Mechanics - Hooke's law and energy conservation

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HopelessStudent
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Homework Statement


One end of a light elastic string of stiffness mg/l and natural length l is attached to a point O. A small bead of mass m is fixed to the free end of the string. The bead is held at O and then released so that it will fall vertically. In terms of find the greatest depth to which it will fall below O.

Homework Equations


Ek=(mv^2)/2 Es=(ex^2)/(2l) Eg=mgh suvat

The Attempt at a Solution


Now i started off by splitting the motion up into 4 parts. Part 1 before it dropped Et = Eg so total energy is mgh which is mg(l+x). Part 2 is as its fell a distance l Et = Eg + Ek. Using suvat i got the speed so i got the equation Et = xmg + mlg. Part 3 will be taken at any time while the mass is moving and extending the string. Part 4 is at the maximum extension and not moving so i got Et=Es which is Et = (mgx^2)/2.

To me that all seems correct but when i try combining the equations to get x i can't seem to get anything that works. Can someone show me where I've gone wrong please? Thanks

I've just noticed I've been using the wrong equations and using modulus of elasticity not k so the final equation should be Et= (mgx^2)2l
 
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