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Completely lost on oscillations problem

  1. Jun 23, 2015 #1
    1. The problem statement, all variables and given/known data
    Two springs are joined and connected to a mass m such that they are all in a straight line. The two springs are connected first and then the mass last so that all three are in a row. If the springs have a stiffness of k1 and then k2, find the frequency of oscillation of m.


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

    3. The attempt at a solution

    so i tried making an F=ma for the mass and spring 1 (which is said was the spring closer to the mass)...

    F=ma system mass

    [itex] F_{el} = ma [/itex]
    [itex] k_1 x_1 = ma [/itex]
    max acceration happens at aplitude:
    [itex] k_1 x_1 = mA \omega ^2 [/itex]

    F=ma system spring 1.

    [itex] F_{el mass} - F _{el 2} = M_{s1} [/itex]
    i am assuming the spring is massless ( i think we can do that)
    so [itex] F_{elmass} = F_{el 2} [itex]
    [itex] k_1 x_1 = k_2 x_2 [/itex]

    i suppose [itex] x_1 + x_2 = A [/itex] when both x's are at maximum. ...
    so [itex] \frac {k_1}{k_1} = k_2 (A-x_1) [/itex]

    [itex] x_1 = \frac {A}{ \frac {k_1}{k_2} + 1 } [/itex]

    go back the the last equation we got in f=ma system mass and subtitute in the x_1 we just found....

    the A's cancel out and after we simply we get:

    [itex] \omega = \sqrt{ \frac{k_1 k_2}{ (k_1 + k_1) m }} [/itex] ////
    and to get F... just divide it by 2pi... right?
    is this even correct?
     
  2. jcsd
  3. Jun 23, 2015 #2
    try it with energy method you would be albe to solve it. if not reply i will post the solution.
     
  4. Jun 23, 2015 #3

    SammyS

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    Be careful about this in the homework section of PF !
     
  5. Jun 23, 2015 #4
    do you see a problem with my solution?
     
  6. Jun 23, 2015 #5

    SammyS

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    toesockshoe,

    I haven't examined your entire solution, but I'm pretty sure that your final answer IS correct !

    Two springs connected in that manner have an effective spring constant of ##\displaystyle\ k_\text{eff}=\frac{k_1\,k_2}{k_1+k_2}\ .##
     
    Last edited: Jun 23, 2015
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