- #1

physics noob

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**yet another "simple" harmonic problem**

1.01kg mass is attached to a horizontal massless spring with k=10N/m The system is set into motion by pulling the mass .2m away from equilibrium, releasing it from rest.

find a time when the mass is at x= -.05m

so i used x=Acos(omega *t)

so to get omega i used MEi = MEf

getting .5(10)(.2)^2 = .5(10)(.05)^2 + .5(1.01)v^2

I am gettin a velocity of .61 which is an angular velocity of

3.05 rad/s i plug it into the aforemention equation

so .05=.20cos(3.05(t)) t = .43 sec which of course is wrong correct answer is .58

where did i go wrong here,,,,, also i have tried several combinations starting Xo at different spots. but i can't seem to get the right answer.......any help is appriciated...THANKS