physics question?

Question

An 0.8-kg object is attached to one end of a spring and the system is set into simple harmonic motion. Please take the values of A and T as: A = 0.091 m and T = 1 s. If the amplitude, A, of the motion is .091m, and angular frequency omega is 6.283 rad/s and the spring constant is 31.582, then
a) Find the speed of the object at t=3s
b) Find the magnitude of the objects acceleration at t=3s

Answer 1

Let

X=Acos(wt) then

v(t)=-Awsin(wt)

a(t)= -Aw^2cos(wt)

w= sqrt(k/m)
w=sqrt(31.582/ 0.8) (I just hope that k is in N/m)
w=6.283 rad/sec [confirmed]

a) v(t)=-Awsin(wt)
v(3)=-.091 x 6.283 sin(6.283 x 3)=0.0688 m/s

b) a(3)=-0.091(6.283)^2cos(6.283 x 3)=-1.72 m/s^2

A few notes
1. You also can start with X=Asin(wt) then
v(t)= Aw cos(wt); a(t)= - A w^2 sin(wt) [just a phase difference)
2. Just a reminder that the angles are in radians.