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Math Help - Springs

  1. #1
    Junior Member
    Joined
    Nov 2008
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    65

    Springs

    Here is the problem:

    A frictionless spring with a 8-kg mass can be held stretched 1.2 meters beyond its natural length by a force of 30 newtons. If the spring begins at its spring-mass equilibrium position, but a push gives it an initial velocity of 1.5 m/sec, find the position of the mass after t seconds.

    So I organized the data and find that

    m = 8 kg
    K = 30N/1.2M = 25 N/m
    and that
    External Force = 1.5 m/s

    How do I approach this problem?

    I think I should use the formula
    mx'' + kx = F0cos(wt) but not sure how to


    Thanks for the help!!
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  2. #2
    Newbie
    Joined
    Oct 2009
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    20
    The wording of this problem is a little tricky. The word "push" is used, and it probably makes you think "push"="force". Instead, you should think of this as an initial value problem (IVP) with an initial velocity undergoing free vibration.

    So, draw your FBD and you'll get:

    mx'' + kx = 0, with x(0)=0 and x'(0)=v0=1.5 m/s

    Let me know if you need more help from there.

    Regards
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  3. #3
    Junior Member
    Joined
    Nov 2008
    Posts
    65
    Well I got

    1/4x''+25x = 0

    and final answer

    x(t) = .15sin(10t)

    but it's wrong
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  4. #4
    Newbie
    Joined
    Oct 2009
    Posts
    20
    Here's a little guidance:

    Assume a solution: x = A*e(lambda*t), where A and lambda are constants.

    You then need to find x'' by differentiating x above twice. Plug those suckers into the equation that you started with:

    mx'' + kx = 0

    You'll get a product on the left hand side with one factor in lambda. That factor must be equal to zero because the other factor (A*e(lambda*t)) cannot be equal to zero (it would imply the trivial solution for x). Use it to solve for lambda (you should get two values for lambda).

    Let me know where you get from that point.
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