Define
κ
T
=
(
1
−
a
)
κ
s
≤
κ
s
{\displaystyle \kappa _{T}=(1-a)\kappa _{s}\leq \kappa _{s}}
If you don't understand this animated gif, look at the next figure The first four terms of the Fourier series of a square wave,
ψ
(
x
)
=
4
π
sin
(
π
x
)
{\displaystyle \psi (x)={\frac {4}{\pi }}\sin(\pi x)}
+
4
3
π
sin
(
3
π
x
)
{\displaystyle +{\frac {4}{3\pi }}\sin(3\pi x)}
+
4
5
π
sin
(
5
π
x
)
{\displaystyle +{\frac {4}{5\pi }}\sin(5\pi x)}
+
4
7
π
sin
(
7
π
x
)
+
…
{\displaystyle +{\frac {4}{7\pi }}\sin(7\pi x)+\ldots }
=
∑
odd
n
=
1
∞
4
n
π
sin
(
n
π
x
)
{\displaystyle =\sum _{{\text{odd }}n=1}^{\infty }{\frac {4}{n\pi }}\sin(n\pi x)}
m
ξ
¨
=
κ
s
ξ
′
′
+
a
κ
s
η
′
η
′
′
{\displaystyle m{\ddot {\xi }}=\kappa _{s}\xi ^{\prime \prime }+a\kappa _{s}\eta ^{\prime }\eta ^{\prime \prime }}
m
η
¨
=
κ
T
η
′
′
{\displaystyle m{\ddot {\eta }}=\kappa _{T}\eta ^{\prime \prime }}
Transverse standing wave:
ω
/
k
=
κ
T
/
m
{\displaystyle \omega /k={\sqrt {\kappa _{T}/m}}}
η
=
A
sin
(
k
x
)
sin
(
ω
t
)
{\displaystyle \eta =A\sin(kx)\sin(\omega t)}
η
′
η
′
′
=
−
k
3
A
2
sin
(
k
x
)
cos
(
k
x
)
sin
2
(
ω
t
)
{\displaystyle \eta ^{\prime }\eta ^{\prime \prime }=-k^{3}A^{2}\sin(kx)\cos(kx)\sin ^{2}(\omega t)}
Define
L
=
m
∂
2
/
∂
t
2
−
κ
s
∂
2
/
∂
X
2
{\displaystyle {\mathcal {L}}=m\partial ^{2}/\partial t^{2}-\kappa _{s}\partial ^{2}/\partial X^{2}}
Second order differential equation with one variable: https://openstax.org/books/calculus-volume-3/pages/7-2-nonhomogeneous-linear-equations
L
ξ
=
−
k
3
A
2
sin
(
k
x
)
cos
(
k
x
)
sin
2
(
ω
t
)
{\displaystyle {\mathcal {L}}\xi =-k^{3}A^{2}\sin(kx)\cos(kx)\sin ^{2}(\omega t)}
ξ
=
ξ
p
(
X
,
t
)
+
ξ
h
(
X
,
t
)
{\displaystyle \xi =\xi _{p}(X,t)+\xi _{h}(X,t)}
ξ
h
{\displaystyle \xi _{h}}
L
ξ
h
=
0
{\displaystyle {\mathcal {L}}\xi _{h}=0}
Link to wikipedia:Fourier series ?
Employ two identities:
sin
(
k
X
)
cos
(
k
X
)
=
sin
(
2
k
X
)
2
{\displaystyle \sin(kX)\cos(kX)={\frac {\sin(2kX)}{2}}}
sin
2
(
ω
t
)
=
1
−
cos
(
2
ω
t
)
2
{\displaystyle \sin ^{2}(\omega t)={\frac {1-\cos(2\omega t)}{2}}}
L
ξ
=
−
a
κ
s
k
3
A
2
sin
(
2
k
X
)
(
1
−
cos
(
2
ω
t
)
4
)
=
−
a
κ
s
k
3
A
2
4
sin
(
2
k
X
)
+
a
κ
s
k
3
A
2
4
sin
(
2
k
X
)
cos
(
2
ω
t
)
{\displaystyle {\begin{aligned}{\mathcal {L}}\xi &=-a\kappa _{s}k^{3}A^{2}\sin(2kX)\left({\frac {1-\cos(2\omega t)}{4}}\right)\\&=-{\frac {a\kappa _{s}k^{3}A^{2}}{4}}\sin(2kX)+{\frac {a\kappa _{s}k^{3}A^{2}}{4}}\sin(2kX)\cos(2\omega t)\end{aligned}}}
To find a particular solution,
ξ
p
,
{\displaystyle \xi _{p},}
L
ξ
1
=
−
a
κ
s
k
3
A
2
4
sin
(
2
k
X
)
{\displaystyle {\mathcal {L}}\xi _{1}=-{\frac {a\kappa _{s}k^{3}A^{2}}{4}}\sin(2kX)}
L
ξ
2
=
a
κ
s
k
3
A
2
4
sin
(
2
k
X
)
cos
(
2
ω
t
)
{\displaystyle {\mathcal {L}}\xi _{2}={\frac {a\kappa _{s}k^{3}A^{2}}{4}}\sin(2kX)\cos(2\omega t)}
Recall
κ
T
=
(
1
−
a
)
κ
s
{\displaystyle \kappa _{T}=(1-a)\kappa _{s}}
a
=
κ
s
−
κ
T
κ
s
{\displaystyle a={\frac {\kappa _{s}-\kappa _{T}}{\kappa _{s}}}}
If
ξ
1
{\displaystyle \xi _{1}}
sin
(
2
k
X
)
{\displaystyle \sin(2kX)}
L
ξ
1
=
4
k
2
κ
s
ξ
1
{\displaystyle {\mathcal {L}}\xi _{1}=4k^{2}\kappa _{s}\xi _{1}}
ξ
1
=
−
a
κ
s
k
A
2
16
κ
T
sin
(
2
k
X
)
{\displaystyle \xi _{1}=-{\frac {a\kappa _{s}kA^{2}}{16\kappa _{T}}}\sin(2kX)}
ξ
1
=
−
k
A
2
16
(
1
−
κ
T
κ
s
)
sin
(
2
k
X
)
{\displaystyle {\color {red}\xi _{1}=-{\frac {kA^{2}}{16}}\left(1-{\frac {\kappa _{T}}{\kappa _{s}}}\right)\sin(2kX)}}
If
ξ
2
{\displaystyle \xi _{2}}
sin
(
2
k
X
)
cos
(
2
ω
t
)
{\displaystyle \sin(2kX)\cos(2\omega t)}
L
ξ
2
=
(
−
4
m
ω
2
+
4
k
2
κ
s
)
ξ
2
{\displaystyle {\mathcal {L}}\xi _{2}=\left(-4m\omega ^{2}+4k^{2}\kappa _{s}\right)\xi _{2}}
ξ
2
=
a
κ
s
16
k
3
A
2
k
2
κ
s
−
m
ω
2
sin
(
2
k
X
)
cos
(
2
ω
t
)
{\displaystyle \xi _{2}={\frac {a\kappa _{s}}{16}}{\frac {k^{3}A^{2}}{k^{2}\kappa _{s}-m\omega ^{2}}}\sin(2kX)\cos(2\omega t)}
ξ
2
=
a
κ
s
k
A
2
16
1
κ
s
−
m
ω
2
/
k
2
sin
(
2
k
X
)
cos
(
2
ω
t
)
{\displaystyle \xi _{2}={\frac {a\kappa _{s}kA^{2}}{16}}{\frac {1}{\kappa _{s}-m\omega ^{2}/k^{2}}}\sin(2kX)\cos(2\omega t)}
ξ
2
=
a
κ
s
k
A
2
16
1
κ
s
−
κ
T
sin
(
2
k
X
)
cos
(
2
ω
t
)
{\displaystyle \xi _{2}={\frac {a\kappa _{s}kA^{2}}{16}}{\frac {1}{\kappa _{s}-\kappa _{T}}}\sin(2kX)\cos(2\omega t)}
ξ
2
=
k
A
2
16
sin
(
2
k
X
)
cos
(
2
ω
t
)
{\displaystyle \xi _{2}={\frac {kA^{2}}{16}}\sin(2kX)\cos(2\omega t)}
cos
(
2
ω
t
)
=
1
−
2
sin
2
ω
t
)
{\displaystyle \cos(2\omega t)=1-2\sin ^{2}\omega t)}
ξ
1
=
k
A
2
16
(
1
−
2
sin
2
ω
t
)
sin
(
2
k
X
)
{\displaystyle {\color {red}\xi _{1}={\frac {kA^{2}}{16}}\left(1-2\sin ^{2}\omega t\right)\sin(2kX)}}
By the linearity of the operator
L
,
{\displaystyle {\mathcal {L}},}
ξ
p
=
ξ
1
+
ξ
2
:
{\displaystyle \xi _{p}=\xi _{1}+\xi _{2}:}
ξ
1
=
k
A
2
8
(
−
sin
2
(
ω
t
)
+
κ
T
2
κ
s
)
sin
(
2
k
X
)
{\displaystyle \xi _{1}={\frac {kA^{2}}{8}}\left(-\sin ^{2}(\omega t)+{\frac {\kappa _{T}}{2\kappa _{s}}}\right)\sin(2kX)}
In these units the speed of a
{
transverse, longitudinal
}
{\displaystyle \left\{{\text{transverse, longitudinal}}\right\}}
{
c
T
=
κ
T
/
m
,
c
s
=
κ
s
/
m
}
{\displaystyle \left\{c_{T}={\sqrt {\kappa _{T}/m}},\,c_{s}={\sqrt {\kappa _{s}/m}}\right\}}
[ 1]
k
L
0
=
π
{\displaystyle kL_{0}=\pi }
ξ
1
=
π
A
2
8
L
0
(
−
sin
2
(
ω
t
)
+
c
T
2
2
c
s
2
)
sin
(
2
k
X
)
{\displaystyle \xi _{1}={\frac {\pi A^{2}}{8L_{0}}}\left(-\sin ^{2}(\omega t)+{\frac {c_{T}^{2}}{2c_{s}^{2}}}\right)\sin(2kX)}
{\displaystyle }
{\displaystyle }
{\displaystyle }
x
x
<
m
a
t
h
>
{\displaystyle {\color {red}xx}<math>}
{\displaystyle }
{\displaystyle }
