Question

A $5$ cm long metal rod ($x$ in cm from $A$ to $B$) is governed by $\dfrac{\partial T{\partial t}=D\,\dfrac{\partial^2 T}{\partial x^2}$ with $D=1.0~\text{cm}^2/\text{s}$ and both ends held at $0^\circ$C. The temperature field is} \[ T(x,t)=\sum_{n=1,3,5,\ldots} C_n \sin\!\left(\frac{n\pi x}{5}\right)e^{-\beta n^{2} t}. \] Find $\beta$ (in s$^{-1}$, rounded to three decimals).

Hint:

In 1-D heat conduction with fixed ends, each sine mode $\sin(n\pi x/L)$ decays as $\exp\!\left[-D(n\pi/L)^2 t\right]$.
If the solution is written as $e^{-\beta n^2 t}$, simply set $\beta = D(\pi/L)^2$.

Answer 1

Step 1: Match the separated-form exponent with the PDE eigenvalue.
For a rod of length $L=5$ cm with zero temperature at both ends, separation of variables gives modes
$\sin\!\left(\dfrac{n\pi x}{L}\right)$ with temporal decay $\exp\!\left[-D\left(\dfrac{n\pi}{L}\right)^2 t\right]$.

Step 2: Identify $\beta$.
Given form uses $\exp(-\beta n^2 t)$, hence equate decay constants:
\[ \beta n^2 = D\left(\frac{n\pi}{L}\right)^2 \ \Rightarrow\ \beta = D\left(\frac{\pi}{L}\right)^2. \]

Step 3: Substitute $D=1.0~\text{cm^2/\text{s}$ and $L=5$ cm.}
\[ \beta = 1.0\left(\frac{\pi}{5}\right)^2 = \frac{\pi^2}{25} \approx \frac{9.8696}{25} = 0.394784~\text{s}^{-1}. \]

Step 4: Rounding.
To three decimals, $\beta \approx \boxed{0.395~\text{s}^{-1}}$.