STEP III 2007 question 9 solution

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The tricky bit of this question is hidden in one little word: is defined as the acute angle determined by . So we're going to have to be careful about what happens when .

Now P.E. at start = P.E. at = .

So we only reach if (i.e. ).

So first consider the easy case where ).

We have K.E. = , so conservation of energy gives . Set to get . Then so .

Thus we have simple harmonic motion with period .

Now we come to the interesting case. Suppose . We consider the motion up until . As before, we have where . So . When t=0, . When t=0, . So we have . So, how long does it take to get to ? Well at this point we have , and so .

Once we get to , we have another 3 pieces of motion, each of which will be as we've just analysed. This would be easier with diagrams, but imagine A,B as positions on a clock. Suppose we start with A = 7'o'clock, B = 5'o'clock. We've just analysed the motion up 'til A=9'o'clock, B=3'o'clock. The motion from there up 'til A=11'o'clock, B=1'o'clock looks exactly the same, only with a time reversal. And then we go back to A=9'o'clock, B=3'o'clock and then A = 7'o'clock, B = 5'o'clock.

So the final period is going to be .

Oscillations will not occur when : instead the two particles will come to rest (in unstable equilibrium) at the point where .

Solution by DFranklin.