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Consider A particle performing Uniform Circular Motion. We know that its projection on diameter performs SHM. Then, if that projection starts SHM on the y axis from mean position, then $y=A\text{sin}(ωt)$ if that projection starts SHM on the y axis from extreme position, then $y=A\text{cos}(ωt)$ if that projection starts SHM on the x axis from mean position, then $x=A\text{cos}(ωt)$ Then What is the equation if that projection starts SHM on the x axis from extreme position?

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  • $\begingroup$ Using polar co-ordinates (see the answer below) is the most general approach. But also note that any SHM that starts from an extreme position has equation $\pm A \cos (\omega t)$ and any SHM that starts from the mean position has equation $\pm A \sin (\omega t)$. $\endgroup$
    – gandalf61
    Mar 19 at 7:07
  • $\begingroup$ Voting to reopen. Clearly a conceptual question not a "do my homework for me" question. $\endgroup$
    – gandalf61
    Mar 19 at 15:32

1 Answer 1

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Start in polar coordinates:

$$ r = A $$ $$ \theta(t) = \omega t + \theta_0 $$

where $\theta_0$ is the phase at $t=0$. Vary $\theta_0$ to create the initial position, and convert to Cartesian coordinates to find $(x(t), y(t))$.

Apply trig identities as needed.

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