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Please help with Simple Harmonic motion!

  1. Feb 10, 2005 #1
    Ok the system which is doing the oscillations is described here

    http://scienceworld.wolfram.com/physics/SpringsTwoSpringsandaMass.html

    however, that differential equation over there gives the equation of motion. What i need is the differential equation for the position of the mass

    ALso given to me is the length of the left spring L1 and the right spring L2. Find th equation of motion

    [tex] m \frac{d^2 y}{dt^2} = -b\frac{dy}{dt}-(k_{1}+k_{2})y[/tex]

    now if this the equatio of moti0n then the equation of posotion would be given by the integral of this??
    so then i sub v = dy/dt and solve?? and assume a new function like [tex]e^{\lambda t} [/tex] for the function y??
     
    Last edited: Feb 10, 2005
  2. jcsd
  3. Feb 10, 2005 #2

    quasar987

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    The system described on the Wolfram page assumes no friction. The equation of motion you wrote has a friction term in it. Please copy/paste the question exactly how it is asked.
     
  4. Feb 10, 2005 #3
    Let k1 be the spring constant for the left spring L1 be its length
    Let k2 be the psirng constant for the right spring L2 be its length
    Let m be the mass
    Let b the damping coefficient
    Write a second order differential equation for the position at time t
    Thats how the question is worded
    Refer to the wolfram page for the diagram only
     
  5. Feb 10, 2005 #4

    dextercioby

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    Yes,then everything is okay. :smile:

    Daniel.
     
  6. Feb 10, 2005 #5
    what do you mean??????
     
  7. Feb 10, 2005 #6
    let [itex] y=Ae^{\lambda t}[/itex]
    just wanna remind you lambda and A are not real

    the real part of y is your solution....
    if you don't wanna deal with complex number...you can try [itex] Ae^{at}sinbt+Be^{at}cosbt [/itex]
     
  8. Feb 10, 2005 #7
    ok i cna solve it but for the lambda i'd get for [tex] y = Ae^{\lambda t} [/tex]

    [tex] \lambda = = \frac{-Ab \pm \sqrt{A^2 b^2 -4Am(k_{1}+k_{2})}}{2Am} [/tex]
    ok so i got the lambda nad all i solved the bloody thing

    but how does the lambda relate to the length of the springs and stuff??
    becuase the book gives an answer
    [tex] m \frac{d^2 y}{dt^2} + b \frac{dy}{dt} + (k_{1}+k_{2})y = k_{1}L_{1} - k_{2} L_{2}+k_{2} [/tex]

    how'd they get that?
     
  9. Feb 10, 2005 #8

    dextercioby

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    The last equation u posted is incorrect.You may wanna post the correct one.

    Daniel.
     
  10. Feb 10, 2005 #9
    which one is wrong?

    this one?? I ttyped it straight out from the textbook [tex] m \frac{d^2 y}{dt^2} + b \frac{dy}{dt} + (k_{1}+k_{2})y = k_{1}L_{1} - k_{2} L_{2}+k_{2} [/tex]

    or the other one in the last post
     
    Last edited: Feb 10, 2005
  11. Feb 10, 2005 #10

    dextercioby

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    Yes,that one.Do you see why it's wrong...?

    Daniel.
     
  12. Feb 10, 2005 #11
    fidning the value of lambda fives me the function y and y is the function of position correct?

    Is ther any initial condition i have to asume in thsi situation ??
     
    Last edited: Feb 10, 2005
  13. Feb 10, 2005 #12
    Your lambda is wrong (just a minor mistake, almost right)... try it again.
    if your lambda are distinct you will have 2 solution:
    [tex] y_1 = A_+e^{\lambda_+ t} [/tex]
    [tex] y_2 = A_-e^{\lambda_- t} [/tex]
    are your answer....
    and any linear combination of y1 and y2 is your answer,too... therefore, the general form of your solution is:
    [tex] y = A_1 y_1 + A_2 y_2 [/tex]

    edit: y,A,lambda are complex number.... for this type of question, complex number is a little bit easier to work with....

    if you want a real solution, use [itex] e^{ix} = cosx+isinx [/itex] and a real initial condition substitude into y, since you have 2 unknown in y, you need 2 equation for the initial condition: ie. y(0)=y_0, y'(0)=y'_0.....
     
    Last edited: Feb 10, 2005
  14. Feb 10, 2005 #13

    dextercioby

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    I was contesting the equation you posted and said to have copied from the book.Maybe it was a typo in the book.

    Daniel.
     
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