Find the increase in the length of the rod

[utkarshakash]

Homework Statement

A metal rod of length L at temperature of 0°C is not uniformly heated such that the temperature is given by the distance x along its length measured from one end when:
$T(x) = T_0 \sin (\pi x/L)$
Accordingly, points at x = 0 and x = L are also zero temperature, whereas at x = L/2, where the argument of sine function is π/2, the temperature have the maximum value T0. The coefficient of linear expansion of the rod is α. Find the increase in the length of the rod in function of α and T0.

The Attempt at a Solution

Let us consider a differential element dx at a distance x from one end of the rod.

$Δ(dx) = dx \alpha dT \\ ΔL = \alpha T_0 \displaystyle \int_0^L \cos \left( \dfrac{\pi x}{L} \right) dx$

But the above equation gives me 0!
I know something's going wrong here.

 

[Doc Al]

$Δ(dx) = dx \alpha dT$

Careful. That should be ΔT, not dT. The temperature difference is a finite function of x, not a differential. (And if it's with respect to 0°, then ΔT = T.)

 

[utkarshakash]

Careful. That should be ΔT, not dT. The temperature difference is a finite function of x, not a differential. (And if it's with respect to 0°, then ΔT = T.)

Thanks!

 

[Tanya Sharma]

What is the answer ?

 

[utkarshakash]

$2L \alpha T_0 / \pi$

 

[Tanya Sharma]

$2L \alpha T_0 / \pi$

Could you please show how you got this answer.

 

[utkarshakash]

Could you please show how you got this answer.

Sure.

Let's consider a differential element dx at a distance x from one end of the rod.
$Δ(dx) = dx \alpha (t(0) - t(x)) \\ =- \alpha T_0 \sin \dfrac{\pi x}{L} dx \\$

Integrating both sides

$ΔL = \dfrac{- L \alpha T_0}{\pi} \left( \cos \dfrac{\pi x}{L} \right)_0^L$

Substitute the values to get the answer.

 

[Tanya Sharma]

Sure.

Let's consider a differential element dx at a distance x from one end of the rod.
$Δ(dx) = dx \alpha (t(0) - t(x)) \\ =- \alpha T_0 \sin \dfrac{\pi x}{L} dx \\$

Integrating both sides

$ΔL = \dfrac{- L \alpha T_0}{\pi} \left( \cos \dfrac{\pi x}{L} \right)_0^L$

Substitute the values to get the answer.

Thanks a lot :)