Question-196785

Question Number 196785 by Spillover last updated on 31/Aug/23

Answered by aleks041103 last updated on 03/Oct/23

Separation of variables Ψ(x,t)=X(x)T(t) ⇒X′′T=−(1/c^2 )XT′′ ⇒((X′′)/X)=−(1/c^2 ) ((T ′′)/T)=−a^2 ⇒X′′+a^2 X=0 ⇒X=sin(ax+b) T ′′ −(ac)^2 T=0⇒T=sinh(act+d) ⇒XT=sin(ax+b)sinh(act+d) X(0)=X(L)=0 ⇒sin(b)=sin(aL+b)=0 ⇒b=0, aL=πk⇒a=((kπ)/L),k∈Z ⇒Ψ(x,t)=Σ_(k=−∞) ^∞ A_k sin(((πkx)/L))sinh(((πkct)/L)+B_k ) where A_k , B_k are constants determined by initial conditions.

$${Separation}\:{of}\:{variables} \\ $$$$\Psi\left({x},{t}\right)={X}\left({x}\right){T}\left({t}\right) \\ $$$$\Rightarrow{X}''{T}=−\frac{\mathrm{1}}{{c}^{\mathrm{2}} }{XT}'' \\ $$$$\Rightarrow\frac{{X}''}{{X}}=−\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\:\frac{{T}\:''}{{T}}=−{a}^{\mathrm{2}} \\ $$$$\Rightarrow{X}''+{a}^{\mathrm{2}} {X}=\mathrm{0} \\ $$$$\Rightarrow{X}={sin}\left({ax}+{b}\right) \\ $$$${T}\:''\:−\left({ac}\right)^{\mathrm{2}} {T}=\mathrm{0}\Rightarrow{T}={sinh}\left({act}+{d}\right) \\ $$$$\Rightarrow{XT}={sin}\left({ax}+{b}\right){sinh}\left({act}+{d}\right) \\ $$$${X}\left(\mathrm{0}\right)={X}\left({L}\right)=\mathrm{0} \\ $$$$\Rightarrow{sin}\left({b}\right)={sin}\left({aL}+{b}\right)=\mathrm{0} \\ $$$$\Rightarrow{b}=\mathrm{0},\:{aL}=\pi{k}\Rightarrow{a}=\frac{{k}\pi}{{L}},{k}\in\mathbb{Z} \\ $$$$\Rightarrow\Psi\left({x},{t}\right)=\underset{{k}=−\infty} {\overset{\infty} {\sum}}{A}_{{k}} {sin}\left(\frac{\pi{kx}}{{L}}\right){sinh}\left(\frac{\pi{kct}}{{L}}+{B}_{{k}} \right) \\ $$$${where}\:{A}_{{k}} ,\:{B}_{{k}} \:{are}\:{constants}\:{determined}\:{by} \\ $$$${initial}\:{conditions}. \\ $$

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Question-196785

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