Part A The position of an object that is oscillating on an ideal spring is given

Question

Part A The position of an object that is oscillating on an ideal spring is given by the equation x=(12.3cm)cos[(1.26s%u22121)t].
(a) At time t=0.815 s, how fast is the object moving?
Part B (b) At time t=0.815 s, what is the magnitude of the acceleration of the object?
Part A The position of an object that is oscillating on an ideal spring is given by the equation x=(12.3cm)cos[(1.26s%u22121)t].
(a) At time t=0.815 s, how fast is the object moving?
Part B (b) At time t=0.815 s, what is the magnitude of the acceleration of the object?

Explanation / Answer

A) d/dt*12.3*cos(1.26*t)=-15.498*sin(1.26*0.815)=10.5251 cm/s

B) d/dt*-10.5251*sin(1.26*0.815)=-13.262*cos(1.26*t)==>using 0.815 for t==>9 cm/s^2

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