Consider the forces on a short fragment of string
F
y
=
T
sin
(
θ
+
Δ
θ
)
−
T
sin
θ
F
y
=
T
sin
(
θ
+
Δ
θ
)
−
T
sin
θ
F
x
=
T
cos
(
θ
+
Δ
θ
)
−
T
cos
θ
F
x
=
T
cos
(
θ
+
Δ
θ
)
−
T
cos
θ
Assume that the displacement in y is small and
T
T
is a constant along the stringthus
θ
θ
and
θ
+
Δ
θ
θ
+
Δ
θ
are smallthen
F
x
≈
0
F
x
≈
0
we can see this by expanding the trig functions
F
x
≈
T
[
1
−
(
θ
+
Δ
θ
)
2
2
−
1
+
θ
2
2
+
…
]
F
x
≈
T
[
1
−
(
θ
+
Δ
θ
)
2
2
−
1
+
θ
2
2
+
…
]
or
F
x
≈
T
θ
Δ
θ
F
x
≈
T
θ
Δ
θ
which is very small.On the other hand
F
y
≈
T
[
θ
+
Δ
θ
−
θ
+
…
]
F
y
≈
T
[
θ
+
Δ
θ
−
θ
+
…
]
or
F
y
≈
T
Δ
θ
F
y
≈
T
Δ
θ
which is not nearly as small. So we will consider the
y
y
component of motion, but approximate there is no x component
F
y
=
T
sin
(
θ
+
Δ
θ
)
−
T
sin
θ
≈
T
tan
(
θ
+
Δ
θ
)
−
T
tan
θ
=
T
(
∂
y
(
x
+
Δ
x
)
∂
x
−
∂
y
∂
x
)
=
T
∂
2
y
∂
x
2
Δ
x
F
y
=
T
sin
(
θ
+
Δ
θ
)
−
T
sin
θ
≈
T
tan
(
θ
+
Δ
θ
)
−
T
tan
θ
=
T
(
∂
y
(
x
+
Δ
x
)
∂
x
−
∂
y
∂
x
)
=
T
∂
2
y
∂
x
2
Δ
x
Also we can write:
F
y
=
m
a
y
F
y
=
m
a
y
m
=
μ
Δ
x
m
=
μ
Δ
x
where
μ
μ
is the mass density
a
y
=
∂
2
y
∂
t
2
a
y
=
∂
2
y
∂
t
2
now have
T
∂
2
y
∂
x
2
Δ
x
=
μ
Δ
x
∂
2
y
∂
t
2
T
∂
2
y
∂
x
2
Δ
x
=
μ
Δ
x
∂
2
y
∂
t
2
∂
2
y
∂
x
2
=
μ
T
∂
2
y
∂
t
2
∂
2
y
∂
x
2
=
μ
T
∂
2
y
∂
t
2
Note dimensions, get a velocity
T
μ
=
v
2
T
μ
=
v
2
∂
2
y
∂
x
2
=
1
v
2
∂
2
y
∂
t
2
∂
2
y
∂
x
2
=
1
v
2
∂
2
y
∂
t
2
The second space derivative of a function is equal to the second
time derivative of a function multiplied by a constant.
Before considering traveling waves, we are going to look at a special case
solution to the wave equation. This is the case of stationary vibrations of a
string.
For example here, lets consider the case where both ends of the string are
fixed at
y
=
0
y
=
0
.
Now we vibrate the string. Every point along the string acts like a little
driven oscillator. So lets assume that every point on string has a time
dependence of the form
cos
ω
t
cos
ω
t
and that the amplitude is a function of distance Assume
y
(
x
,
t
)
=
f
(
x
)
cos
ω
t
y
(
x
,
t
)
=
f
(
x
)
cos
ω
t
then
∂
2
y
∂
t
2
=
−
ω
2
f
(
x
)
cos
ω
t
∂
2
y
∂
t
2
=
−
ω
2
f
(
x
)
cos
ω
t
∂
2
y
∂
x
2
=
∂
2
f
∂
x
2
cos
ω
t
∂
2
y
∂
x
2
=
∂
2
f
∂
x
2
cos
ω
t
Substitute into wave equation
∂
2
y
∂
x
2
=
1
v
2
∂
2
y
∂
t
2
∂
2
y
∂
x
2
=
1
v
2
∂
2
y
∂
t
2
∂
2
f
∂
x
2
cos
ω
t
=
−
ω
2
v
2
f
(
x
)
cos
ω
t
∂
2
f
∂
x
2
cos
ω
t
=
−
ω
2
v
2
f
(
x
)
cos
ω
t
Then every
f
(
x
)
f
(
x
)
that satisfies:
∂
2
f
∂
x
2
=
−
ω
2
v
2
f
∂
2
f
∂
x
2
=
−
ω
2
v
2
f
is a solution of the wave equation
A solution is (requiring
f
(
0
)
=
0
f
(
0
)
=
0
since ends fixed)
f
(
x
)
=
A
sin
(
ω
x
v
)
f
(
x
)
=
A
sin
(
ω
x
v
)
Another boundary condition is
f
(
L
)
=
0
f
(
L
)
=
0
so get
A
sin
(
ω
L
v
)
=
0
A
sin
(
ω
L
v
)
=
0
Thus
ω
L
v
=
n
π
ω
L
v
=
n
π
ω
=
n
π
v
L
ω
=
n
π
v
L
Be careful with the equations above:
v
v
is the letter vee and is for velocity. now we introduce the frequency
ν
ν
which is the Greek letter nu.
recall
ν
=
ω
/
2
π
ν
=
ω
/
2
π
so
ν
n
=
n
v
2
L
=
n
2
L
(
T
μ
)
1
2
ν
n
=
n
v
2
L
=
n
2
L
(
T
μ
)
1
2
This is a very important feature of wave phenomena. Things can be quantized.
This is why a musical instrument will play specific notes. Note, that we
must have an integral number of half sine waves
λ
n
=
2
L
n
λ
n
=
2
L
n
end up with
f
n
(
x
)
=
A
n
sin
(
2
π
x
λ
n
)
f
n
(
x
)
=
A
n
sin
(
2
π
x
λ
n
)
leading to
y
n
(
x
,
t
)
=
A
n
sin
(
2
π
x
λ
n
)
cos
ω
n
t
y
n
(
x
,
t
)
=
A
n
sin
(
2
π
x
λ
n
)
cos
ω
n
t
where
ω
n
=
n
π
L
(
T
μ
)
1
2
=
n
π
L
v
=
n
ω
1
ω
n
=
n
π
L
(
T
μ
)
1
2
=
n
π
L
v
=
n
ω
1
ω
1
ω
1
is the fundamental frequency