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The position of a 0.30-kg object attached to a spring is described by x = (0.28 m) cos(08nt). (Assume t is in seconds.) (a) Find the amplitude of the motion. 0.28 (b) Find the spring constant. 1.89 N/m (c) Find the position of the object at t 0.36 s. 0.173 d) Find the object's speed at t- 0.36 s. Enter a number differs from the correct answer by more than 10%. Double check your calculations. m/ Need Help? Read It

Answer 1

Please note that

$v= \frac{dx}{dt}$

$v= -0.28sin(0.8 \pi t)$ m/s

Now, put t=0.36 s, to get object's speed

$v= -0.28sin(0.8 \pi \times 0.36) m/s$

$v= -0.22 m/s$