0
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ 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
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.