Finding Tension in an Elastic String? I know that this is a homework type question and I'm not asking a particular physics question, but I'm really desperate for help.
Here's the question:

I tried to divide the string to 2 parts with $O$ as the mid-point of $AB$. 
Let $AO$=$T_{1}$ and $OB$=$T_{2}$, then $T_{2}-T_{1}=m\ddot x $. I don't know what to do next.
Here's the marking scheme: 

 A: Six years later:
We are given that the natural length of the string is $2l$.
When the mass is suspended at the midpoint, $O$, of the string, the extension perpendicular to $AB$ is $x$. Let $M$ be the position of the mass at perpendicular extension $x$.
Then the extended length of the string from $A$ to $M$ (the hypotenuse of a triangle with sides $a$ and $x$) is:
$$AM=\sqrt(a^{2}+x^{2})$$
Since the string is extended at its mid-point, this is the same as the extension from M to B. Hence, the total length, $AMB$, of the extended string is
$$L=2\sqrt(a^{2}+x^{2})$$
The extension of the string from its natural length is
$$\delta {l}=2\sqrt(a^{2}+x^{2})-2l=2\left[{\sqrt(a^{2}+x^{2})-l} \right]$$
Using the formula for Tension in a string and substituting for $\delta {l}$:
$$\begin{align} T &=\lambda \frac{\delta {l}}{l} \\
&=\lambda \frac{2\left[{\sqrt(a^{2}+x^{2})-l} \right]}{2l} \\
&=\lambda \frac{\left[{\sqrt(a^{2}+x^{2})-l} \right]}{l}\end{align} $$
Finally, if  $x^{2} \ll 1$, this gives:
$$\begin{align} T &=\lambda \frac{\left[{\sqrt(a^{2})-l} \right]}{l} \\
 &=\lambda \frac{(a-l)}{l}
\end{align}$$
