Assume that simple harmonic motion of a spring is described by the equation S=4 cos πt/ 2 where s is in centimetres and t is in seconds. When during the time 0<=t<=8 is the spring passing through the origin?

Question

I attempted this question but the graph doesn't pass through the origin.

Answer 1

you are correct.

However, S=0 when t=1,3,5,7
That may have been what they intended to ask.

Answer 2

you are right,
S = 4 cos (πt/2) does not pass through the origin.
Perhaps you meant, where does it cross the t - axis?

in that case 4cos(πt/2) = 0
we know cos(π/2) = 0
so πt/2 = π/2
t = 1 or t = -1

we also know that the period of your function is
2π/(π/2) = 4
so adding or subtracting 4 from any answer will yield another solution
1+4 = 5,
-1+4 = -3
1 - 4 = -3
So it will cross again at 5 , 9 , -3 , 3 etc
Also because of the symmetry of the function about the S axis, the times it crosses the t-axis for your given intervals are:
-5, -3, -1, 1, 3, and 5

confirmation:
http://www.wolframalpha.com/input/?i=y+%3D+4cos%28%CF%80t%2F2%29+from+-8+to+8

Answer 3

oops, forgot the 7, thanks Steve