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Simple Harmonic Motion and Springs

  1. Apr 15, 2012 #1
    An object of mass m is supended from a vertical spring of force constant 1269 N/m. When the object is pulled down 0.068 m from equilibrium and released from rest, the object oscillates at 10 Hz. The mass of the object is ____ kg.



    τ= 2∏*sqrt(M/K)
    M = mass
    K = spring constant
    1/f = τ


    3. The attempt at a solution
    1/10 = .1
    .1/2∏ = .15708
    .157082 = M/K
    .024674*K = M
    .024674*1269 = 31.3113

    and I keep getting told i have the wrong answer...help?
     
  2. jcsd
  3. Apr 15, 2012 #2
    Well, the energy in a spring is 1/2Kx^2. This then relates to the kinetic energy, as at the top the potential energy in the spring is all converted into gravitational potential energy. Now I think, i could be wrong on this, that the amplitude above the equilibrium point is the same as the amplitude below, ie how far below the spring is initially below the equilibrium point.

    You can set these two equal to each other, and use h in mgh as 2x. Rearrange and you can calculate the force. This, if its correct gives an answer of 4.39kg
     
  4. Apr 15, 2012 #3
    umm i understand what your trying to say, but 4.39kg is wrong aswell
     
  5. Apr 15, 2012 #4
    Better way, your way works, mine is more complicated, though since T=2*pi*Sqrt(m/K) 0.01*1269/(4*pi^2) which gives 0.321441kg
     
  6. Apr 15, 2012 #5
    you have made some calculator errors.
    0.1/2pi = 0.0159 not 0.157
    You have calculated (0.1/2) x pi
     
  7. Apr 15, 2012 #6
    ahh i see, I have a ti-89 calculator, and i did my grouping wrong so it was doing things a little out of what i wanted it to lol, bad grouping on my part, but thanks a whole bunch!
     
  8. Apr 15, 2012 #7
    An object is hung on the end of a vertical spring and is released from rest with the spring unstressed. If the object falls 0.17 m before coming to rest, the period of the resulting oscillatory motion will be _____ s.

    im am uttlerly lost on this...
     
  9. Apr 15, 2012 #8
    Are you given anything else? Like the mass?
     
  10. Apr 15, 2012 #9
    Not convinced by this, but try it neverthe less. At equilibrium F=kx =mg, for it to be in equilibrium, by that meaning, m/k = x/g. so putting that into the period equation. T=2pi*sqrt(m/k) is the same as T=2pi*sqrt(x/g) which is 0.827 seconds.
     
  11. Apr 15, 2012 #10
    nope...
    but ive been looking around and i think the equation yo = mg/k
    the one to find the difference in the equilibrium points of when there is no mass and when there is a mass.
    but i dont know if my algebra is bad but i rearrange and get y*g = k/m which is ω
    but i also have seen some do g/y = k/m
    but i think it could also be y*g = m/k
    and with the right algebraic rearrangement you could use T = 2∏/ω
     
  12. Apr 15, 2012 #11
    Well its g/y=k/m which is the same thing as i said, as im taking x to be y, as there is no other info supplied. By the fact that omega is 2pif, then by the t equation you used in the first part, omega squared =m/k, from that you can get the time period.
     
  13. Apr 15, 2012 #12
    tried your solution, still saying incorrect answer
     
  14. Apr 15, 2012 #13
    odd, using what you said to work it out, i got T=.108994, so something is wrong if we got different answers doing the same thing...

    nvm i forgot to sqrt the g/y, i did get the same value as you
     
  15. Apr 15, 2012 #14

    Doc Al

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    Staff: Mentor

    Do this over. You're off by a factor of 10. (And don't round off pi when doing the arithmetic.)

    Edit: Oops... I see that truesearch already pointed this out in post #5. And I see that you are piling several problems into the same thread.
     
    Last edited: Apr 15, 2012
  16. Apr 15, 2012 #15
    Good :) glad that it helped
     
  17. Apr 15, 2012 #16

    Doc Al

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    Staff: Mentor

    As the mass falls, what's conserved?

    You can also use the equilibrium position to determine the ratio m/k, but where is that point?
     
  18. Apr 15, 2012 #17
    the potential energy of the spring?
     
  19. Apr 15, 2012 #18

    Doc Al

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    Staff: Mentor

    The potential energy by itself is not conserved--it changes as the mass drops. But what is conserved?
     
  20. Apr 15, 2012 #19
    if your refering to what does not change...g,m,k,T,f
     
  21. Apr 15, 2012 #20

    Doc Al

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    Staff: Mentor

    Stick to energy. What other forms of energy are involved as the mass falls?
     
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