Mass attached to a horizontal spring

  • Thread starter pvpkillerx
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  • #1
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A particle with a mass of 0.210kg is attached to a horizontal spring with a force constant of 0.84 N/m. At the moment t=0, the particle has its maximum speed of 5m/s and is moving to the left. (Assume that the positive direction is to the right.)

Determine the particle's equation of motion, specifying its position as a function of time (use the following as necessary: t)

F = -kx
F = ma




What I did was:
-0.84x = 0.210a
-0.84x = 0.210[(v - 5)/t)]
-0.84x = 0.210[(x/t - 5)/t]


I still can't isolate for x, and i dont know if what i am doing is right. Please help.
 

Answers and Replies

  • #2
22
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nvm, i figured it out, its 5/2Sin(2t+pi)
 

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