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Mass and springs problem

  1. Oct 3, 2004 #1
    suppose that the two springs have different spring constants k1=280 N/m and k2=260 N/m and the mass of the object is m=14 kg. Find the frequency of oscillation of the block in Hz.

    the first picture on this site is what the problem looks like. Its a mass between two springs that are attached to walls.

    i know that angular frequency equals the square root of k/m. but i don't know how to do it when there are two springs involved. Please help Asap. Thanks so much
  2. jcsd
  3. Oct 3, 2004 #2


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    Welcome to PF!
    Let your origin be at where both springs has their rest length.
    If you displace the box a distance "x", what are the forces (with direction) acting on it from either box?
  4. Oct 3, 2004 #3
    So if i have F1= -kx = -280x and F2= 260x do i then add them ? if so i would get Ft= -20x ...then should i use -20 as my value for k? and substitute it into the equation that i said before...where frequency equals the square root of (k divided by m)?

    THanks so much for your help by the way...this forum is a wonderful idea and i am definately going to spread the word!
  5. Oct 4, 2004 #4


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    No that is incorrect!
    Let the rest lengths be [tex]L_{1},L_{2}[/tex]
    For clarity, "x" be a positive number.
    Then, the new length of spring 1 is [tex]L_{1}+x[/tex]
    Hence, [tex]F_{1}=-k(L_{1}+x-L_{1})=-kx[/tex]
    Let's look at spring 2:
    If spring 2 had been lengthened by a positive amount "y" (dragged out to the left), then, the force from it would drag the block to the right (the positive direction).
    Hence, for positive displacement of spring 2 "y", [tex]F_{2}=2ky[/tex]
    Now, setting y=-x (spring 2 is actually shortened), we get:

    Hence, total force F on block is:
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