Answer to Question #183952 in Differential Equations for Santhi

Question

 string is stretched and fastened 

 to two points 

l

cm apart. Motion is started by displacing the string into the form of the curve 

 

l

x

l

x

y

 2

cos

3

2sin

and then releasing it from this position at time 

t 0.

Do not use the 

 symbol 

l

but its actual value should be used in all of your calculations (steps). Find the 

 displacement function 

y(x,t).

Answer 1

The displacement of the point of the string at a distance x from the left end 0 at

time t is given by the equation-

$\dfrac{d^2y}{dt^2}=a^2\dfrac{d^2y}{dx^2}$

Since the ends of the string x=0 and x=l are fixed, they do not undergo any displacement

at any time.

$\text{ Hence } y(x,t)=0, for t\ge 0 ……. (2)\\ \text{and }y(l,t)=0 for t\ge 0 ……. (3)$

Since the string is released from rest initially, that is , at t=0, the initial velocity of every

point of the string in the y-direction is zero.

$\dfrac{dy}{dt}(x,0)=0 for 0\le x\le l$

Since the string is initially displaced in to the form of the curve , t0he coordinates 

The solution to the above problem is-

$y(x,t)=(Acospx+Bsinpx)(Ccospat+Dsinpat)$

Where A, B, C, D and p are arbitrary constants that are to be found out by using the

boundary conditions. 

Using Boundary condition we can calculate the values of Arbitary constant-

$A=0,B=0,D=0,p=\dfrac{n\pi}{l}$

The general solution is-

$y(x,t)=\sum_{n=1}^{\infty}(c_nk)sin\dfrac{n\pi}{l}cos\dfrac{n\pi at}{l}$

Again using Boundary condition we have-

$\sum_{n=1}^{\infty}\lambda_n sin\dfrac{n\pi x}{l}=f(x) for 0\le x\le l$

As $f(x)=ksin^3\dfrac{\pi x}{l}$

Now calculating f(x) and equating with above equation we get-

$y(x,t)=\dfrac{3k}{4}sin\dfrac{\pi x}{l}cos\dfrac{\pi at}{l}-\dfrac{k}{4}sin\dfrac{3\pi x}{l}cos\dfrac{3\pi at}{l}$