M4 Tension DE Question

With these questions I'm struggling to figure out exactly which value to use as $x$ in the $T = \frac{\lambda x}{l}$ formula

Because the string is already stretched, how does this factor into the value for $x$? Does the extension of moving the particle P $\frac{1}{2}L$ from where it is at rest (it is attached to the centre) compound with the initial extension to give $x_1 = x + L$?

Many thanks

Question is from Edexcel M4 January 2004 Q4

Tagged some people who may be able to help: @ghostwalker @tiny hobbit @Gome44

link doesnt work

This one should work:

Draw a diagram. The length of the string from P to B is 1.5L-x. The natural length of this string is L, thus extension is 0.5L-x

So yes you do factor in the initial extension

Thanks for explaining, I've got that tension right now.

Just checked the solution and I can't see why they have included a second tension $T_2$ on their answer though?



Surely if you pull this particle on a taut string to the right there will only be a tension to the left opposing that extension, why is there a tension $T_2$ in this diagram?

Edit: Is this about the initial tension from the stretch from A to B. We assume that the string is in two parts, one with extension $\frac{1}{2} L$ and another with the same extension? Hence when we factor in the tension to one string we have to do the same for the other?

Once the particle is moving, both halves of the string will be longer than their natural length, so both will have a tension.