The head of a piston moves with SHM about a centre $O$. When $1$ metre from $O$ its speed is $7.2 \, \text{ms}^{-1}$, and when $2.4$ metres from $O$ its speed is $3 \,\text{ms}^{-1}$. Find the amplitude of the motion.
To solve this, I set up a pair of simultaneous equations using the result $v^2 = \omega^2(a^2-x^2)$:
$$ w^2(a^2-1) = 7.2^2$$ $$w^2(a^2 - 2.4^2) = 3^2$$
Which gives $a = 2.6$ when solved, the correct answer.
However the solutions simply give the working: $a^2 = 2.4^2 + 1 \implies a =2.6$. This seems like some much easier 'shortcut' method to get to the answer. I can see that $2.4$ and $1$ are the distances involved, but I can't understand why this works and why the relevant velocities don't seem to be needed at all.
Is there a standard result or much easier method to solve this?
It just happens to be a coincidence in this example that $$\frac{7.2^2}{3^2}=2.4^2$$
One would usually divide the two equations to eliminate $w^2$ and rearrange to get $a^2$, so on this occasion this leads to $$a^2-1=2.4^2(a^2-2.4^2)$$ Hence $$a^2(2.4^2-1)=2.4^4-1=(2.4^2+1)(2.4^2-1)$$
and hence the result.
There is no special short-cut in general