Question

The initial b, and simplify Find the Find the temperature at th emiddle of the rod after w seconds using the first two terms of the A.

Answer 1

Solution: II

A bar with thermal diffusivity 25 and length \pi has both ends at zero temperature. The initial temperature of the bar is f(x) =x. The solution of the corresponding heat equation is given by

$u(x,t)=\sum_{n=1}^{\infty}b_{n}e^{(-n^2\pi^2kt)/L^2}\sin\left ( \frac{n\pi}{L} x\right )~~~~~~~~~~(i)$'

A.

The initial condition(IC) is given at t=0,

bn sin (-r) = f(x) = n= 1

The boundary condition(BC) is given by at x=0 and at x=L;

$u(0,t)=\sum_{n=1}^{\infty}b_{n}e^{(-n^2\pi^2kt)/L^2}\sin\left ( \frac{n\pi}{L} .0\right )=0$

and

n= 1 n=1

Therefore initial and boundary conditions are given by:

B. We know that

n = m rl

n Sin

Multiplying both sides by
Sin
, and integrating from 0 to L, we have

in_T

x sin(ー2) dr 〉.bn n=1 sin(ーー2) sin

$\Rightarrow \int_{0}^{L}x\sin\left ( \frac{m\pi}{L}x \right )dx=\sum_{n=1}^{\infty}b_{n} \left\{\begin{matrix} 0~~~n\neq m\\ \frac{L}{2},~~ n=m \end{matrix}\right.$

$\Rightarrow \int_{0}^{L}x\sin\left ( \frac{m\pi}{L}x \right )dx=b_{m}\frac{L}{2}$

$\Rightarrow b_{m}= \frac{2}{L}\int_{0}^{L}x\sin\left ( \frac{m\pi}{L}x \right )dx$

$\Rightarrow b_{n}= \frac{2}{L}\int_{0}^{L}x\sin\left ( \frac{n\pi}{L}x \right )dx$

$\Rightarrow b_{n}= \frac{2}{L} \left [ -\frac{xL}{n\pi}\cos\left ( \frac{n\pi}{L}x \right ) \right ]_{0}^{L}+ \frac{2}{L}\frac{L}{n\pi}\int_{0}^{L}\cos\left ( \frac{n\pi}{L}x \right )dx$

$\Rightarrow b_{n}= \frac{2}{L} \left [ -\frac{xL}{n\pi}\cos\left ( \frac{n\pi}{L}x \right ) \right ]_{0}^{L}+ \frac{2}{L}\frac{L^2}{n^2\pi^2}\left [\sin\left ( \frac{n\pi}{L}x \right ) \right ]_{0}^{L}$

$\Rightarrow b_{n}= \frac{2}{L} \left [ -\frac{L.L}{n\pi}\cos\left ( \frac{n\pi}{L}L \right ) \right ]+ 0$

$\Rightarrow b_{n}=- \frac{2L}{n \pi } \cos(n\pi)$

$\Rightarrow b_{n}=- \frac{2L}{n \pi } (-1)^{n}$