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SHM(mass suspended vertically on spring)

  1. Jul 4, 2009 #1
    1. The problem statement, all variables and given/known data

    An object is hung on the end of a vertical spring and is released from rest with the spring unstressed. If the object falls 3.42 cm before coming to rest, find the period of the resulting oscillatory motion.

    2. Relevant equations



    3. The attempt at a solution

    This was a homework question given and the answer is in the back of the book and it says
    T = .262 seconds. Now i have no clue how they got this because I thought that the period was related to the spring constant and the mass but not to the amplitude so how can we find T if neither k nor m is given?
     
  2. jcsd
  3. Jul 4, 2009 #2

    cepheid

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    Hi nrweis, welcome to PF!

    You can deduce something about them from the restoring force that is required to balance the object's weight.
     
  4. Jul 4, 2009 #3

    LowlyPion

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    Welcome to PF.

    You don't necessarily need m and k, because don't you also know that

    a(t) = ω2*x(t)

    Won't a(t) = 9.8 at x/2 from the dead weight equilibrium point, which is at the peak of its oscillation?

    Since ω = 2πf = 2π/T ...
     
  5. Jul 4, 2009 #4
    Ok I thought about it for a while and thought I was on the right track but still got it wrong. This is what i did:

    F=k∆x
    mg=k∆x

    √(g/∆x)= √(k/m)=ω=2π(1/T)

    √(g/∆x)= 2π(1/T)

    T=2π√(∆x/g)

    and I got like .37 seconds.
    Was I on the right track?
     
  6. Jul 4, 2009 #5
    now I'm all confused. I was thinking that delta x was the amplitude
     
  7. Jul 4, 2009 #6
    ok I just misunderstood the question. When I divided the 3.42 cm by 2 and reentered it to the formula I got the right answer. Thanks guys.
     
  8. Jul 4, 2009 #7

    LowlyPion

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    Yes. Δx is the amplitude, but the x-position function which expresses x as a function of time is only x/2 distant from the mid point.

    Congrats on figuring it out.
     
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