# How can I express velocity as a function of position in a damped oscillation? [closed]

In a damped oscillation that obeys $$x(t)=Ae^{-bt/2m}\cos(ωt)$$ which shows the position of the oscillating object as a function of time, how can I express the velocity of the oscillating object as a function of position? I tried differentiating $$x(t)$$ with respect to $$t$$ and replacing terms with $$x$$ but kept failing to completely eliminate $$t$$.

• Related meta discussion: physics.meta.stackexchange.com/q/12972/2451 Commented Jun 26, 2020 at 19:50
• I've removed a number of comments that were attempting to answer the question and/or responses to them. Commenters, please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. Commented Jun 26, 2020 at 20:23

A quick plot shows that this is generally messy/impossible to do. I took $$b/2m=1,\omega=4$$. To write $$v$$ as a function of $$x$$ you would either have to split up $$v$$ in many pieces or do some trick. But even with a trick you won't get $$v(x)$$ for the entire domain.
Here's what I tried \begin{align} x(t)&=Ae^{-\alpha t}\cos{\omega t}\\ &=\frac A 2e^{-\alpha t}\left(e^{i\omega t}+e^{-i\omega t}\right)\\ &=\frac A 2\left(e^{(i\omega-\alpha )t}+e^{(-i\omega-\alpha )t}\right)\\ &\equiv\frac A 2\left(e^{zt}+e^{\bar zt}\right)\\ &=A\,\text {Re}\!\left(\,e^{zt}\right) \end{align} At this point it goes wrong. By taking the real part you throw away all the information in the imaginary part. Many different values of $$t$$ could lead to the same value of $$x$$ (which is indeed what happens). To still be able to invert the relation you can define $$X(t)=Ae^{zt},\quad V(t)=Aze^{zt}\\ \bar X(t)=Ae^{\bar zt},\quad\overline V(t)=A\bar ze^{\bar zt}$$ and you can show that $$x(t)=\tfrac 1 2 (X(t)+\bar X(t)),\ v(t)=\tfrac 1 2 (V(t)+\overline V(t))$$. It is then easy to show that $$V(X)=zX$$ and $$\overline V(\bar X)=\bar z\bar X$$. So finally we get \begin{align} v(X,\bar X)&=\tfrac 1 2 (zX+\bar z\bar X)\\ &=-\frac \alpha 2 (X+\bar X)+\frac {i\omega}2(X-\bar X)\\ &=-\alpha\,\text{Re}X-\omega\,\text{Im}X\\ &=-\frac b{2m}x-\omega\,\text{Im}X \end{align} So this is still pretty useless because there is no way to define $$X$$ in terms of $$x$$ but I hope this gave you at least some insight on why this fails.