1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Spring Harmonic Oscilation in Gravitational Field

  1. Jun 25, 2009 #1
    1. A mass M is attached to a fixed spring of constant K and oscillating with an amplitude A, the spring is hung from the ceiling in a gravitational field with constant free-fall acceleration g.



    2. What is the oscillation frequency? Is it affected by gravity?
    Mx''=-Kx-mg => ??




    3. I figured it should be simply \omega = sqrt(k/m), but it looks like I'm wrong. This question is part of a longer calculation in a multiple choice exam. The only solution I have for comparison is the final result of the computation, but I don't understand the physical explanation. From what they say the frequency is not sqrt(k/m)
     
    Last edited: Jun 25, 2009
  2. jcsd
  3. Jun 25, 2009 #2

    jgens

    User Avatar
    Gold Member

    Why would you assume that the "angular velocity" omega is equal to the frequency?
     
  4. Jun 25, 2009 #3

    rock.freak667

    User Avatar
    Homework Helper

    At equilibrium

    mg=kx => m/k = x/g
     
  5. Jun 25, 2009 #4
    When I said frequency I was actually referring to the angular velocity \omega=2pi*f.
    They are practically equivalent.

    OK, how do I get information about the frequency/angular velocity from here?
     
  6. Jun 26, 2009 #5
    So I asked around irl, looks like the answer sheet is wrong.
    The oscillation frequency remains unchanged by gravity.
     
  7. Jun 26, 2009 #6

    rock.freak667

    User Avatar
    Homework Helper

    If

    [tex]T =2 \pi \sqrt{\frac{m}{k}} \ and \ \frac{m}{k}=\frac{x}{g}[/tex]

    then doesn't

    [tex]T = 2\pi \sqrt{\frac{x}{g}}[/tex]


    since T depends on g, doesn't g affect frequency,f (=1/T)?
     
  8. Jun 26, 2009 #7

    Doc Al

    User Avatar

    Staff: Mentor

    No, since x (the equilibrium displacement) is proportional to g. (x = mg/k) If x were a constant independent of g, then your reasoning would be correct.
     
  9. Jun 26, 2009 #8

    jgens

    User Avatar
    Gold Member

    Certainly not. Think about it this way: Once I hang a spring in a gravitational field the string will strech until it reaches equilibrium at some distance x0 and kx0 = mg. At this point, the net force acting on the block of mass M is zero (just as if we suspended the block in the absence of a gravitational field). Since the net force on the block is zero, the only force that we need to consider is the additional restoritive force acting on the block when we strech the string some distance A from its equilibrium position at x0. Hence, once the spring is at equilibrium our equation for the net force acting on the block becomes F = kx where x is the distance from the blocks equilibrium position at x0. Because of this, the period of oscillation does not depend on the presence of the graviational field.
     
  10. Jun 26, 2009 #9
    Try and look at T(g).

    T(g) = 2π*√(x/g) = 2π*√(m/k) which is independent of g.

    What is the ratio, x/g? It is m/k. And that is a constant value determined only by the spring and the mass of the object attached to it.
     
  11. Jun 26, 2009 #10

    rock.freak667

    User Avatar
    Homework Helper

    I understand better now.

    So even if you do this on the moon where g is different. x will extend in relation to g such that m/k will be the same to produce the same period then?
     
  12. Jun 26, 2009 #11

    Doc Al

    User Avatar

    Staff: Mentor

    Exactly. The period has nothing to with gravity. Even if the spring and mass were oriented horizontally, the period of oscillation would be the same.
     
  13. Jun 27, 2009 #12

    rock.freak667

    User Avatar
    Homework Helper

    :biggrin: thanks..I always used to get this type of question wrong even though I thought my reasoning was correct and no one ever told me the correct thing.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook