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Velocity for Spring Constant with Amplitude

  1. Apr 21, 2010 #1
    1. The problem statement, all variables and given/known data
    A 79.8 g mass is attached to a horizontal
    spring with a spring constant of 2.66 N/m
    and released from rest with an amplitude of
    39.5 cm.
    What is the velocity of the mass when it
    is halfway to the equilibrium position if the
    surface is frictionless?
    Answer in units of m/s.

    2. Relevant equations
    v = [tex]\pm[/tex][tex]\sqrt{}[/tex]k/m(A2-x2


    3. The attempt at a solution
    So I used the above equation, and got 2.280533. The answer is wrong. Did I use the correct equation? Thanks.
     
  2. jcsd
  3. Apr 22, 2010 #2
    the answer required is in m/s ..

    I tried myself to use the equation you wrote and i obtained different answer than the one you got .. so I would suggest you to look more carefully at the units ..

    convert gram to kilogram .. cm to m ..
     
  4. Apr 22, 2010 #3
    I converted everything, and got the same answer.
     
  5. Apr 23, 2010 #4
    can you please write the steps you followed in detials so i can check where you did wrong..
     
  6. Apr 23, 2010 #5
    [tex]\sqrt{}(2.66N/m / 0.0798 kg)(0.395)^2[/tex] = 2.2805

    Thanks.
     
  7. Apr 23, 2010 #6
    the first thing you missed is x? .. you have in the equation you supplied us with a term for x^2 .. I can not see it in your solution ..

    note: your question is asking you to find the velocity when it reaches half way to its equilibrium position .. what you calculated is (wA) which I believe is the maximum velocity..
     
  8. Apr 23, 2010 #7
    No, it's not x2. It is the square root of the whole thing, and inside the square root, I square the 0.395. And I believe it's not the maximum velocity because vmaxhas a totally different equation. Thanks.
     
  9. Apr 23, 2010 #8
    MissPenguins, please note that if the equation is valid, the vmax would be given by setting x equal to 0. Also, I'm fairly sure that bigstar is correct: you forgot to subtract x2 from A2.
     
    Last edited: Apr 23, 2010
  10. Apr 23, 2010 #9
    so are you saying that you got for (A^2 - x^2) term a value of 0.395^2 ? .. this still different from what i got?? ..

    What value you substitute for x in the equation?
     
  11. Apr 23, 2010 #10
    Isn't x2=0 since x is not given?
     
  12. Apr 23, 2010 #11
    I thought x is 0 since it's not given. Am I wrong?
     
  13. Apr 23, 2010 #12
    noooooo .. Read the question carefully ..

    it said find the velocity when it is halfway from the equilibrium .. If it said when it reaches the equilibrium then you put x=0 ..

    Try again, and think more deeply about what value for x you should use ..
     
  14. Apr 23, 2010 #13
    Miss Penguins read bigstar's post and keep in mind that just because a value for a quantity is not given (in this case it is given, just not explicitly), you can't assume that the quantity's value is 0.

    BTW, I don't have to be a mentor to give help, do I? (I'm new)
     
  15. Apr 23, 2010 #14

    no you dont have to .. But you should follow the rules of this forum .. You are not allowed to solve the problem just few hints can be given....
     
  16. Apr 24, 2010 #15
    When it says halfway, which is something divide by 2, correct?
     
  17. Apr 24, 2010 #16
    okay that is a good start .. Can you figure out what should this something be ..

    Hint: try to make a simple drawing of the problem, things may get clearer for you regarding the choice of that ''something'' ..
     
  18. Apr 24, 2010 #17
    Would x be half the Amplitude? I have this same problem and only one chance left to get the correct answer.
     
  19. Apr 24, 2010 #18
    You tell me, why do you think that x is half the amplitude?
     
  20. Apr 24, 2010 #19
    The amplitude is how far the spring moves in either direction from equilibrium. The question wants to know how fast it is moving when it is half the distance from the equilibrium which would be one half of the amplitude.
     
  21. Apr 24, 2010 #20

    Hmm, well .. You convinced me with your answer :) , lets see whether the last answer you give would give you the right solution ..
     
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