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 }} , and 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)} where ξ h {\displaystyle \xi _{h}} is the solution to the homogeneous equation, i.e., solution to 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}}} and 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},} to (?) we first consider two different inhomogeneous equations:
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}} is proportional to sin ( 2 k X ) {\displaystyle \sin(2kX)} , then L ξ 1 = 4 k 2 κ s ξ 1 {\displaystyle {\mathcal {L}}\xi _{1}=4k^{2}\kappa _{s}\xi _{1}} , and: 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}} is proportional to sin ( 2 k X ) cos ( 2 ω t ) {\displaystyle \sin(2kX)\cos(2\omega t)} , then 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}} and: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)} . Now use 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}},} we see that a particular solution to (?) is the sum of 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\}} wave is { 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\}} . This permits us to write an expression that does not depend on the choice of units.[1] Relating the wavenumber of the lowest order mode to string length by 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)}
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Other identities
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