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STEP III 2007 question 9 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP III 2007 question 9 solution

The tricky bit of this question is hidden in one little word: \theta is defined as the acute angle determined by 2\theta =\angle AOB. So we're going to have to be careful about what happens when \theta = \frac{\pi}{2}.

Now P.E. at start = mk^2a^2(\beta - \alpha)^2 P.E. at \theta = \frac{\pi}{2} = mk^2a^2(\pi/2 - \alpha)^2.

So we only reach \theta = \frac{\pi}{2} if \alpha - \beta > \pi/2 - \alpha (i.e. \beta < 2\alpha - \pi/2).

So first consider the easy case where \beta > 2\alpha - \pi/2).

We have K.E. = ma^2\dot{\theta}^2, so conservation of energy gives \dot{\theta}^2+k^2(\theta-\alpha)^2=const. Set \phi = \theta - \alpha to get \dot{\phi}^2 + k^2\phi^2 = \text{const}. Then \dot{\phi}\ddot{\phi} =-k^2\phi\dot{\phi} so \ddot{\phi}=-k^2\phi.

Thus we have simple harmonic motion with period \frac{2\pi}{k}.

Now we come to the interesting case. Suppose \beta < 2\alpha - \pi/2. We consider the motion up until \theta = \pi/2. As before, we have \ddot{\phi}=-k^2\phi where \phi = \theta - \alpha. So \phi = A \cos(kt+\epsilon). When t=0, \dot{\phi} = 0 \implies \epsilon = 0. When t=0, \phi =\beta - \alpha \implies A = \beta - \alpha. So we have \phi = (\beta-\alpha)\cos kt. So, how long does it take to get to \theta = \pi/2? Well at this point we have \phi = \pi/2 - \alpha \implies \cos kt = \frac{\pi/2-\alpha}{\beta-\alpha}, and so t = \cos^{-1}\left(\frac{\pi/2-\alpha}{\beta-\alpha}\right).

Once we get to \theta=\pi/2, 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 4 \cos^{-1}\left(\frac{\pi/2-\alpha}{\beta-\alpha}\right).

Oscillations will not occur when \beta = 2\alpha - \pi/2: instead the two particles will come to rest (in unstable equilibrium) at the point where \theta = \pi/2.

Solution by DFranklin.

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