Question

The position of an object that is oscillating on an ideal spring is given by the equation x=(12.3cm)cos[(1.26s−1)t]. (a) at time t=0.815 s, how fast is the object moving?

Answer 1

x=((12.3/100)m)cos[(1.26s^−1)t]
 v= dx/dt = -((12.3/100)*1.26)sin[(1.26s^−1)t]
 v=-((12.3/100)*1.26)sin[(1.26s^−1)t]=-((12.3/100)*1.26)sin[(1.26s^−1)*(0.815)]
 v= -0.13261622 m/s
 the object moving at  0.13 m/s at time t=0.815 s