Question

A string is stretched between two fixed points at a distance 2l cm and the points of the string are given initial velocities (v) is given by v(x, 0) = {kxl, 0<x<lk(2l−x)l, l<x<2l, x being the distance from an end point. Find the displacement of the string at any time.

Answer 1

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

$$\frac{\delta^2y}{ \delta t^2}=a^2 \frac{\delta^2y}{ \delta x^2}------------(1)$$

The suitable solution of (1) is given by

$y(x,t) = (A cos lx + B sin lx)(C cos lat + D sin lat) -------(2)$

The boundary conditions are

$(i)\ y(0,t) = 0,\ for\ t\ge0. \\ (ii)\ y(ℓ,t) ,\ for\ t\ge0. \\ (iii)\ y(x,0)=0\ for\ all\ 0<x<i\\ (iv)\ [\frac{\delta y}{\delta t}]_{t=0}=g(x)= \begin{cases} \frac{kx}{l} & 0<x<l \\ \frac{k(2l-x)}{l} & l<x<2l \end{cases}$

Using    (i) and (ii) in (2) , we get

$0 = A(Ccoslat + Dsinlat)$ , for all $t \ge 0.$

Therefore,           $A = 0$

Hence equation (2) becomes

$y(x,t) = B sinlx(Ccoslat + Dsinlat) ------------ (3)$

Using (ii) in (3), we get

 

$0 = Bsinlℓ (Ccoslat+Dsinlat)$ , for all  $t \ge 0$ , which gives $lℓ = n\pi$

Hence, $l=\frac{n \pi}{ℓ}$ , n being an integer.

Thus

$y(x,t)=Bsin \frac{n \pi x}{l}[Ccos \frac{n \pi at}{l}+Dsin \frac{n \pi at}{l}]-----------(4)$

Using (iii) in (4), we get

$0=Bsin \frac{n \pi x}{l}C$

Therefore, $C = 0$

Hence $y(x,t)=Bsin \frac{n \pi x}{l} \cdot Dsin \frac{n \pi at}{l}=B_1sin \frac{n \pi x}{l} \cdot Dsin \frac{n \pi at}{l}$

Where $B_1=BD$

The most general solution is

$y(x,t)=\sum_{n=1}^\infin B_nsin \frac{n \pi x}{l} \cdot sin \frac{n \pi at}{l}---------(5)$

Differentiating (5) partially w.r.t t, we get

$\frac{\delta y}{\delta t}=\sum_{n=1}^\infin B_nsin \frac{n \pi x}{l} cos \frac{n \pi at}{l} \cdot \frac{n \pi a}{l}$

Using condition (iv) in the above equation, we get

$y_t(x,0)=g(x)= \begin{cases} \frac{kx}{l} & 0<x<l \\ \frac{2l-x}{l} & l<x<2l \end{cases}$

Taking $\frac{kx}{l},\ 0<x<l$ we get

$\frac{kx}{l}=\sum_{n=1}^\infin B_n \frac{n \pi a}{l} \cdot sin\frac{n \pi x}{l}\\ i.e\ B_n \frac{n \pi a}{l}=\frac{2}{l}\int_0^1 f(x)\cdot sin\frac{n \pi x}{l}dx\\ i.e\ B_n =\frac{2}{n \pi a}\int_0^1 f(x)\cdot sin\frac{n \pi x}{l}dx=\frac{2}{n \pi a}\int_0^1 \frac{kx}{l}\cdot sin\frac{n \pi x}{l}dx= \frac{2}{n \pi a}\int_0^1 \frac{kx}{l}d[\frac{-cos\frac{n \pi x}{l}}{\frac{n \pi x}{l}}]\\ = \frac{2}{n \pi a}\{\frac{kx}{l}d[\frac{-cos\frac{n \pi x}{l}}{\frac{n \pi x}{l}}]-\frac{k}{l}[\frac{-sin\frac{n \pi}{l}}{\frac{n^2 \pi^2 }{l}}]\}=\frac{2}{n \pi a}\{\frac{-2cos\ n\pi+2}{\frac{n^3 \pi^3 }{l^3}}\}=\frac{2}{n \pi a} \cdot \frac{2l^3}{n^3 \pi^3}\{1-cos\ n\pi\}\\ i.e\ B_n =\frac{4l^3}{n^4 \pi^4 a}\{1-(-1)^n\}\\ or\\ B_n= \begin{cases} \frac{8l^3}{l} & if\ n\ is\ odd \\ 0 & if\ n\ is\ even \end{cases}\\ \therefore\ solution\ is\\ y(x,t)=\frac{8l^3}{n^4a}\sum_{n=1}^\infin \frac{1}{(2n-1)^4}\sin\ \frac{(2n-1)\pi a t}{l}\sin\ \frac{(2n-1)\pi x}{l},\ \forall\ 0<x<l--->Answer$

Taking $\frac{k(2l-x)}{l},\ l<x<2l$ we get

$\frac{2kl-kx}{l}=\sum_{n=1}^\infin B_n \frac{n \pi a}{2l} \cdot sin\frac{n \pi x}{2l}\\ i.e\ B_n \frac{n \pi a}{l}=\frac{2}{l}\int_l^{2l} g(x)\cdot sin\frac{n \pi x}{2l}dx\\ i.e\ B_n =\frac{2}{n \pi a}\int_l^{2l} g(x)\cdot sin\frac{n \pi x}{2l}dx=\frac{2}{n \pi a}\int_l^{2l} \frac{2kl-kx}{l}\cdot sin\frac{n \pi x}{2l}dx= \frac{2}{n \pi a}\int_l^{2l} \frac{2kl-kx}{l}d[\frac{-cos\frac{n \pi x}{2l}}{\frac{n \pi x}{2l}}]\\ = \frac{2}{n \pi a}\{\frac{2kl-kx}{l}d[\frac{-cos\frac{n \pi x}{2l}}{\frac{n \pi x}{2l}}]-\frac{2kl-k}{l}[\frac{-sin\frac{n \pi}{2l}}{\frac{n^2 \pi^2 }{2l}}]\}=\frac{2}{n \pi a}\{\frac{-2cos\ n\pi+2}{\frac{n^3 \pi^3 }{4l^3}}\}=\frac{2}{n \pi a} \cdot \frac{8l^3}{n^3 \pi^3}\{1-cos\ n\pi\}\\ i.e\ B_n =\frac{16l^3}{n^4 \pi^4 a}\{1-(-1)^n\}\\ or\\ B_n= \begin{cases} \frac{32l^3}{l} & if\ n\ is\ odd \\ 0 & if\ n\ is\ even \end{cases}\\ \therefore\ solution\ is\\ y(x,t)=\frac{32l^3}{n^4a}\sum_{n=1}^\infin \frac{1}{(2n-1)^4}\sin\ \frac{(2n-1)\pi a t}{l}\sin\ \frac{(2n-1)\pi x}{l},\ \forall\ l<x<2l-->Answer$