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Oscillator equation

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An oscillator with period 2.7 ms passes through equilibrium at t = 5.4 ms with velocity v = -6.8 m/s. The equation of the oscillator's motion is

x(t) = (  ?   )  cm cos ( ( (  ?  )  /s ) t +   (  ?  )  )

For the second question mark I did 1/2.7 (the period) to get the frequency, and then I multiplied it by 2pi to get the angular velocity.

My answer for that question mark was 2327, but I'm not sure what to do about the other two. I tried using the time and the angular velocity to go back and solve for amplitude, question mark 1, but that didn't work.

 

 

I still don't get it. I mean your solution makes solving for the first question mark clear, but I have no idea how to solve for the third.

asked Oct 26, 2013 in Intro College Physics by anonymous
retagged Oct 27, 2013

1 Answer

+1 vote
Now you have ||\omega||, nice.

At equilibrium, the velocity has its maximum magnitude ||v=\omega A||, which gives you A based on the numbers given in the question.

Then, since at t=5.4 ms, the oscillator is at equilibrium, you would have ||x(t=5.4 ms)=0||. Plug in the t, A, and ||\omega||, you can find ||\phi _0||.
answered Oct 26, 2013 by PhysicsDude (21,750 points)
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