Lets calculate the energy in a wave of a string:
Consider a fragment of string so small it can be considered straight, as is
shown in the figure
The the kinetic energy is
K
=
1
2
m
K
=
1
2
m
v
2
2
for the string fragment
m
=
μ
ⅆ
x
m
=
μ
ⅆ
x
Why is this
ⅆ
x
ⅆ
x
and not
ⅆ
s
ⅆ
s
?
Lets consider that the string is not perturbed, then it is horizontal and has
mass as given. When the string is perturbed it stretches a little bit - but
the mass does not increase.
So we have
K
=
1
2
μ
ⅆ
x
(
∂
y
∂
t
)
2
K
=
1
2
μ
ⅆ
x
(
∂
y
∂
t
)
2
and using this we can define the energy per unit length, ie. the kinetic
energy density:
ⅆ
K
ⅆ
x
=
1
2
μ
(
∂
y
∂
t
)
2
ⅆ
K
ⅆ
x
=
1
2
μ
(
∂
y
∂
t
)
2
When the string segment is stretched from the length
ⅆ
x
ⅆ
x
to the length
ⅆ
s
ⅆ
s
an amount of work
=
T
(
ⅆ
s
−
ⅆ
x
)
=
T
(
ⅆ
s
−
ⅆ
x
)
is done. This is equal to the potential energy stored in the stretched string
segment. So the potential energy in this case is:
U
=
T
(
ⅆ
s
−
ⅆ
x
)
U
=
T
(
ⅆ
s
−
ⅆ
x
)
Now
ⅆ
s
=
(
ⅆ
x
2
+
ⅆ
y
2
)
1
/
2
=
ⅆ
x
[
1
+
(
∂
y
∂
x
)
2
]
1
/
2
ⅆ
s
=
(
ⅆ
x
2
+
ⅆ
y
2
)
1
/
2
=
ⅆ
x
[
1
+
(
∂
y
∂
x
)
2
]
1
/
2
Recall the binomial expansion
(
1
+
A
)
n
=
1
+
n
A
+
n
(
n
−
1
)
A
2
2
!
+
n
(
n
−
1
)
(
n
−
2
)
A
3
3
!
+
…
(
1
+
A
)
n
=
1
+
n
A
+
n
(
n
−
1
)
A
2
2
!
+
n
(
n
−
1
)
(
n
−
2
)
A
3
3
!
+
…
so
ⅆ
s
≈
ⅆ
x
+
1
2
(
∂
y
∂
x
)
2
ⅆ
x
ⅆ
s
≈
ⅆ
x
+
1
2
(
∂
y
∂
x
)
2
ⅆ
x
U
=
T
(
ⅆ
s
−
ⅆ
x
)
≈
1
2
T
(
∂
y
∂
x
)
2
ⅆ
x
U
=
T
(
ⅆ
s
−
ⅆ
x
)
≈
1
2
T
(
∂
y
∂
x
)
2
ⅆ
x
or the potential energy density
ⅆ
U
ⅆ
x
=
1
2
T
(
∂
y
∂
x
)
2
ⅆ
U
ⅆ
x
=
1
2
T
(
∂
y
∂
x
)
2
To get the kinetic energy in a wavelength, lets start with
y
=
A
sin
(
2
π
x
λ
−
ω
t
)
y
=
A
sin
(
2
π
x
λ
−
ω
t
)
∂
y
∂
t
=
−
ω
A
cos
(
2
π
x
λ
−
ω
t
)
∂
y
∂
t
=
−
ω
A
cos
(
2
π
x
λ
−
ω
t
)
Lets evaluate it at time 0.
∂
y
∂
t
|
t
=
0
=
−
ω
A
cos
(
2
π
x
λ
)
∂
y
∂
t
|
t
=
0
=
−
ω
A
cos
(
2
π
x
λ
)
so
ⅆ
K
ⅆ
x
=
1
2
μ
ω
2
A
2
cos
2
(
2
π
x
λ
)
ⅆ
K
ⅆ
x
=
1
2
μ
ω
2
A
2
cos
2
(
2
π
x
λ
)
now integrate
K
=
∫
0
λ
ⅆ
K
ⅆ
x
ⅆ
x
=
1
2
μ
ω
2
A
2
∫
0
λ
cos
2
(
2
π
x
λ
)
ⅆ
x
K
=
∫
0
λ
ⅆ
K
ⅆ
x
ⅆ
x
=
1
2
μ
ω
2
A
2
∫
0
λ
cos
2
(
2
π
x
λ
)
ⅆ
x
In order to do this integral we use the following trig identity:
cos
2
A
=
cos
2
A
+
1
2
cos
2
A
=
cos
2
A
+
1
2
so we get
K
=
1
2
μ
ω
2
A
2
[
x
2
+
λ
8
π
sin
4
π
x
λ
]
|
x
=
0
λ
=
1
4
μ
λ
ω
2
A
2
K
=
1
2
μ
ω
2
A
2
[
x
2
+
λ
8
π
sin
4
π
x
λ
]
|
x
=
0
λ
=
1
4
μ
λ
ω
2
A
2
In similar fashion the potential energy can be found to be
U
=
1
4
μ
λ
ω
2
A
2
.
U
=
1
4
μ
λ
ω
2
A
2
.
Deriving this will be assigned as a homework problem
So
E
=
K
+
U
=
1
2
μ
λ
ω
2
A
2
E
=
K
+
U
=
1
2
μ
λ
ω
2
A
2
Power
P
=
Δ
E
Δ
t
=
1
2
μ
λ
ω
2
A
2
τ
=
1
2
μ
ω
2
A
2
v
P
=
Δ
E
Δ
t
=
1
2
μ
λ
ω
2
A
2
τ
=
1
2
μ
ω
2
A
2
v
Where I have used
τ
=
1
/
ν
τ
=
1
/
ν
and
λ
ν
=
λ
ν
=
v
thus
τ
=
λ
/
v
τ
=
λ
/
v
