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Spring's force constant

  1. Mar 31, 2004 #1
    hello everybody:
    here is my question:

    In a physics lab, you attach a 0.200-kg air-track glider to the end of an ideal spring of negligible mass and start it oscillating. The elapsed time from when the glider first moves through the equilibrium point to the second time it moves through that point is 2.60 s.
    what is the spring's force constant?
    how can i calculate that? how can i get the period ?
    do i use the formula
    Code (Text):
    omega = sqrt (k/m)
    ? i'm confused between frequency and angular frequency. can u tell me how to use those formulas for this problem.
    thanks :wink:
  2. jcsd
  3. Mar 31, 2004 #2


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    Science Advisor
    Homework Helper

    The behavior of a spring is a bit like seeing a spinning wheel head on - you only see one of the components of circular motion.

    You can describe the position of the mass as [tex]p(t) = r \sin {\omega} t[/tex] You know that [tex]p(0)=p(2.60)=0[/tex], so you should be able to find [tex]\omega[/tex] and use your formula from there.
  4. Mar 31, 2004 #3
    The motion is going to be described by the equation

    [tex]x = A\cos{(\omega t)}[/tex]

    where x is the displacement off of the equilibrium position, t is the time, and omega ([itex]\omega[/itex]) is as you said it was.

    We immediately see that the motion is periodic. Whenever the expression cosine is operating on is equal to

    n is any integer,

    then the cosine will be zero and the displacement will be zero, i.e. it will be at the equilibrium position. So the first time it passes through the equilibrium point is whenever what's inside the cosine is equal to pi/2, the second time whenever what's inside the cosine is equal to 3pi/2.

    Now you know that the time in the first one is some convenient moment in time, say zero, and the time in the second one is 0 + 2.60s. So now we have an equation and you can solve for the spring constant.

    It takes 2.6 seconds to go from the equilibrium position and back, and that's half a period, so what do you think the period is?

    Frequency is the one over the period. Angular frequency is omega.


    Edit: Just so my post doesn't seem to conflict with Nate's post (they don't), the generalized description of motion of a spring is

    [tex]x = A\cos{(\omega t + \phi)}[/tex]

    note that for a special choice of [itex]\phi[/itex] you go from the cosine formula to the sine formula. Since we don't have any initial conditions anyway, we can choose [itex]\phi[/itex] to be whatever we want in this problem, which is why both formulas are acceptable.
    Last edited: Mar 31, 2004
  5. Mar 31, 2004 #4

    Doc Al

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    Staff: Mentor

    Several things:
    (1) The period is the time for one complete cycle: it would be the time from when the glider moves through the equilibrium point to the next time it moves through that point in the same direction.
    (2) [itex]\omega[/itex] is angular frequency, measured in radians/sec; frequency is cycles/sec. One cycle = [itex]2 \pi[/itex] radians. So [itex]\omega = \frac{2 \pi}{period}[/itex]

    Depending on what you measured (see #1), your period would be either 2.60 sec or twice that. Find the angular frequency and solve for k.

    Note: while I was writing this I see that NateTG and cookiemonster have answered you. Oh well, what's one more? :smile:
  6. Mar 31, 2004 #5
    thanks to all 3 of you!!! I understand much better now. I was getting confused as to how to set the period thing , i get confused everytime they talk about the equilibrium position and so on. :tongue:
    i'll bother some more later.
    thanks again
  7. Mar 31, 2004 #6
    If you'd have thrown in the Acos(wt) + Bsin(wt) formula in there, Doc Al, we would've had a clean sweep!

  8. Mar 31, 2004 #7

    Doc Al

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    Staff: Mentor

    I can always go back and edit it in! :wink:
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