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
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.
1 answer
1 answer