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A Few Questions About SHM

  1. Feb 18, 2008 #1
    I have been doing quite a few SHM problems, and I just have a few questions in general. A lot of questions evolved from one particular problem type: A mass attached to the end of a vertical spring of spring constant k.

    My questions:
    1. How can we prove that we can use the equation w=(k/m)^(1/2) for this problem. Normally, you can just go:
    ma=-ky
    a=-k/m * y
    a=-w^2*y
    y's cancel out
    w=(k/m)^(1/2)
    but in this case you should have to account for the mg force, but in most solutions, I do not seem mg accounted for?

    In one problem, I was asked to solve for the maximum amplitude the shm could have in order to not surpass a certain acceleration. Once again, all answers were along the lines:
    ma=-kA
    mg=-kA
    A=-mg/k
    Once again, how can you neglect the mg force?

    My only idea is that since we determine the equilibrium point for most of these problems at the beginning - the point where the spring force matches the gravitational force - that they treat this equilibrium point like the spring's equilibrium point and can somehow, magically, neglect the spring force?
     
  2. jcsd
  3. Feb 18, 2008 #2
    I have been doing problems for the last two hours, and still haven't really gotten much further on figuring this out..
     
  4. Feb 19, 2008 #3
    you didn't prove that you could use w=(k/m)^(1/2). you have a second order differential equation y'' = -k/m * y
    To solve this just try y = a sin (b *t) as a solution and then find out what a and b are.

    if you would include an mg force then your new differential equation would become

    y'' = -k/m * y - mg. try to prove that if y=F(t) is a solution of the first differential equation, that y = F(t) - mg/k is a solution of the second one.
     
  5. Feb 19, 2008 #4

    rock.freak667

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    Homework Helper

    When the spring is hanging vertically in equilibrium: Tension =Weight i.e ke=mg

    When you displace it a small distance,x, Resultant force,F =mg-k(e+x)...use F=ma now.
     
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