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Homework Help: SHM Question

  1. May 5, 2005 #1
    Q: The prong of a tuning fork moves back and forth when it is set into vibration. The distance the prong moves between its extreme positions is 2.29 mm. If the frequency of the tuning fork is 440.2 Hz, what are the maximum velocity and the maximum acceleration of the prong? Assume SHM.


    I was just hoping somebody could tell me if I did this problem correctly :smile:

    For Vmax I have : (2л) x (440.2Hz) x (0.0029cm) = 8.02 m/s

    For amax I have : [(2л x 440.2Hz)^2] x (0.0029cm) = 22185 m/s2


    I would appreciate any input! Thanks!
     
  2. jcsd
  3. May 5, 2005 #2

    OlderDan

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    Not quite- What is the amplitude of the motion?

    Plus- you will need to be a lot more careful with the units
     
    Last edited: May 5, 2005
  4. May 5, 2005 #3
    Oh, okay..... so then:

    Vmax : (2л) x (440.2Hz) x (0.00229m) = 6.33 m/s

    amax : [(2л x 440.2Hz)^2] x (0.00229m) = 17518 m/s2

    Does that look about right? Or am I still way off?
     
  5. May 5, 2005 #4

    Doc Al

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    The amplitude is still not correct.
     
  6. May 5, 2005 #5
    Well, I think I'm just stuck then. Can anybody give me a clue as to what the correct amplitude shoudl be? I would really appreciate it!
     
  7. May 5, 2005 #6

    Doc Al

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    You'll smack yourself in the head when you realize it:
    Remember that amplitude is the maximum displacement from the equilibrium position, not the distance between the extreme positions.
     
  8. May 5, 2005 #7
    You're right. I have no idea why I didn't realize that! Thank you so much for you help.
     
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