STEP III 2007 question 9 solution

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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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