Matt1234: You can see, at t = t1 = 0, x1 = 0.10 m. In other words, you need to fully deflect the mass, then release it, in order for it to vibrate. Therefore, at t1 = 0, we have,
x1 = 0.10 m = (0.10 m)*cos(4*t1 + phi)
cos(4*t1 + phi) = 1
4*t1 + phi = acos(1)
phi = acos(1) - 4*t1
phi = 0 - 4*0
phi = 0 rad
Therefore, we have,
x(t) = A*cos(omega*t + phi)
x(t) = A*cos(omega*t + 0)
x(t) = A*cos(omega*t)
Likewise, v(t) = -A*omega*sin(omega*t).
(c) Let x(t2) = 0.06 m. Therefore,
x(t2) = A*cos(omega*t2)
0.06 = 0.10*cos(4*t2)
Hint 1: Can you solve for t2? After that, can you compute v(t2)?
(d) Hint 2: Can you compute a(t2)?
(e) Hint 3: Let x(t3) = 0 m. Can you solve for t3? Hint 4: Let x(t4) = -0.08 m. Can you solve for t4?
Try again. Also, please do not post wide images directly to the forum page. Just post a text link to wide images.