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Mass connected to two springs

  1. Feb 8, 2012 #1
    1. The problem statement, all variables and given/known data

    EDIT: I have removed the second setup, since I solved it. I still can't figure out this one though.

    A mass m is connected to two springs, with spring constants k[itex]_{1}[/itex] and k[itex]_{2}[/itex].

    The first setup has both springs connected in series, connecting the mass horizontally to the wall, with the mass resting on the floor. Show that the period for this configuration is given by

    T=2[itex]\pi[/itex][itex]\sqrt{m(\frac{1}{k_{1}}+\frac{1}{k_{2}})}[/itex]

    Hopefully I described the diagram sufficiently.

    3. The attempt at a solution

    Not really sure where to start. My first thought was to try and come up with an equation of motion, but honestly we haven't learned differential equations yet, so I don't think that's how I'm supposed to do it. Anyway, I can't seem to get it in terms of x regardless, since each spring presumably stretches a different length. I know that the two lengths will add up to the total displacement of the mass, but that only lets me get rid of one variable. So finding an equation of motion doesn't seem to be working for me. And even if I found it, I don't know how much good it would do me unless it turned out to be a real easy differential equation.

    Anyway, this is all I got:

    F = -k[itex]_{1}[/itex]x[itex]_{1}[/itex]-k[itex]_{2}[/itex]x[itex]_{2}[/itex]

    x[itex]_{1}[/itex]+x[itex]_{2}[/itex] = x
     
    Last edited: Feb 9, 2012
  2. jcsd
  3. Feb 9, 2012 #2
    One way of approaching the problem is to ask yourself, if I was to replace the two spring system with a single spring (for which I know how to obtain the period), what should be the spring constant of the single spring?

    Hint: The magnitude of force at each junction (wall-spring1, spring1-spring2, spring2-mass) is the same.
     
  4. Feb 9, 2012 #3
    Well, if I assume the force is the same at each junction, that makes the problem much easier. But how do we know that? Is it because the springs are massless, so the force at one end has to be the same as the force at the other?
     
  5. Feb 9, 2012 #4
  6. Feb 9, 2012 #5
    Great. That clears it up. Thanks a lot!
     
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