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Equation of motion for the system, and determine the frequen

  1. Oct 1, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-10-1_2-29-12.png
    A small block of mass m is suspended from the top of a box by a mass-less string of length L. Two identical springs, each with spring constant k, extend from the block to the sides of the box, as shown in the diagram to the right. The length of the springs is such that they are not stretched when the block is in its equilibrium position.Find the equation of motion for the system, and determine the frequency of small oscillations. Briefly explain your reasoning.
    Hint: Think about what the restoring force on the block would be, if springs were not there. For small x, this force is approximately proportional to x. You should make this small-x approximation, before you add the force due to the springs.

    2. Relevant equations
    F= -mgx/l
    F=-kx? or F= 1/2 kx^2?
    3. The attempt at a solution
    so far, I got
    Fnet = -kx+ (-kx) -(mgx/l)

    mX= -x (kx+mg/l)
    ω= √( (kx+ mg/l) /m)
    f= ω/2 π
    = √( (kx+ mg/l) /4π^2 m)

    but they seems very wrong...
     
  2. jcsd
  3. Oct 1, 2016 #2

    Simon Bridge

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    What makes you think this "seems very wrong"?

    Please show your reasoning along with your working.
    ie: what does this mean: "mX= -x (kx+mg/l)" and where did it come from?
     
  4. Oct 1, 2016 #3

    haruspex

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    Good so far.
    Not sure what you have done here. I'm guessing X stands for ##\ddot x##, but you had two kx terms and now you have one kx2 term instead.
     
  5. Oct 1, 2016 #4
    yes, that X means ##\ddot x##
    so F= m##\ddot x##= -x (k+mg/l)?

    f= ω/2 π
    = √( (k+ mg/l) /4π^2 m)
    is it all done? I can't go any furhter after this
     
    Last edited: Oct 1, 2016
  6. Oct 1, 2016 #5
    Sorry
    it should be
    F= m##\ddot x##= -x (2k+mg/l)?

    f= ω/2 π
    = √( (k+ mg/l) /2π^2 m)
     
  7. Oct 1, 2016 #6

    haruspex

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    You are still making algebraic errors. 2k+mg/l cannot easily turn into k+mg/l.
     
  8. Oct 1, 2016 #7
    because I put 2π into the "√" so it turn to √(4π^2)
    then √( (2k+ mg/l) /4π^2 m) => √( (k+ mg/l) /2π^2 m) isn't it?
     
  9. Oct 1, 2016 #8

    Simon Bridge

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    You have been asked to do this:
    ... you are having difficulty with algebra and the bit in bold.

    The way to minimize algebra errors is to do the algebra one step at a time... and include little notes about why you did each step. That is what "Breifly explain your reasoning" means (in this context).

    So lets see: that ##m\ddot x = -(2k + mg/l)x## you got, is that the equation of motion you were asked for? If so then say so.

    How do you get from there to your expression for ##\omega =##?? What is your reasoning?

    Having got an expression for ##\omega## you proceeded to ##f=\omega /2\pi## and explained your final expression thus:
    No it isn't. Do it one step at a time...
    ... next step is either to separate the numerator so you have two fractions or shift the "2" outside the parentheses in the numerator, or recognize that the numerator is itself a fraction and put that fraction over a common denominator... there's a lot of choice but you need to pick one, like this:
    -- separating the fraction:
    $$\sqrt{\frac{2k+mg/l}{4\pi^2m}} = \sqrt{ \frac{2k}{4\pi^2m}+\frac{mg/l}{4\pi^2m}} = \cdots$$
    -- putting the "2" outside the parentheses:
    $$\sqrt{\frac{2k+mg/l}{4\pi^2m}} = \sqrt{ \frac{2(k+mg/2l)}{4\pi^2m}} = \cdots$$
    -- common denominator
    $$\sqrt{\frac{2k+mg/l}{4\pi^2m}} =\sqrt{\frac{2kl+mg}{l}\frac{1}{4\pi^2m}}=\cdots$$

    ... pick one and finish the algebra.
    Don't forget to show your reasoning as you go.
     
  10. Oct 2, 2016 #9
    oh yes! I have to factor that "2" out first
    thank you!!
     
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