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Find the Maximum speed

  1. Dec 6, 2009 #1
    The motion of a piston in an automobile engine is nearly simple harmonic. If the 1 kg piston travels back and forth over a total distance of 10 cm, what is the maximum speed (magnitude of velocity) when the engine is running at 3000 rpm? (Hint: the 10 cm distance has something to do with amplitude; 3000 rpm needs to be converted to rad/sec)


    I did convert the 3000 rpm, and I got 314.159 rad/sec.


    I need some help with this problem a.s.a.p. I don't know where to go with it. Thanks for any help you can give me.
     
  2. jcsd
  3. Dec 6, 2009 #2

    rock.freak667

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    What formula do you think you will need here?
     
  4. Dec 6, 2009 #3
    I don't know, and I don't know any way to find A. I keep thinking that A has something to do with the 3000 rpm, but I am not sure.
     
  5. Dec 6, 2009 #4

    diazona

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    What formula(s) do you know that would allow you to find the maximum velocity?
     
  6. Dec 6, 2009 #5

    rock.freak667

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    A is the amplitude. If the piston moves, from minimum to maximum, a distance of 10cm, what distance does it move when it goes from 0 to maximum?
     
  7. Dec 6, 2009 #6
    5 cm? Cause 10 cm would be the total length, while +A is half, and -A is the other half.
     
  8. Dec 6, 2009 #7

    rock.freak667

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    Right good. Now can you formulate an equation for speed given kinetic energy is 1/2mv2 and potential energy is 1/2kx2. What is the formula for the total energy?
     
  9. Dec 6, 2009 #8
    E = K + U

    (K is kinetic energy, u is potential energy)

    But using the equations you gave, I am missing most of the quantities. How can I use them?
     
  10. Dec 6, 2009 #9

    rock.freak667

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    I told you K and U, if you get an expression for E, you will get what you need with the quantities you know.
     
  11. Dec 6, 2009 #10
    But for K, I need a velocity, which I don't have.

    And for U, I need a spring constant value for k, but we have x (10 cm).
     
  12. Dec 6, 2009 #11

    rock.freak667

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    you don't need to plug in all the values at one time. Just formulate and then see if you can get the terms needed.

    what is E in terms of k and A?
     
  13. Dec 6, 2009 #12
    Well since I don't have all the values I need, I get:

    K = 1/2(1 kg)(v)2

    and

    U = 1/2(k)(.01 m)2

    and then we still have E = K + U
     
  14. Dec 6, 2009 #13

    rock.freak667

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    do not put in the values as yet.

    [tex]E=\frac{1}{2}mv^2+\frac{1}{2}kx^2[/tex]

    Do you know another way to re-write E in terms of k and A?
     
  15. Dec 6, 2009 #14
    I think what is being asked for is an equation that resembles the other oscillators.


    x(t)=A Sin(argument in t) where your argument has you hitting both extremes at a rate of 50 times a second, say by multiplying t by something like 100 * pi (314.15)
    Now remember to get the max velocity, you need to know when acceleration is zero.
    The velocity is -wA sin(wt). sin(wt) is at most 1.0. Hope this helps, I've been at this most of the day and am burnt to a crisp.
     
  16. Dec 6, 2009 #15
    So do I use the equation:

    vmax = wA ? Or what? I am really confused on this one...
     
  17. Dec 6, 2009 #16
    YES! Vmax=wA. Good job. Sorry if I'm a little fried and testy.
     
  18. Dec 6, 2009 #17

    rock.freak667

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    yes, had you derived it, you would get [itex]v_{max}=\omega A[/itex]
     
  19. Dec 6, 2009 #18
    No, don't be. I owe you big time for all of your help today, and I would be the same way if I was helping someone like me.

    Now that I got that right, where do I get my w and A values?
     
  20. Dec 6, 2009 #19

    rock.freak667

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    A is the amplitude that you found, and w is the angular velocity you converted in the first post.
     
  21. Dec 6, 2009 #20
    Okay, and I got:

    vmax = (314.159 rad/s)(.05 m)
    = 15.71 m/s


    Thanks a lot, both of you, for your help. I appreciate it more than you know. :redface:
     
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