An object is undergoing SHM with period 0.340 s and amplitude 5.85 cm . At t=0...
An object is undergoing SHM with period 0.340 s and amplitude 5.85 cm . At t=0 the object is instantaneously at rest at x= 5.85 cm . Calculate the time it takes the object to go from x= 5.85 cm to x= -1.50 cm .
Homework Answers
T = 0.340s
A = 5.85cm
At t = 0, x = 5.85cm and v = 0
so x = Acos(2pi*t/T)
now -1.5 = 5.85cos(2*3.14*t/0.34)
=> -0.25641 = cos(18.47t)
=> 1.8291 = 18.47t
t = 0.099 s
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