Answer to Question #265567 in Differential Equations for Gowtham

Question

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?

Answer 1

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

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

A suitable solution of (1) is given by;

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

The boundary conditions are;

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

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

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

(iv)

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

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

A=0

$\lambda=\frac{nπ}{l}$

Therefore;

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

Differentiate to obtain;

$\frac{\partial y}{\partial t}=Bsin\frac{nπx}{l}(-Csin\frac{nπat}{l}.\frac{nπt}{l}+Dcos\frac{nπat}{l}.\frac{nπa}{l})$

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

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

Since $B\neq0$ ,then D=0

Equation (2) reduces to;

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

Rewrite;

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

(Note that B₁=BC )

Using condition (iv) ,we get that;

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

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

$B_n=\frac 2l\int_0^lf(x)sin\frac{nπx}{l}dx$

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

Rewrite;

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

Integrating on the given limits;

$=\frac{4h}{l^2}[[\frac{-xcos\frac{nπx}{l}}{\frac{nπ}{l}}+\frac{sin \frac{nπx}{l}}{\frac{n^2π^2}{l^2}}]_0^{\frac L2}+[\frac{(x-l)cos\frac{nπx}{l}}{\frac{nπ}{l}}+\frac{sin\frac{nπx}{l}}{\frac{n^2π^2}{l^2}}]_{\frac l2}^l]$

Applying the limits ;

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

Hence, the solution is;

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