I came across this question in a textbook and after a full day of going over it and consulting friends none of us can figure out why a particular approach to this question doesn't yield the correct answer.
The question
A light elastic string of natural length $0.2m$ has its ends attached to two fixed points $A$ and $B$ which are on the same horizontal level with $AB = 0.2m$. A particle of mass $5$kg is attached to the string at the point $P$ where $AP = 0.15m$. The system is released and P hangs in equilibrium below AB with $\angle{APB} = 90^{\circ}$. If $\angle{BAP} = {\theta}$, show that the ratio of the extension of $AP$ and $BP$ is $$\frac{4cos{\theta}-3}{4sin{\theta}-1}$$
The correct method
Let the extension of $AP$ be $x_1$ and the extension of $BP$ be $x_2$
$x_1 = 0.2cos{\theta} – 0.15$
$x_2 = 0.2sin{\theta} – 0.05$
$\frac{x_1}{x_2} = \frac{0.2cos{\theta} – 0.15}{0.2sin{\theta} – 0.05}$
$\therefore \frac{4cos{\theta}-3}{4sin{\theta}-1}$
Our method (seemingly incorrect)
With the same symbols for $AP$ and $BP$ and the tensions in $AP$ and $BP$ being $T_1$ and $T_2$ respectively.
Since the system is in equilibrium resolving horizontally gives:
$T_1cos{\theta} = T_2sin{\theta}$
$\therefore \frac{T_1}{T_2} = tan{\theta}$
Since for an elastic string $T = \frac{{\lambda}x}{l}$
$T_1 = \frac{{\lambda}x_1}{0.15}$
$T_2 = \frac{{\lambda}x_2}{0.05}$
$\therefore \frac{x_1}{x_2} = 3\frac{T_1}{T_2}$
$\therefore \frac{x_1}{x_2} = 3tan{\theta}$
The discrepancy here is not that I do not understand the first solution, but why the second method does not yield a correct result, the steps followed seem logical and correct to me.
Best Answer
Both are correct, and the equality between the two identities will allow you to find $\theta$ and the other geometric parameters.
Note in fact that P will not move along the vertical from the original position, but will move somewhat towards B, since PB is more rigid than PA.