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Spring constant and oscillation expression? Help.

  1. Jan 17, 2012 #1
    1. The problem statement, all variables and given/known data

    Here is the question:

    sevtpe.png
    2. Relevant equations



    3. The attempt at a solution

    I know that SHM is: accel = -(constant) (displacement)

    Linear from my book says: Ax = Ftotal/m (dont quite get this)

    Any help? THanks!
     
  2. jcsd
  3. Jan 17, 2012 #2
    I would say they are identical for a given displacement.
    F = 2kX in each case
     
  4. Jan 17, 2012 #3
    Can you show me how?

    I need to be able to write an expression for the net force on block 1, and use it to derive an expression for the period of oscilation, in terms of k and m
     
  5. Jan 17, 2012 #4
    In each case I think the force on the block is 2kx (x is the displacement)
    I would use this to write an expression for the acceleration (= F/m)
    then acc =ω^2x to get an expression for ω^2 then get the time period expression.
    get back to me if you cannot do this
     
  6. Jan 17, 2012 #5
    Having issues with this!

    Ok FIRST for a) Write an expression for the net force on block 1 - Lets start with this ok?

    Would it not be just: 2kx --- is that correct? Its because there is 2 spring constants.

    ---->NOW if that is right, lets move onto " Use it to derive an expression for the period of oscillation, in terms of k and m"

    I dont know how to get this...?
     
  7. Jan 17, 2012 #6
    So if the force = 2kx then acceleration = F/m = 2kx/m
    Acceleration = ω^2 x
    so ω^2 = 2k/m
    Can you get an expression for frequency or time period from this?
    ask if you are not sure..... but you know the 'rules' of this place... I prefer not to give you the answer directly..... you are almost there.
     
  8. Jan 17, 2012 #7
    K wait... acceleration is 2kx/m

    Where did "Acceleration = w^2x, so w^2 = 2k/m come from? Can you explain this to me?
     
  9. Jan 17, 2012 #8
    Umm: Period = 2pi sqrt(m/2k)

    Is that what you mean?

    and the frequency would be 1/Period
     
  10. Jan 17, 2012 #9
    No problem.... be patient
    SHM is when F is proportional to displacement,x,
    that means F = kx.... should be F = -kx because displacement is a vector but forget that for now
    This means that F = ma = kx so a = kx/m... looks better like this ... (k/m)x
    By analysing SHM and its links with circular motion you get a = ω^2 x, so ω^2 = k/m
    In this problem, because there are 2 springs k = 2k
    so ω^2 = 2k/m.... are you OK now?
    I will explain in more detail if you do not get it !!!!! Don't panic .... it is tricky
     
  11. Jan 17, 2012 #10
    you have got it:approve:
    well done
     
  12. Jan 17, 2012 #11
    Oh cool. So,

    Umm: Period = 2pi sqrt(m/2k)

    Is that what you mean?

    and the frequency would be 1/Period


    This answers a) completly?
     
  13. Jan 17, 2012 #12
    I think so
     
  14. Jan 17, 2012 #13
    it would be good to have confirmation from another contribution but I am happy with my answer
     
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