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Simple harmonic motion problem

  1. Jun 3, 2012 #1
    1. The problem statement, all variables and given/known data

    A mass is oscillating on a spring with simple harmonic motion. Its amplitude is 0.80m and its maximum speed is 1.5m/s, at the point of equilibrium. What is the speed of the mass at 0.60m?

    Is there enough information to answer this question? The answer given is 0.75m/s, but I can't figure out how to get there.

    2. Relevant equations

    Ep = kx2/2
    Ec = mv2/2
    Em = Ep + Ec

    3. The attempt at a solution

    I've made many attempts at this, all completely wrong. I always end up trying to factorize (my teacher told us that the answer involves factorizing) but it goes absolutely haywire.

    Em = Em'
    Epmax = Ecmax
    kxmax2/2 = mvmax2/2 (the 1/2 cancels for each formula)
    vmax/xmax = √k/m (which should be constant)
    (1,5m/s)2/(0.80m)2 = √k/m
    1.875(not sure what unit to put here) = √k/m ∴ k/m = 3.516

    kx2 + mv2 = mvmax2 (again the 1/2 cancels)
    mvmax2 - mv2 = kx2
    m(vmax2 - v2) = kx2
    vmax2 - v2 = kx2/m
    k/m = (vmax2 - v2)/x2

    If the k/m stays constant, which I believe it should, it should be an easy substitution to find the v I'm looking for.

    After manipulating the formula I got, I'm left with

    v = √-(k/m - vmax2/x2)(x2)

    After substituting, I get 0.99m/s.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 3, 2012 #2

    Doc Al

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    Staff: Mentor

    Instead of all that, think in terms of fractions or ratios. At the point in question, what fraction of the total energy is Ep and what fraction is KE?
     
  4. Jun 3, 2012 #3
    Is it possible the system is hanging?


    I got the same answer of .99m/s using a horizontal system and energy conservation principals.
     
    Last edited: Jun 3, 2012
  5. Jun 3, 2012 #4
    Me too.
    v=VmaxSin(ArcCos(6/8))
    V=1.5 X sin(41.41°)
    v=0.99 m/s
    http://img818.imageshack.us/img818/3757/shmw.jpg [Broken]
     
    Last edited by a moderator: May 6, 2017
  6. Jun 3, 2012 #5

    Doc Al

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    Staff: Mentor

    I get the same answer.
     
  7. Jun 3, 2012 #6
    Do you mean x2/xmax2? = .36/.64 = 0.5625? So 56% of the mechanical energy is potential energy? Which would mean that 44% is kinetic energy. Not sure if that's what you mean or not, or if I just did some completely useless calculations. If that's what you meant, I don't see how that will help me :/
     
  8. Jun 3, 2012 #7
    Just an observation, but [itex]1.5\sqrt{\frac{.8 - .6}{.8}} =.75[/itex], maybe that can help lead to an answer.

    Like (v1/v2)2 = d1/d2, where d is the distance from the maximum amplitude, but how to prove this?
     
    Last edited: Jun 3, 2012
  9. Jun 3, 2012 #8

    Doc Al

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    That's exactly what I meant. It's just a quicker way of getting the (same) answer:
    KEx=0.6 = .4375*KEmax
    1/2mV2 = .4375(1/2mVmax2)
     
  10. Jun 3, 2012 #9
    But if you do that (mass cancels, 1/2 cancels), you get 0.99m/s again!
     
  11. Jun 3, 2012 #10
    Another formula
    ω=Vmax/R
    v=ω√(A2-x2)
    v=(1.5/0.8)√(0.82-0.62)
    v=0.99 m/s
     
  12. Jun 3, 2012 #11

    Doc Al

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    Of course. I'd say that was the correct answer.

    What book are you using? Are you sure you didn't just read the wrong answer?
     
  13. Jun 3, 2012 #12
    It's the Alberta high school physics textbook. It says 0.75m/s for sure. If all of you, who are experts in physics, agree with me and say it should be 0.99m/s, should I risk it and go with that answer? Or make up a way to get to 0.75m/s (as the rest of my class did) and just get partial points?
     
  14. Jun 3, 2012 #13
    Maybe you can show us your friend's work(rest of the class) that lead to the answer v=0.75 m/s
     
  15. Jun 3, 2012 #14
    All they did was t=d/v, using 0.80m for d. Then they manipulated the formula d=((Vi + Vf)/2)t to find Vf, using 0.60 as d. They got the answer by fluke, because velocity and acceleration are always changing in a spring.
     
  16. Jun 4, 2012 #15

    Doc Al

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    Well, that 'solution' is clearly bogus. I say stick with your correct solution and trust the instructor to realize that the given answer is incorrect.
     
  17. Jun 4, 2012 #16
    It turns out that 0.99m/s is right! Thanks everyone for all your help!
     
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