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

Paxton Weiss

2 May, 18:33

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?

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Answers (1)

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Rachel French · 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