Question

A string is stretched and fastened to two points x = 0 and x = l apart. Motion is started by

displacing the string into the form y = k(lx − x2

) from which it is released at time t = 0. Find

the displacement of any point on the string at a distance of x from one end at time t.

Hint: From this problem, we have the following boundary conditions:

y(0,t) = 0 for all t > 0

y(l,t) = 0 for all t > 0

∂y

∂t

(x, 0) = 0 (initial velocity is zero)

y(x, 0) = k(lx − x2

)

Answer 1

$y=k(lx-x^2)$

Boundary conditions:

1. $y(0,t)=0$ For all t>0

2. $y(1,t)=0$ For all t>0

3. $\frac{dy}{dt}(y,0)=0$

4. $y(x,0)=k(lx-x^2)$

The displacement y(x,t) is given by the equation

$\frac{d^2y}{dt^2}=a^2\frac{d^2y}{dx^2}$ ..........(i)

The suitable solution of (i) is given by

$y(x,t)=(Acos\ lx+Bsin\ lx)(Ccos\ lat+Dsin\ lat)$......(ii)

Using the first and second boundary conditions in (ii) we get,

A=0 and $\lambda=\frac{nπ}{l}$

$\therefore y(x,t)=Bsin\frac{nπx}{l}(Ccos \frac{nπat}{l}+Dsin\frac{nπat}{l}.........(iii)$

Now $\frac{dy}{dt}=Bsin\frac{nπx}{l}(-Csin\frac{nπat}{l}\cdot\frac{nπa}{l}+Dcos\frac{nπat}{l}\cdot\frac{nπa}{l})$

Using third boundary condition,

0=$Bsin\frac{nπx}{l}D\frac{nπa}{l}$

Here, B can not be 0, hence,D=0

Thus equation (iii) becomes;

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

$=B_1sin\frac{nπx}{l}\cdot cos\frac{nπat}{l}$ ,where B₁=Bc

The most general solution is;

$y(x,t)=\displaystyle\sum_{n=1}^{\infty}Bsin\frac{nπx}{l} cos\frac{nπat}{l}$ ....(iv)

Using the forth boundary condition;

$k(lx-x^2)=\displaystyle\sum_{n=1}^{\infty}Bsin\frac{nπx}{l}$ .....(v)

The RHS of (v) is the half range Fourier sine series of the LHS function

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

$=\frac{2k}{l}\int_0^l(lx-x^2)d(\frac{-cos\frac{nπx}{l}}{\frac{nπ}{l}})$

$=\frac{2k}{l}[(lx-x^2)d(\frac{-cos\frac{nπx}{l}}{\frac{nπ}{l}})-(l-2x)(\frac{-sin\frac{nπx}{l}}{\frac{n^2π^2}{l^2}})+(-2)(\frac{cos\frac{nπx}{l}}{\frac{n^3π^3}{l^3}}]_0^l$

$=\frac{2k}{l}(\frac{-2cosnπ}{\frac{n^3π^3}{l^3}})+(\frac{2}{\frac{n^3π^3}{l^3}})$

$=\frac{2k}{l}\cdot\frac{2l^3}{n^3π^3}(1-cosnπ)$

$i.e, B_n=\frac{4kl^2}{n^3π^3}(1-(-1)^n)\ or\ B_n= \begin{cases} \frac{8kl^2}{n^3π^3} &\text{if } n\ is\ odd \\ 0 &\text{if } n\ is\ even \end{cases}$

$\therefore y(x,t)=\displaystyle\sum_{n=odd}^{\infty}\frac{8kl^2}{n^3π^3}cos\frac{nπat}{l}sin\frac{nπx}{l}$

Or $y(x,t)=\frac{8k}{π^3}\displaystyle\sum_{n=1}^{\infty}\frac{1}{(2n-1)^3}cos\frac{(2n-1)πat}{l}sin\frac{(2n-1)πx}{l}$