In a spring test, Eddie releases the spring from rest at a height of 20 cm beyond the spring's natural length (i.e. y(0)=20). He finds that the resulting vibration has a period of 1 s and that the amplitude of the vibration halves after 3 s ("half-life"). From his knowledge of damped springs, he also knows that the spring should behave according to the equation \(y(t) = Ae^{-bt}cos(ct)\)



Solve the function paramaters A, b and c


I've solved A = 20 by setting time = 0

And as the period = 1s then C must = 2pi (at least I believe)


just having trouble finding B. I know I have to set the function to half (y(t)/2) however that still leaves me with an unknown answer and an unknown variable.


Any help is much appreciated!

 Mar 5, 2018



\(\text{Period } = \dfrac{2\pi}{c} = 1, \text{so } c = 2\pi\\ \text{Then } x = 20e^{(-bt)}* cos(2\pi*t)\\ \text{When t = 3, then x = 10}\\ \Rightarrow 10 = 20e^{(-3b)}cos(6\pi)\\ 0.5 = e^{(-3b)} * 1\\ Ln(0.5) = (-3b)\\ -3b = Ln(0.5)\\ -3b = -0.69314718\\ b= 0.23104906018\\ \text { }\\ 20e^{(-0.23104906018*3)}cos(6\pi*3)= 10 \text{ (check)}\\ a=20, \; b=0.23104906018, \; c=2\pi \)


Edit: Cosmetic


 Mar 5, 2018
edited by GingerAle  Mar 6, 2018

Thank you!


Ended up getting there in the end, but wanted to keep the question up so I could get my answer confirmed.


Thanks for your help.

 Mar 6, 2018

16 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.