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Simple Harmonic Motion - Mass and massive spring

  1. Nov 29, 2011 #1
    1. The problem statement, all variables and given/known data

    A block of mass M is connected to a spring of mass m and oscillates in simple harmonic motion on a horizontal, frictionless track (Figure P15.62). The force constant of the spring is k and the equilibrium length is l. Assume that all portions of the spring oscillate in phase and that the velocity of a segment dx is proportional to the distance x from the fixed end; that is,

    [itex]v_{x}=\frac{x}{l}v[/itex]

    Also, note that the mass of a segment of the spring is

    [itex]dm=\frac{m}{l}dx[/itex]

    (a) Find the kinetic energy of the system when the block has a speed v, using m, M, and v as necessary.

    (b) Find the period of oscillation, using m, M, and k as necessary.


    2. Relevant equations

    The above two equations. Also,

    [itex]K=\frac{1}{2} \int{v^{2}}dm[/itex]

    [itex]v=-A \sqrt{\frac{k}{m}} sin(t \sqrt{\frac{k}{m}})[/itex]

    [itex]T=2 \pi \sqrt{\frac{m}{k}}[/itex]


    3. The attempt at a solution

    (a) I simply plugged and chugged using the two given equations and the kinetic energy equation. The only tricky part was realizing that when the velocity and kinetic energy are at their maximum, x=l , meaning, the distance the from the wall to the mass is the same as the distance from the wall to the equilibrium point.

    (b) Not so simple. I can't find any equations for period other than the one above, and that one isn't really helping me. I could just plug in M and m into it and add the resulting equations, but I have a feeling the term with m in it would have some sort of denominator, like the term with m in my correct equation for kinetic energy.

    Suggestions? Is my intuition failing me, or is there something I'm missing?
     
  2. jcsd
  3. Nov 29, 2011 #2

    ehild

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    Concerning the vibration of the spring, there are two variables. One is the position x of a little piece of mass dm along the length of the unstretched spring, and the other variable is Δx, the displacement of that piece of string from its equilibrium position. All pieces move in phase, with velocity v(x)=x/L V: V is the velocity of the object of mas M at the end of the spring. The equilibrium position of that object is X=L, and its displacement and velocity are ΔL=Acos(ωt), V=-Aωsin(ω).

    Write out the KE of a small piece of string as function of x, than integrate for the whole length of the spring to get the KE of the whole spring. Add the KE of the object of mass M.
    The potential energy (elastic energy) is that of the spring, of spring constant k and stretched by ΔL. KE+PE= const, so the kinetic energy at the equilibrium position is the same as the potential energy of the spring when ΔL is at maximum. You get the angular frequency ω from this condition.

    ehild
     
  4. Nov 30, 2011 #3
    Ahahaha! I got it. Man, physics/math at it's finest. Thanks, those last two sentences really helped me out. Once I figured out how to get rid of the sine/cosine terms in the conservation of energy terms, it was a piece of cake.
     
  5. Nov 30, 2011 #4

    ehild

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    What is cosine/sine at maximum displacement/maximum speed?

    ehild
     
    Last edited: Nov 30, 2011
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