1
$\begingroup$

In engineering class we got the following task and I am stuck at solving for a given variable.

The equation that determines the position of my spring pendulum looks like this: $$y\left(t\right)=A\cdot e^{-\frac{t}{2\tau}}\cdot\cos\left(w_d\cdot t\right)$$ $$\tau\ =\frac{m}{k}$$ $$w_d=\sqrt{w_0^2-\left(\frac{1}{2\tau}\right)^2}$$ $$w_0=\sqrt{\frac{D}{m}}$$

$D$ and $m$ are given, therefor I can calculate $w_0$. My task is to calculate k, so that after $\cos\left(w_d\cdot t\right)$ ran 5 times (5 periods), the amplitude should have reduced to $$A \cdot 0.1$$

My attempt was the following:

$$e^{-\frac{t}{2\tau}}=\ 0.1$$ $$e^{-\frac{tk}{2m}}=\ 0.1$$ $$-\frac{tk}{2m}=\ln\left(0.1\right)$$ $$\Rightarrow k=-ln(0.1)\cdot 2m\cdot \frac{1}{t}$$ $$\Rightarrow t=-ln(0.1)\cdot 2m\cdot \frac{1}{k}$$

Then I deal with the $cos$ part: $$5\ =\ w_d\cdot2\pi\cdot t$$ Now I am stuck. When plugging $$\Rightarrow t=-ln(0.1)\cdot 2m\cdot \frac{1}{k}$$ into the formula above I am not able to find k. I even tried to use Wolfram Alpha but it did not give me a solution for k. What can I do?

$\endgroup$
2
  • $\begingroup$ But it's not t = 5 $\endgroup$ – Finn Eggers Nov 5 '17 at 13:07
  • $\begingroup$ It is t = 5T ... $\endgroup$ – Finn Eggers Nov 5 '17 at 13:08
1
$\begingroup$

First of all, you have $t = 5T = 5(2\pi/\omega_d) = 10\pi/\omega_d$ (you seem to have the $2\pi$ on the wrong side). So now from $$ t = \frac{2m\ln 10}{k} = \frac{10\pi}{\omega_d} $$ if you cross-multiply and square both sides, then substitute the expressions you have for $\omega_d$ and $\omega_0$, you get $$ 25\pi^2k^2 = (\ln 10)^2(mD - k^2/4) $$

I really don't see what's stopping you from rearranging this to solve for $k$.

$\endgroup$

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.