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M2 - Hooke's Law

Heyy. Can anyone just check this? It's a really wordy question, with an odd answer. So I'm not sure if it's right.

A particle of mass, m, is attached to two light elastic springs AC and BC of natural lengths 0.35m and 0.45m respectively. The other ends of the strings are attached to two points A and B which are in a horizontal line 80cm apart. The particle rests in equilibrium 30cm below the midpoint of AB.

Find the modulus of elasticity of each string in terms of m.

My diagram;

diagram mech.png

So I worked out the angles as 53.1. And as it's a standard 3,4,5 triangle we can work out x.

Resolving vertically;

Tcos53.1 = mg

Therefore;

T = \frac{mg}{cos53.1}

So sub all the values into the Hooke's Law formula and re-arrange.
Which gave me the answer of;

\lambda = \frac{68.6m}{1.801}

That's the answer I got for the first string (AC) but it's a bit of an odd number - can anyone confirm whether that's what they got?

E: But if I have two lots of the tension, would I have to re-arrange

2Tcos53.1 = mg

?

And I've just noticed that there's some unit conversion to do - to begin the problem, it's all in cm at the moment. But should it not be in m? In which case, I should convert it to m ..before (?) doing the final Hooke's Law equation?

E2: If the above edit is all true. Then the new final answer I have is;