Asked by Tricia
A spring is attached to the ceiling and pulled 17 cm down from equilibrium and released. After 3 seconds the amplitude has decreased to 16 cm. The spring oscillates 13 times each second. Assume that the amplitude is decreasing exponentially. Find an equation for the distance, D the end of the spring is below equilibrium in terms of seconds, t.
Answers
Answered by
Steve
since the spring starts at (0,-17), we know that
D(t) = -17 e^-at cos(kt)
since the period is 1/13, k = 26π and we have
D(t) = -17 e^-at cos(26πt)
Now, we know that the amplitude has decreased to (16/17)^(t/3)
That means that a = -1/3 ln(16/17) = .02, and
D(t) = -17 e^-0.02t cos(26πt)
You can see a plot at
http://www.wolframalpha.com/input/?i=plot+y+%3D+-17+e^%28-0.02t%29+cos%2826%CF%80t%29%2C+y%3D16+for+t%3D0+to+3
D(t) = -17 e^-at cos(kt)
since the period is 1/13, k = 26π and we have
D(t) = -17 e^-at cos(26πt)
Now, we know that the amplitude has decreased to (16/17)^(t/3)
That means that a = -1/3 ln(16/17) = .02, and
D(t) = -17 e^-0.02t cos(26πt)
You can see a plot at
http://www.wolframalpha.com/input/?i=plot+y+%3D+-17+e^%28-0.02t%29+cos%2826%CF%80t%29%2C+y%3D16+for+t%3D0+to+3
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.