Answer to Question #199556 in Differential Equations for Rajkumar

Question

A tightly stretched string with fixed end points x = 0 and x = l is initially in a position

given by y = y₀ sin³(πx/l) . It is released from rest from the initial position. Find the

displacement y(x, t) .

Answer 1

Wave equation:

$y_{xx}=y_{tt}/c^2$

Separate the variables:

$y(x,t)=X(x)Y(t)$

$\frac{X''}{X}=\frac{Y''}{c^2Y}=-k^2$

$X''+k^2X=0$

$X(x)=c_1sin(kx)+c_2cos(kx)$

$Y''+c^2k^2Y=0$

$Y(t)=c_3sin(ckt)+c_4cos(ckt)$

$y(x,t)=(c_1sin(kx)+c_2cos(kx))(c_3sin(ckt)+c_4cos(ckt))$

The boundary conditions:

$y(0,t)=0$ . fixed node

$y(l,t)=0$ . fixed node

$y_t(x,0)=0$ , string is only released at t=0

$y(x,0)=y_0sin^3(\pi x/l)$ , initial shape

For the 1st condition:

$c_2(c_3sin(ckt)+c_4cos(ckt))=0$

For the 2nd condition:

$c_1sin(kl)(c_3sin(ckt)+cos(ckt))=0$

$k=n\pi/l$ , $n=1,2,...$

Now, the solution:

$y(x,t)=c_1sin(n\pi x/l)(c_3sin(n\pi ct/l)+c_4cos(n\pi ct/l))$

For the 3rd condition:

$c_1sin(n\pi x/l)(c_3n\pi c/l)=0\implies c_3=0$

$y(x,t)=c_1c_4\displaystyle{\sum^{\infin}_1b_nsin(n\pi x/l)cos(n\pi ct/l)}$

For the 4th condition:

$y(x,0)=y_0sin^3(\pi x/l)=y_0(\frac{3}{4}sin(\pi x/l)-\frac{1}{4}sin(3\pi x/l))=$

$=\displaystyle{\sum^{\infin}_1b_nsin(n\pi x/l)}$

Comparing coefficients, we get:

$b_1=3y_0/4$

$b_2=b_4=0$

$b_3=-y_0/4$

The final solution:

$y(x,t)=\frac{3}{4}y_0(\pi x/l)cos(\pi ct/l)-\frac{1}{4}y_0(3\pi x/l)cos(3\pi ct/l)$