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Homework Help: Maximum velocity of a piston

  1. Apr 24, 2010 #1
    1. The problem statement, all variables and given/known data
    A piston in an automobile engine is in simple harmonic motion. Its amplitude of oscillation from the equilibrium (centered) position is ±4.73 cm and its mass is 1.733 kg. Find the maximum velocity of the piston when the auto engine is running at the rate of 4380 rev/min.
    Answer in units of m/s.


    2. Relevant equations

    Vmax=Aw

    3. The attempt at a solution
    I attempted to take the Amplitude (.0437 m) and multiply it by the angular frequency. I first converted it into revolutions per second and then divided 2pi by the answer I got.
     
  2. jcsd
  3. Apr 24, 2010 #2
    Hello Bearbull,w=2pif.
     
  4. Apr 24, 2010 #3
    And f=1/T and T=4380 rev/min= 73 rev/s?
     
  5. Apr 24, 2010 #4
    Got it wrong again,

    Vmax=(.0437m)(2pi(1/73rev/s))
     
  6. Apr 24, 2010 #5
    f=4380r/min=4380/60 revs/second=73r/s
     
    Last edited: Apr 24, 2010
  7. Apr 24, 2010 #6
    This part is wrong. Remember 1 rev = 2pi.
    So you multiply by 2pi, and then 1 min = 60 seconds. So you divide by 60 seconds. I did not get 73 rev/s. I got another number.
     
  8. Apr 24, 2010 #7
    w=2pitimes73
     
  9. Apr 24, 2010 #8
    I did 2pi*73 rev/s and then multiplied it by my amplitude (.0437 m). Got it wrong.
     
  10. Apr 24, 2010 #9
    What answer did you get and what should the answer be?
     
  11. Apr 24, 2010 #10
    I got 20.043989 and I have no idea what the correct answer is supposed to be. I only have one more chance to get it correct
     
  12. Apr 24, 2010 #11
    so you carried out the following calculation:
    2 times 3.142(pi)times73times 0.0473.I picked up my calculator and did it and got an answer close to yours but not the same.Try again and report back here before you submit your answer.
     
  13. Apr 24, 2010 #12
    Haha. Had the numbers mixed up. Thanks
     
  14. Apr 24, 2010 #13
    I'm always doing that:smile:
     
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