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Amplitude question

  1. Jul 5, 2008 #1
    Hello, I am having trouble finding the A in this problem. I found T=1/2=.5, I found K =31.55n/m. I tried using the formula KA2=MV2+KX2 to find amplitude but ended up with A=0.0435m
    This just does'nt seem right . I used .05m for a value of X. Does anybody know what I am doing wrong?



    A 200 g mass attached to a horizontal spring oscillates at a frequency of 2.0 Hz. At one instant, the mass is at x=5.0cm and has Vx=.30 cm/s determine the period, Vmax, Amplitude, and the total energy.
     
  2. jcsd
  3. Jul 5, 2008 #2
    The V should be Vx=-30cm/s
     
  4. Jul 5, 2008 #3

    alphysicist

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    Hi nicksim117,

    Can you list the details of what you did to get 0.0435m? It looks to me like you have a calculation error (or maybe you're interpreting the equation incorrectly?). Did you square the velocity?
     
  5. Jul 5, 2008 #4

    Redbelly98

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    Everything in the equation is a positive quantity, since v and x get squared. There is just no way for A to end up being less than x.
     
  6. Jul 5, 2008 #5
    k looks ok

    use energy methods E=K+U

    you know v,x and k solve for A. the rest is pretty straight forward.
     
  7. Jul 6, 2008 #6
    1. The problem statement, all variables and given/known data

    Hello, I am having trouble solving this problem trying to find A I keep getting an amplitude which is less than my distance can anybody see what i am doing wrong?

    A 200g mass attached to a horizontal spring oscillates at a frequency of 2.0 hz. At one instant, the mass is at X=5cm and has a V=-30cm/s Determine:
    The period T=.5
    The amplitude?
    the Vmax
    the total energy






    2. Relevant equations

    KA2=MV2+KX2 or A2=(MV2+KX2)/K

    3. The attempt at a solution

    solve for A using 1/2KA2=1/2MV2+1/2KX2= (.5)(31.55)A2=
    > (.5)(.2)(-.3m/s)+(.5)(31.55)(.05m)2
    > ==15.77A2=-.009+.0394375==.0304375/15.77=A2===.001930==A=.04393m
    > V=-30cm/s
    > K=31.55n/m
    > X=5cm
    > M=.2kg

    A=.04393m
     
  8. Jul 6, 2008 #7
    I don't know why i keep getting that answer. I posted the equation worked out as a new thread.

    [Edit by Doc Al: I merged the two threads. Do not create multiple threads on the same problem.]​
     
    Last edited by a moderator: Jul 6, 2008
  9. Jul 6, 2008 #8

    Doc Al

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    Your approach is fine, but recheck your arithmetic.
     
  10. Jul 6, 2008 #9

    alphysicist

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    You squared the velocity, so it should be positive 0.009
     
  11. Jul 6, 2008 #10

    Doc Al

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    There ya go! :wink:
     
  12. Jul 6, 2008 #11
    Thank you!!
     
  13. Jul 6, 2008 #12
    For the same problem i am ask to report total energy in the system. When I use the formula and value at that instantE=K+U=1/2MV2+1/2KX2 I get .0484J. When I use the same formula but Vmax and A are used in place of V and X I get .0954J Should these values be the same because energy is conserved.
     
  14. Jul 6, 2008 #13

    Doc Al

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    Do v=Vmax and x=A occur at the same point? :wink:

    Energy is most definitely conserved--that should tell you that something is wrong with your thinking.
     
  15. Jul 6, 2008 #14
    Ok, that make sense, so If energy is conserved and I use the values given in the problem and plug them into E=K+U=1/2MV2+1/2KX2 to find total energy in the system they should equal the value of either Umax 1/2 KA2or Kmax 1/2MV2 not both added together.
     
  16. Jul 6, 2008 #15

    Doc Al

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    Exactly.
     
  17. Jul 6, 2008 #16
    thanks for your help!
     
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