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!

vest4R  Mar 5, 2018

2+0 Answers




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


Edit: Cosmetic


GingerAle  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.

vest4R  Mar 6, 2018

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