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?

Question

Physics

Davon Nixon

9 October, 13:45

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?

+5

Answers (1)

Know the Answer?

Cale Marks · 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