# 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 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. 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.