I am working on an exercise for my physics class. In order to retrieve the solution I have to solve the following equation for $t$ depending on $g$ ($\Rightarrow t(g)$). This is where I am stuck.

$$e^{-\frac{g\cdot t}{2m}}\cdot\cos\left(w\cdot t-\frac{\pi}{2}\right)=0,1$$

or because $w$ in itself depends on $g$ and $m=1500$ is a known value

$$e^{-\frac{g\cdot t}{3000}}\cdot\cos\left(\sqrt{4-\frac{1}{9\cdot10^6}\cdot g^2}\cdot t-\frac{\pi}{2}\right)=0,1$$

I have tried other methods and avoid this equation, but is seems that there is no way around it. So how can I solve an equation like this with an exponential term as well as a trigonometrical one?

I have read about differential functions. Is this an example to them?

Thank you for any helpful contribution.


  • $\begingroup$ i think this is impossible, since $t$ stands in the argumant of the exponential function and in the $\cos$ function $\endgroup$ Nov 5 '17 at 13:11
  • $\begingroup$ you will need a numerical method $\endgroup$ Nov 5 '17 at 13:12
  • $\begingroup$ is g given by $9.81ms^{-2}$ ? $\endgroup$ Nov 5 '17 at 13:14
  • $\begingroup$ No. $g$ shall be the constant of friction which is unknown. I will try using Desmos $\endgroup$
    – Philipp
    Nov 5 '17 at 13:17

Well, in general we have:

$$\exp\left(\text{a}\cdot t\right)\cdot\cos\left(\omega\cdot t+\varphi\right)=0\tag1$$

Now, we get two possible solutions:

  1. Solution $1$: $$\exp\left(\text{a}\cdot t\right)=0\tag2$$

But equation $\left(2\right)$ does not have a solution.

  1. Solution $2$: $$\cos\left(\omega\cdot t+\varphi\right)=0\space\Longleftrightarrow\space t=-\frac{\pi+2\cdot\varphi-2\pi\cdot\text{n}}{2\cdot\omega}\tag3$$

When $\omega\ne0$ and $\text{n}\in\mathbb{Z}$

So, in your problem $\omega=\sqrt{4-\frac{1}{9\cdot10^6}\cdot\text{g}^2}$ and $\varphi=-\frac{\pi}{2}$:


When $\text{g}^2\ne36\cdot10^6$


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