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Speed of the Mass

  1. Dec 5, 2009 #1
    A mass of .4 kg, hanging from a spring with a constant of 80 N/m, is set into an up-and-down simple harmonic motion. What is the speed of the mass when moving through a point at .05 m displacement? The starting displacement of the mass is .10 m from its equilibrium position.

    * zero
    * 1.4 m/s
    * 1.7 m/s
    * 1.2 m/s



    I am confused with this one. I don't know where to begin. Any help would be appreciated. I need to have this done by Monday. Thanks.
     
  2. jcsd
  3. Dec 5, 2009 #2

    ideasrule

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    Homework Helper

    Have you tried the conservation of energy?
     
  4. Dec 5, 2009 #3
    How do I do that? I mean I know what it is, but what numbers go where?
     
  5. Dec 6, 2009 #4
    This is a bit like the pendulum problem we just did. The relevant eqns are F=-ky where k is the force constant of the spring (given) Also, if we are to take ideasrule tip, we know that the energy (potential) stored in the spring is 1/2ky^2.

    When the spring-mass is set into motion some of the potential energy in the spring will be converted to kinetic energy but the sum is always the same, ie

    1/2mv^2+1/2ky(t)^2=1/2k(y')^2 where y(t) is the location as a function of time and relative to the equilibrium position and y' is the equilibrium position with the mass. I believe this approach allows us to ignore the potential energy from gravity. Try it and let me know. In this case, y'=0.1 and y(t)=0.05
     
  6. Dec 6, 2009 #5
    I see what you're saying, but how do I find k for all of the equations? I know I probably have to use F = -ky, but how do I use it?
     
  7. Dec 6, 2009 #6

    The K is given. This is the 80N/m. Using energy conservation, you should not have to use F=-ky. Sorry if I misled you, I didn't realize that K was given and thought we had to solve for it using that.
     
  8. Dec 6, 2009 #7
    Just to make sure I am doing this right, I am using the equation you gave me before to find v?
     
  9. Dec 6, 2009 #8
    yes, the one that has elements of potential and kinetic energy.
     
  10. Dec 6, 2009 #9
    Okay, I finally got .38729 as my velocity.
     
  11. Dec 6, 2009 #10
    show some work, I'll check in the meantime.
     
  12. Dec 6, 2009 #11
    Okay, try your best to follow along. It's hard to type something like this:

    v = [tex]\sqrt{}(1/2(80 N/m)(.1 m)2 - 1/2(80 N/m)(.05 m)2)/ 1/2(4 kg)[/tex]

    Which I got:

    v = [tex]\sqrt{}(.4 - .1)/2)[/tex]

    which led me to that answer. (SUP means exponent, so (SUP 2 SUP) means squared. It messed that up for some reason.
     
  13. Dec 6, 2009 #12
    Hmmm, I get a different answer than given as well, but it is 3.87m/s. Let me think about this for a minute, I'm wondering if he is giving the equilibrium of the spring w/o the mass.
     
  14. Dec 6, 2009 #13
    Well i did the math again, and got 1/2mv^2=1/2 k(0.01-0.0025)/m

    v^2=80(0.0075)/.4=1.5 so I'm sticking with 1.2 for the answer. But no guarantees.
     
  15. Dec 6, 2009 #14
    But 1.5 wasn't an answer. Are we gonna stick with 1.2? And why 1.2?
     
  16. Dec 6, 2009 #15
    So should I just say 1.2? And how do I show how I arrived at that answer?
     
  17. Dec 6, 2009 #16
    1.5 was v^2 so v=1.22 Your work was fine, just recheck the math.
     
  18. Dec 6, 2009 #17
    Yes! I found my mistake. And you were right. Thanks a lot.
     
  19. Dec 6, 2009 #18
    Yes! You were right, and I checked it and found my error. Thank you! :biggrin:
     
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