Question #284508

If a string of length l is initially at rest in equilibrium position and each of its points is given the velocity dy/dt= b sin^3πx/l find the displacement





1
Expert's answer
2022-01-04T13:42:28-0500

The displacement y(x,t) is given by the equation:

ytt=a2yxxy_{tt}=a^2y_{xx}

boundary conditions:

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

yt(0)=bsin(3πx/l)y_t(0)=b sin(3πx/l)

y(x,0)=0y(x,0)=0


solution:

y(x,t)=Bsin(nπx/l)(Ccos(nπat/l)+Dsin(nπat/l))y(x,t)=Bsin(n\pi x/l)(Ccos(n\pi at/l)+Dsin(n\pi at/l))

then:

y(x,0)=Bsin(nπx/l)C=0    C=0y(x,0)=Bsin(n\pi x/l)C=0 \implies C=0

y(x,t)=Bsin(nπx/l)Dsin(nπat/l)y(x,t)=Bsin(n\pi x/l)Dsin(n\pi at/l)

the most general solution:

y(x,t)=Bnsin(nπx/l)sin(nπat/l)y(x,t)=\sum B_nsin(n\pi x/l)sin(n\pi at/l)


then:

yt=Bn(nπa/l)sin(nπx/l)sin(nπat/l)y_t=\sum B_n(n\pi a/l)sin(n\pi x/l)sin(n\pi at/l)


yt(0)=bsin(3πx/l)=Bn(nπa/l)sin(nπx/l)sin(nπat/l)y_t(0)=b sin(3πx/l)=\sum B_n(n\pi a/l)sin(n\pi x/l)sin(n\pi at/l)


b=B3(3πa/l),B1=B2=B4=...=0b=B_3(3\pi a/l),B_1=B_2=B_4=...=0


B3=bl3πaB_3=\frac{bl}{3\pi a}


y(x,t)=B3sin(3πx/l)sin(3πat/l)=bl3πasin(3πx/l)sin(3πat/l)y(x,t)=B_3sin(3\pi x/l)sin(3\pi at/l)=\frac{bl}{3\pi a}sin(3\pi x/l)sin(3\pi at/l)


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