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Graph of a particle's position

  1. Apr 19, 2015 #1
    1. The problem statement, all variables and given/known data

    2WGbXQi.png
    2. Relevant equations


    3. The attempt at a solution
    the particle is oscillating

    so we know at 1.5 seconds, it is at the equilibrium point , where speed is highest. (the speeds are 0 at the highest and lowest points of the graph.)

    we see that it goes from -6m to 6m, which is 12 meters in 1 second.

    so is the answer 12 meters per second? since that is the highest it can go?
     
  2. jcsd
  3. Apr 19, 2015 #2
    Are you familiar with the equations that govern simple harmonic motion?
     
  4. Apr 19, 2015 #3
    no sir, i will google it right now.
     
  5. Apr 19, 2015 #4

    SammyS

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    That's just the average speed from t = 1 second to t = 2 seconds .
     
  6. Apr 19, 2015 #5
  7. Apr 19, 2015 #6
  8. Apr 20, 2015 #7
  9. Apr 20, 2015 #8
    Yes.
     
  10. Apr 20, 2015 #9
    how do we find angular velocity? Is the particle moving around in a circle? I thought it was more like a spring, going back and forth.
     
  11. Apr 20, 2015 #10
    It is.

    No, but its horizontal displacement is the same as if it were moving in uniform circular motion. I would look at a textbook or online resource for a more complete explanation.

    The distance between two crests (high points) on the graph you provided is the amount of time it takes the mass to make a complete "revolution" (2π radians). Angular velocity is measured in radians per second... do you see where I am going with this?
     
  12. Apr 20, 2015 #11
    yes, so it takes 2 seconds to do 1 rev.

    so 2π radians in 2 secs , 1 radian per sec.

    ω = 1 radian per sec.

    A = 3 meters

    and cos ω t = cos ((1r/s) (2 second)) = .99939



    v = ω A cos ωt = 1 x 3 x .999 = 3 m/s

    correct?
     
  13. Apr 20, 2015 #12
    On a bit of a physics binge, eh? Don't worry, PF has you covered.
     
  14. Apr 20, 2015 #13

    BvU

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    Draw a straight line with slope 3 m/s through the point (1.5s, 0 m) to check...
     
  15. Apr 20, 2015 #14
    yes, silly me always doing homework when everyone is asleep.
     
  16. Apr 20, 2015 #15

    BvU

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    Well done !
     
  17. Apr 20, 2015 #16
    whoops, yes.

    and cos ω t = cos ((π r/s) (2 second)) = .9984

    answer still came out to be 3 m/s.
     
  18. Apr 20, 2015 #17
    like this?
    VLjU4HS.png

    I have no idea how to confirm if that is suppose to correct or not.
     
  19. Apr 20, 2015 #18
    No, I think BvU meant more like this SnapshotSHM.jpg
     
  20. Apr 20, 2015 #19
    i think he was trying to tell me my answer was wrong.

    by the looks of it, my speed should be -3m/s instead of 3 m/s.

    correct?
     
  21. Apr 20, 2015 #20
    You just drew a tangent line at t = 2.5 instead of t = 1.5. At t = 2.5 the mass has a negative velocity. I believe BvU was just pointing out that you could draw a line tangent to the graph in order to check your answer.
     
    Last edited: Apr 20, 2015
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