A weight is attached to an elastic spring that is suspended from a ceiling. If the weight is pulled 1 inch below its rest position and released its displacement in inches after t seconds is given by x(t)=2cos(5ðt+ð/3). Find the first two times for which the displacement s 1.5 inches.

Answer 1

what's the problem? You have the formula:

x(t) = 2cos(5pi t + pi/3)
just solve

2cos(5pi t + pi/3) = 3/2
cos(5pi t + pi/3) = 3/4

cos .722 = .75

so,
5pi t + pi/3 = .722 or 2pi-.722=5.560
t = (.722-pi/3)/5pi = -.021 or 6.262
t = (5.560-pi/3)/5pi = .287

see the graph at

http://www.wolframalpha.com/input/?i=solve+2cos%285pi+t+%2B+pi%2F3%29+%3D+3%2F2

You need to find the next occurrence after 0.287 by noting that the period is 0.4