Question

The length of an elastic string is $L_1$ when the tension is $4N$, and $L_2$ when the tension is $5N$. What is the length of the string when the tension is $7N$?

Answer 1

Let $L_1=L+x_1,L_2=L+x_2$

$F=YA\frac{x}{L}$

$L_1=\frac{YA{x}_1}{L}$

$4=\frac{YA{x}_1}{L}$

$5=\frac{YA{x}_2}{L}$

$\frac{4}{5}=\frac{x_1}{x_2}$

$\frac{4}{5}=\frac{L_1-L}{L_2-L}$

$4L_2-4L=5L_1-5L$

$L=5L_1-4L_2$

Now, $F=7N$

$7=\frac{YA}{L}.x_3$

$\frac{YA}{L}=\frac{5}{x_2}$

$=\frac{5}{L_2-L}$

$=\frac{5}{L_2-(3L_1-4L_2)}$

$\frac{5}{5(L_2-L_1)}$

$=\frac{1}{L_2-L_1}$

$7=\frac{1}{L_2-L_1}.x_3$

$x_3=L_3-L$

$x_3=7(L_2-L_1)$

$7L_2-7L_1=L_3-(5L_1-4L_2)$

$L_3=3L_2-2L_1$