Question

A mass attached to a spring oscillates with a period of 1.71 s and an amplitude of 0.041 m. If the mass starts at x = 0.041 m at time t = 0, where is it at time t = 5.61 s

Answer 1

$x(t) = Asin(\omega t)+0.041$

$\omega = 2\pi/1.71 = 3.674$

At t=0, x=0.041

At t=5.61

Answer 2

calculate how much it travels within that time.

there are 5.61/1.71 = 3.28 periods. we only care about the .28 as each period will bring it back to 0.041m mark. Assume unfirom movement, then the position will be at .28 of a period. At .5 of a period the position is 0m, so we also need to multiply by 2.
.28* 2* 0.041m = 0.02296 m.