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)
Timeaveraged 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
−
=
.
Timeaveraged 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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 Spring '08
 WORMER
 Energy, Kinetic Energy, Potential Energy, Work, Cos, Ekin

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