Question #279760

A string of length l has it's ends x=0 and x=l fixed. It is released from its rest in position y=[4 lamda x(l-x)]/l². Find an expression of the displacement of the string at any subsequent time.

1
Expert's answer
2021-12-15T16:13:51-0500

wave equation:

ytt=λ2yxxy_{tt}=\lambda^2 y_{xx}


solution:

y(x,t)=sin(nπx/l)(ancos(λπnt/l)+bnsin(λπnt/l))y(x,t)=\sum sin(n\pi x/l)(a_ncos(\lambda \pi nt/l)+b_n sin(\lambda \pi nt/l))


where


an=2l0lf(x)sin(nπx/l)dxa_n=\frac{2}{l}\int^l_0 f(x) sin(n\pi x/l)dx


bn=2λnπ0lg(x)sin(nπx/l)dxb_n=\frac{2}{\lambda n \pi}\int^l_0 g(x) sin(n\pi x/l)dx


we have:

f(x)=y(x,0)=4λx(lx)/l2f(x)=y(x,0)=4\lambda x(l-x)/l^2

g(x)=yt(x,0)=0g(x)=y_t(x,0)=0

y(0,t)=y(l,t)=yt(0,t)=yt(l,t)=0y(0,t)=y(l,t)=y_t(0,t)=y_t(l,t)=0


then:

bn=0b_n=0


an=2l0l(4λx(lx)/l2)sin(nπx/l)dx=8λl3l3(πnsin(πn)+2cos(πn)2)π3n3=a_n=\frac{2}{l}\int^l_0(4\lambda x(l-x)/l^2) sin(n\pi x/l)dx=-\frac{8\lambda}{l^3}\frac{l^3(\pi nsin(\pi n)+2cos(\pi n)-2)}{\pi^3n^3}=


=8λ(πnsin(πn)+2cos(πn)2)π3n3=-\frac{8\lambda(\pi nsin(\pi n)+2cos(\pi n)-2)}{\pi^3n^3}


y(x,t)=8λ(πnsin(πn)+2cos(πn)2)π3n3sin(nπx/l)cos(λπnt/l)y(x,t)=-\sum\frac{8\lambda(\pi nsin(\pi n)+2cos(\pi n)-2)}{\pi^3n^3}sin(n\pi x/l)cos(\lambda \pi nt/l)


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