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Formula for G using s.h.m

  1. Apr 9, 2007 #1
    ok, im trying to measure thevalue of g using a mass attatched to a spring. i have the period of bounce, the mass of the..well...mass, and the elastic constant of the spring. i have been informed its relatively easy to find g using this setup but im struggling. iv spent the last 3 nights either rearranging formulae or serching the internet and my brains dead and i really need a break from this. anyone? any help like atall is massivley appreciated. though as usual the answers probably pretty obvious and im jst being blind. but thanks anyway.
  2. jcsd
  3. Apr 9, 2007 #2


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    The period of motion is not dependent on g and so cannot give a value for it.
    The motion of a weight on a spring, assuming it is hanging, is given by the differential equation
    [tex]m\frac{d^2 x}{dt^2}= -kx- mg[/tex]
    [tex]\frac{d^2 x}{dt^2}= -\frac{k}{m}- g[/tex]

    The general solution to that is
    [tex]x(t)= C_1 cos(t\sqrt{\frac{k}{m}})+ C_2 sin(t\sqrt{\frac{k}{m})- frac{g}{k}[/tex]
    The period is given by
    with no dependence on g.

    Of course, you have mg= kx where x is the "stretch" when you hang the weight on the spring initially so g= kx/m. That's done with the spring motionless.
    Last edited by a moderator: Apr 9, 2007
  4. Apr 9, 2007 #3
    ...but the period is dependant on the mass which can be affected by altering gravity?
    i know im grasping at nothing but im pretty sure my friend managed this and im pretty stubborn about not failing when i've wasted so much on this
  5. Apr 9, 2007 #4


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    How about

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

    EDIT: Whoops. Never mind. If I passed basic reading comprehension I would have noticed that you said a mass on a spring, not a string. Don't mind me.
    Last edited: Apr 9, 2007
  6. Apr 9, 2007 #5
    would this work? what could i use for l?
    i still have mass and spring elastic to use up.
    i didnt really measure anything else :frown:
  7. Apr 9, 2007 #6

    Doc Al

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    Which is for a simple pendulum, not a mass on a spring (as FredGarvin certainly knows).
  8. Apr 9, 2007 #7
    simple pendulum indeed, however ill use anything. i am definetley that desperate.
  9. Apr 9, 2007 #8
    surley it would be possible somehow to use the elastic constant of the spring to find the restoring force which would be somewhat the same as the force of gravity acting on the mass? no? anyone?
  10. Apr 9, 2007 #9
    Maybe the catch is to use the period [tex]T[/tex]of the oscillation and the lengthening [tex]\ell[/tex] of the string:
    [tex] g = \ell\left({2\pi\over T}\right)^2[/tex]
    (if I didn't did a mistake)
  11. Apr 9, 2007 #10
    omg :surprised thankyou im getting aroundabout 9 for my value of g, and since my technique can be considered fairly crude i will have to assume that this is the best formulae ever concived. thankyou much, and my mum thanks you as i hae stopped sulking about not being able to work this out. thanks to you all though, i realise your time is important so thanks for even reading the post, but to the guy who posted that formulae i owe my life. :!!)
    ... any chance of a proof for this? :rofl:
    but really thankyou
    im so happy:biggrin:
    my easter holidays might not be ruined afterall
  12. Apr 9, 2007 #11
    ......................no. What he told you is wrong and you got a close answer by dumb luck. The period of oscillations has nothing to do with g. Read the very first post and what HallsofIvy told you.
  13. Apr 9, 2007 #12
    :( i feared this day would come
    are you sure?
    i know someone who has done this before and im convinced that there's a way. sure it was dumb luck? i tried it on all of my readings and they were all pretty much right. its late where i am and im tired and i cant work out if they all would be right from just luck
    could this not work then since the mass in the spring constant cancels with the mass in the shm equation so that i can match the extension to the period o the same masses? i really dont know. ill be quiet now.
  14. Apr 9, 2007 #13
    Read HallsofIvys post. Gravity does not play a role in SHM of a spring. Are you talking about a spring or a string?
    Last edited: Apr 9, 2007
  15. Apr 9, 2007 #14
    ...yeh i read that and was disheartened, however
    t=2pi root m/k
    w^2 = k/m = g/e
    so cant we change this to
    t=2pi root e/g
    ...which means that the period can be dependant on gravity?
    i know hallsofivy said that it didnt but why couldnt this work?
    thanks for your patience
    and the fast response that was awesome cheers

    and yeh sorry definetly a sPring
  16. Apr 9, 2007 #15
    w^2 = k/m = g/e

    where did this come from?

    The answer is no. The spring and mass does not depend on gravity, ever. You can take it to the moon and you will get the same answer. You can turn gravity off and still get the same answer.
  17. Apr 9, 2007 #16
    ...ok... erm...
    i have it written in a book, and i think its because the whole hookes law thing gives f = kx and f = ma gives xk = ma
    a as the acceleration can substitute for g
    and e for extension because its easier that way
    so we have
    mg = ke
    and then we rearrange or something to get
    k/m = g/e
  18. Apr 9, 2007 #17
    I do not know if I make it clear.
    When you hang a mass m the spring lengthens of [tex]\ell[/tex]:
    If you put the mass and the spring in oscillation the period is:
    [tex]T=2\pi\sqrt{m\over k} [/tex]
    [tex]{k\over m}= \left({2\pi\over T}\right)^2[/tex]
    From the first formula:
    [tex] g=\ell{k\over m}[/tex]
    [tex]g = \ell \left({2\pi\over T}\right)^2 [/tex]
    Last edited: Apr 9, 2007
  19. Apr 9, 2007 #18
    This is not a pendulum lpfr. (And please delete one of your double posts.)
  20. Apr 9, 2007 #19
    Do I say it was?
    It is a mass at the extremity of a spring and it oscillates VERTICALLY!
  21. Apr 9, 2007 #20
    I see what your doing now, but its still rather meaningless because your calling [tex]\ell[/tex], what should really be called [tex]\Delta L[/tex] and that is a STATIC measurement of the physical system. Its exactly measuring the displacement of the mass. You dont need SHM to do that. All you have to do is put the mass on and measure the deflection. Its a static experiment.

    see what I mean?
    Last edited: Apr 9, 2007
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