Question

An isosceles triangle is formed with a rod of length $l_1$ and coefficient of linear expansion $\alpha_1$ for the base and two thin rods each of length $l_2$ and coefficient of linear expansion $\alpha_2$ for the two pieces, if the distance between the apex and the midpoint of the base remain unchanged as the temperatures varied show that

$\frac{l_1}{l_2} = 2\sqrt{\frac{\alpha_2}{\alpha_1}}$.

Answer 1

The derivation shows that $\frac{l_1}{l_2} = 2\sqrt{\frac{\alpha_2}{\alpha_1}}$.

Answer 2

Let the isosceles triangle have base length $l_1$ and equal side lengths $l_2$. Let $h$ be the height from the apex to the midpoint of the base. The relationship between these lengths is given by the Pythagorean theorem: $h^2 + \left(\frac{l_1}{2}\right)^2 = l_2^2$ The problem states that the height $h$ remains unchanged as the temperature varies. This means $h$ is a constant.

Let the initial lengths be $l_{1,0}$ and $l_{2,0}$. When the temperature changes by $\Delta T$, the lengths become: $l_1 = l_{1,0}(1 + \alpha_1 \Delta T)$ $l_2 = l_{2,0}(1 + \alpha_2 \Delta T)$

Since $h$ is constant, the relation $h^2 + (l_1/2)^2 = l_2^2$ must hold at all temperatures. Let's express $h^2$ using the initial lengths: $h^2 = l_{2,0}^2 - (l_{1,0}/2)^2$. Substituting the expressions for $l_1$ and $l_2$ into the Pythagorean theorem: $h^2 + \frac{1}{4} [l_{1,0}(1 + \alpha_1 \Delta T)]^2 = [l_{2,0}(1 + \alpha_2 \Delta T)]^2$ Substitute the expression for $h^2$: $l_{2,0}^2 - \frac{l_{1,0}^2}{4} + \frac{l_{1,0}^2}{4}(1 + \alpha_1 \Delta T)^2 = l_{2,0}^2(1 + \alpha_2 \Delta T)^2$ For small temperature changes, we can approximate $(1+x)^2 \approx 1+2x$ and neglect terms of order $(\Delta T)^2$: $(1 + \alpha_1 \Delta T)^2 \approx 1 + 2\alpha_1 \Delta T$ $(1 + \alpha_2 \Delta T)^2 \approx 1 + 2\alpha_2 \Delta T$ The equation becomes: $l_{2,0}^2 - \frac{l_{1,0}^2}{4} + \frac{l_{1,0}^2}{4}(1 + 2\alpha_1 \Delta T) \approx l_{2,0}^2(1 + 2\alpha_2 \Delta T)$ $l_{2,0}^2 - \frac{l_{1,0}^2}{4} + \frac{l_{1,0}^2}{4} + \frac{l_{1,0}^2}{4}(2\alpha_1 \Delta T) \approx l_{2,0}^2 + l_{2,0}^2(2\alpha_2 \Delta T)$ Substituting $h^2 = l_{2,0}^2 - \frac{l_{1,0}^2}{4}$: $h^2 + \frac{l_{1,0}^2}{4}(1 + 2\alpha_1 \Delta T) \approx l_{2,0}^2(1 + 2\alpha_2 \Delta T)$ $l_{2,0}^2 - \frac{l_{1,0}^2}{4} + \frac{l_{1,0}^2}{4} + \frac{l_{1,0}^2}{4}(2\alpha_1 \Delta T) \approx l_{2,0}^2 + l_{2,0}^2(2\alpha_2 \Delta T)$ $\frac{l_{1,0}^2}{4}(2\alpha_1 \Delta T) \approx l_{2,0}^2(2\alpha_2 \Delta T)$ Canceling $2\Delta T$ from both sides: $\frac{l_{1,0}^2}{4}\alpha_1 \approx l_{2,0}^2\alpha_2$ Rearranging to find the ratio of initial lengths: $\frac{l_{1,0}^2}{l_{2,0}^2} \approx \frac{4\alpha_2}{\alpha_1}$ Taking the square root of both sides gives: $\frac{l_{1,0}}{l_{2,0}} = 2\sqrt{\frac{\alpha_2}{\alpha_1}}$ The question implies $l_1$ and $l_2$ are the initial lengths.