONAL
MATERIALS SCIENCE
An Introduction




by

June Gunn Lee

** Immediate Download
** Swift Response
** All Chapters included

, Homework and Solutions



Table of Contents

Chapter 2 01

Chapter 3 07

Chapter 4 20

Chapter 5 25

Chapter 6 29

Chapter 7 33

Chapter 8 37



Chapter 2


Homework 2.1

Consider a one-dimensional harmonic oscillator of mass, m, and spring constant of k that follows

Hooke's law, F = −kx , in an isolated system (E = const.). Then the Newton's second law

becomes the second-order ordinary differential equation as:

k
a (t ) = − x(t )
m

Let us assume that the spring is fixed at one side and is ideal with zero mass. Given oscillator

=
frequency of ω (=
k m)1/2 1 and initial conditions of x0 = 5 and v0 = 0 , find and discuss the

following:

• Write down the potential expression for the system

• Analytical solution for ball positions with time

1

, • Numerical solution for ball positions with time by taking the first two terms in the

Taylor series expansion (the first-order Euler method) with ∆t = 0.2 for 10 timesteps

• Compare the two methods above on a position-time plot and discuss what will happen if

the numerical calculation proceeds further for more timesteps and how to improve its

accuracy



Solution 2.1

Model




An ideal ball-spring system in one dimension.

Potential expression

F = −kx

1
U =− ∫ Fdx =− ∫ −kxdx = kx2 : Parabolic curve
2




2

, Analytical solution:

F = −kx= ma , 0 = −kx= ma at equilibrium

This second-order differential equation has the general solutions as:

v0 cos(ωt ) − ω x0 sin(ωt ) =
v(t ) = −5sin(t )

v0
x(t ) = x0 cos(ωt ) + sin(ωt ) = 5cos(t )
ω



Numerical solution:

v(t + ∆=
t ) v(0) + a (0)∆t , …..

x(t + ∆=
t ) x(0) + v(0)∆t , …..

−kx= ma , k m = 1 → a = − x



Numerical calculation for one-dimensional harmonic oscillator

t ∆t x a = −x v
0.0 - x0 = 5 a 0 = -5 v0 = 0
x1 = x0 + v0∆t v1 = v0 + a 0∆t
0.2 0.2 a 1 = -5
=5+0=5 = 0 - 1 = -1
0.4 0.2 x2 = x1 + v1∆t a 2 = -4.8 v2 = v1 + a 1∆t

3

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