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Suppose we are given the following equations $$\dot x=p_x-ky\\\dot y=p_y+kx\\\dot p_x=-k(p_y+kx)\\ \dot p_y=-k(-p_x+ky)$$

I can see that it describes circular motion. but why it it true that for a circle with origin $(a,b)$ the following equation stands:$$2a+{p_y\over k}=x$$?

I suspect it's something rather obvious, but which has managed to elude me...

Please?

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1 Answer 1

Just take the time derivative of both sides of the equation: $$\frac{\mathrm{d}}{\mathrm{d}t}\left(2a + \frac{p_y}{k}\right) = \frac{\mathrm{d}x}{\mathrm{d}t}\\ \iff \dot p_y = k \dot x\\ \iff -k(-p_x + ky) = k(p_x - ky)$$ So your equation is valid if we set the initial values correctly. this can be done by chosing $x(0) = 2a$ and $p_y=0$, say, which corresponds to a circle around your specified origin with radius $a$.

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