Answer to Question #265567 in Differential Equations for Gowtham

Question #265567

A tightly stretched string of length"l"has it's ends fastened at x=0andx=l. The midpoint of the string is taken the height h and released. find the initial displacement function at timet=0?

1
Expert's answer
2021-11-19T06:53:51-0500

Displacement of a string y(x,t) given by;

2yt2=a22yx2\frac{\partial^2y}{\partial t^2}=a^2\frac{\partial^2y}{\partial x^2} .....(1)

A suitable solution of (1) is given by;

y(x,t)=(Acoslx+Bsinlx)(Ccoslat+Dsinlat).....(2)y(x,t)=(Acoslx+Bsinlx)(Ccoslat+Dsinlat).....(2)

The boundary conditions are;

(i)y(0,t)=0(i)y(0,t)=0

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

(iii)(yt)t=0=0(iii)(\frac{\partial y}{\partial t})_{t=0}=0

(iv)

(y(x,0)={2hxlif 0xl22h(lx)lif l2xl(y(x,0) = \begin{cases} \frac{2hx}{l} &\text{if } 0\leq x\leq\frac l2 \\ \frac{2h(l-x)}{l} &\text{if } \frac l2\leq x\leq l \end{cases}

Using conditions (i) and (ii) in (2),we get;

A=0

λ=nπl\lambda=\frac{nπ}{l}

Therefore;

y(x,t)=Bsinnπxl[Ccosnπatl+Dsinnπatl]y(x,t)=Bsin\frac{nπx}{l}[Ccos\frac{nπat}{l}+Dsin\frac{nπat}{l}]

Differentiate to obtain;

yt=Bsinnπxl(Csinnπatl.nπtl+Dcosnπatl.nπal)\frac{\partial y}{\partial t}=Bsin\frac{nπx}{l}(-Csin\frac{nπat}{l}.\frac{nπt}{l}+Dcos\frac{nπat}{l}.\frac{nπa}{l})

Using conditions (iii) in the above equation we get;

0=BsinnπxlDsinnπal0=Bsin\frac{nπx}{l}Dsin\frac{nπa}{l}

Since B0B\neq0 ,then D=0

Equation (2) reduces to;

y(x,t)=BsinnπxlCcosnπatly(x,t)=Bsin\frac{nπx}{l}Ccos\frac{nπat}{l}

Rewrite;

y(x,t)=B1sinnπxlcosnπatl....(4)y(x,t)=B_1sin\frac{nπx}{l}cos\frac{nπat}{l}....(4)

(Note that B1=BC )

Using condition (iv) ,we get that;

y(x,0)=n=0Bnsinnπxl..(5)y(x,0)=\displaystyle\sum_{n=0}^{\infin}B_nsin\frac{nπx}{l}..…(5)

The RHS of equation (5) is the half range Fourier sine series of the LHS function, therefore;

Bn=2l0lf(x)sinnπxldxB_n=\frac 2l\int_0^lf(x)sin\frac{nπx}{l}dx

=2l[0L22hxlsinnπxldx+l202h(lx)lsinnπxldx]=\frac2l[\int_0^{\frac L2}\frac{2hx}{l}sin\frac{nπx}{l}dx+\int_{\frac l2}^0\frac{2h(l-x)}{l}sin\frac{nπx}{l}dx]

Rewrite;

=4hl2[0L2xsinnπxldx+L20(lx)sinnπxldx]=\frac{4h}{l^2}[\int_0^{\frac L2}xsin\frac{nπx}{l}dx+\int_{\frac L2}^0(l-x)sin\frac{nπx}{l}dx]

Integrating on the given limits;

=4hl2[[xcosnπxlnπl+sinnπxln2π2l2]0L2+[(xl)cosnπxlnπl+sinnπxln2π2l2]l2l]=\frac{4h}{l^2}[[\frac{-xcos\frac{nπx}{l}}{\frac{nπ}{l}}+\frac{sin \frac{nπx}{l}}{\frac{n^2π^2}{l^2}}]_0^{\frac L2}+[\frac{(x-l)cos\frac{nπx}{l}}{\frac{nπ}{l}}+\frac{sin\frac{nπx}{l}}{\frac{n^2π^2}{l^2}}]_{\frac l2}^l]

Applying the limits ;

4hl2[sinnπ2n2π2l2]=4hsinnπ2n2π2\frac{4h}{l^2}[\frac{sin\frac{nπ}{2}}{\frac{n^2π^2}{l^2}}]=\frac{4hsin\frac{nπ}{2}}{n^2π^2}

Hence, the solution is;

y(x)=n=14hsinnπ2n2π2cosnπatlsinnπxly(x)=\displaystyle\sum_{n=1}^{\infin}\frac{4hsin\frac{nπ}{2}}{n^2π^2}cos\frac{nπat}{l}sin\frac{nπx}{l}

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