The position of a 0.30-kg object attached to a spring is described by x = (0.29

Question

The position of a 0.30-kg object attached to a spring is described by

x = (0.29 m) cos(0.8t). (Assume t is in seconds.)

(a) Find the amplitude of the motion.
.29 m

(b) Find the spring constant.
1.89 N/m

(c) Find the position of the object at t = 0.34 s.
.201   Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. m

(d) Find the object's speed at t = 0.34 s.

.398 Your response differs from the correct answer by more than 10%. Double check your calculations. m/s

Need help with c & d, please.

Explanation / Answer

c) X =(0.29 m) cos(0.8pi*t)

For t= 0.34

= (0.29m) cos(0.8*pi*0.34)

= 0.28996 m

d) speed = dx/dt = (0.29 m)*(0.8p)* (-sin(0.8pt))

= -(0.29 m)*(0.8pi)*(sin(0.8pi*0.34))

= -0.0108

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