An air-track cart is attached to a spring and completes one oscillation every 5.67 s in simple harmonic motion. at time t = 0.00 s the cart is released at the position x = +0.250 m. what is the position of the cart when t = 29.6

The cart is moving by simple harmonic motion, and its position at time t is described by

where
A is the amplitude of the oscillation
is the angular frequency

The amplitude of the oscillation corresponds to the maximum displacement of the spring, which corresponds to the initial position where the spring was released: 
A=0.250 m

The period of the motion is T=5.67 s, and the angular frequency is related to the period by


Therefore now we can calculate the position of the system at the time t=29.6 s:

Then the answer would be d then