SOLUTIONS MANUAL FOR
COMPUTATIONAL
MATERIALS SCIENCE
An Introduction
by
June Gunn Lee
** Immediate Download
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** 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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