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

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.

Philipp

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

$$t=-\frac{\pi+2\cdot\left(-\frac{\pi}{2}\right)-2\pi\cdot\text{n}}{2\cdot\sqrt{4-\frac{1}{9\cdot10^6}\cdot\text{g}^2}}=\frac{\pi\cdot\text{n}}{\sqrt{4-\frac{\text{g}^2}{9\cdot10^6}}}\tag4$$

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