Displacement of a string y(x,t) given by;
∂t2∂2y=a2∂x2∂2y .....(1)
A suitable solution of (1) is given by;
y(x,t)=(Acoslx+Bsinlx)(Ccoslat+Dsinlat).....(2)
The boundary conditions are;
(i)y(0,t)=0
(ii)y(l,t)=0
(iii)(∂t∂y)t=0=0
(iv)
(y(x,0)={l2hxl2h(l−x)if 0≤x≤2lif 2l≤x≤l
Using conditions (i) and (ii) in (2),we get;
A=0
λ=lnπ
Therefore;
y(x,t)=Bsinlnπx[Ccoslnπat+Dsinlnπat]
Differentiate to obtain;
∂t∂y=Bsinlnπx(−Csinlnπat.lnπt+Dcoslnπat.lnπa)
Using conditions (iii) in the above equation we get;
0=BsinlnπxDsinlnπa
Since B=0 ,then D=0
Equation (2) reduces to;
y(x,t)=BsinlnπxCcoslnπat
Rewrite;
y(x,t)=B1sinlnπxcoslnπat....(4)
(Note that B1=BC )
Using condition (iv) ,we get that;
y(x,0)=n=0∑∞Bnsinlnπx..…(5)
The RHS of equation (5) is the half range Fourier sine series of the LHS function, therefore;
Bn=l2∫0lf(x)sinlnπxdx
=l2[∫02Ll2hxsinlnπxdx+∫2l0l2h(l−x)sinlnπxdx]
Rewrite;
=l24h[∫02Lxsinlnπxdx+∫2L0(l−x)sinlnπxdx]
Integrating on the given limits;
=l24h[[lnπ−xcoslnπx+l2n2π2sinlnπx]02L+[lnπ(x−l)coslnπx+l2n2π2sinlnπx]2ll]
Applying the limits ;
l24h[l2n2π2sin2nπ]=n2π24hsin2nπ
Hence, the solution is;
y(x)=n=1∑∞n2π24hsin2nπcoslnπatsinlnπx
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