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Been doing some advanced mechanics questions and stumbled upon one i can't wrap my head around. It goes as follows:

One end of a light elastic string of stiffness /l and natural length is attached to a point O. A small bead of mass 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.

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

e 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

HoplessStudent
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2 Answers2

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The force $F$ in the string, when the bead is $x$ metres below it's length $l$, is given by $$F=\dfrac{mg}{l}x$$

So now it is natural to consider the following:

  • What two forces act on the bead after it goes beyond its natural length?

  • What is the relation between these forces at the lowest point?

For the second bullet you should be able to formulate an equation that allows you to find the extension $x$

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Assume the initial energy is (all potential energy): $$E_i = mgl$$

The final Energy is (potential and elastic): $$E_f = mg(l-\Delta l)+\frac12k \Delta l^2$$

Thus, equating the energy:

$$E_i=E_f \implies mgl=mgl-mg\Delta l+\frac12 \frac{mg}{l} \Delta l^2$$

$$\Delta l^2-2l\Delta l=0 \implies \Delta l(\Delta l-2l)=0\implies \Delta l=0 \quad,\Delta l=2l$$

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