The figure shows a possible arrangement of an isochronic pendulum, that is, a pendulum that does not change in frequency with changes in ambient temperature. The three light vertical bars, of length $ L $, have coefficients of linear thermal expansion $ \alpha_1 $ and the two dark vertical bars, of length $ L_2 $ have coefficient $ \alpha_2 $. Determine $ L_1 $ and $ L_2 $ so that the length $ L $ remains constant. Suppose the joints and supports are fixed and rigid.
Can someone help me? I think it is possible to solve ... but how? It does not make it very clear what is fixed and what is not
The bars $L_1$ expand up while the bars $L_2$ expand down. The number of bars is irrelevant, just the configuration of two bars expanding up and one expanding down is relevant.
At temperature t:
$L_1(1+\alpha_1 t)-L_2(1+\alpha_2 t)+L_1(1+\alpha_1 t)=L$
$(2L_1\alpha_1 -L_2\alpha_2)t=L-2L_1+L_2$
Since t is a variable representing temperature then the left side of equal appears to be variable, while the right side of equal is a constant. This leads to the necessity that the left side of equal to be a constant as well, leading to:
$\left\{ {\begin{array}{rr} 2L_1\alpha_1-L_2\alpha_2=0\\ L-2L_1+L_2=0\\ \end{array} }\right.$
This is a linear system with solution $L_1=\frac{L}{2}\cdot \frac{\alpha_2}{\alpha_2-\alpha_1}, L_2=L\cdot \frac{\alpha_1}{\alpha_2-\alpha_1}$