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Simple Harmonic Question

  1. Feb 24, 2012 #1
    1. The problem statement, all variables and given/known data
    A 300g mass at the end of a spring oscillates such that it is a 2.0cm above the top of a table at its lowest point and 16cm above at its highest point. Its period is 4s. Find the Speed & acceleration of the mass when 9 cm above the tabletop?


    2. Relevant equations
    F=Kx
    A (amplitude) = X(max)
    V= Aω
    a = xω (square)


    3. The attempt at a solution
    When its 9cm above the tabletop it should be at its equilibruim position. So both of them should be zero. But the answer sheet said Velocity is .11m/s and Acceleration is 0m/s
    I am really confused, Could someone help me out? Please :)

    In my Work i found out that the a) Amplitude is .07m and b) K ( constant) is .74

    My work :
    http://img197.imageshack.us/img197/2286/photonxs.jpg [Broken]

    Thanks in advance :)
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Feb 24, 2012 #2
    Your handwriting is also very neat.. everyone that posts here seems to have nice handwriting

    Regarding your question, why should velocity be zero at the equilibrium position?
     
  4. Feb 24, 2012 #3
    Haha thanks :)
    I think because it's at equilibruim state which is its original position. So It should have no velocity or maybe i got the whole concept wrong. Reason i'm asking here is because my next class is on tuesday i don't want to wait that long :)
     
  5. Feb 24, 2012 #4
    If the mass has a period of 4s and s≠0 then the mass must be moving.

    It would have no velocity if you started the mass off at the equilibrium position, that is true, but not if you start anywhere else or start with a velocity.

    If you go and look at a spring (you're a student of the physical sciences you should have at least 50 different calibrated springs on your person at any one time) and start it moving, you can see that it doesn't have zero veolcity at the equilibrium position, at least it doesn't before friction dampens its motion out.
     
  6. Feb 24, 2012 #5
    Oh wow i got it haha, Its V=Aω ---> 1.57(.07) --> .11 m/s
    Now comes the acceleration part ---> a=ω(square) times x ---> 1.57(square) times (0) = 0

    Great !!! Am i correct?
     
  7. Feb 24, 2012 #6
    I was making this question wayy too complicated for myself, turned out its a super easy question haha
     
  8. Feb 24, 2012 #7
    Another question will the velocity be same if it was at 14cm above?
     
  9. Feb 24, 2012 #8
    Since the period is the same no matter what height you drop the mass from, what would this tell you about it's velocity at the equilibrium position?
     
  10. Feb 24, 2012 #9
    That it's starting to slow down, once it reaches -A it will have 0 velocity and then it will start gaining velocity again. correct ?
     
  11. Feb 24, 2012 #10
  12. Feb 24, 2012 #11
    Can you please show me how to solve it? I wanna know how can the solve if its at 14cm above table top
     
  13. Feb 25, 2012 #12
    do you know how to solve the equations of motion?
     
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