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?

  • $\begingroup$ 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

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 $k$.


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