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Laplace Transforms

  1. May 3, 2015 #1
    1. The problem statement, all variables and given/known data
    tFewRWs.png

    2. Relevant equations
    Laplace Transforms

    3. The attempt at a solution

    Using basic physics knowledge I got
    m1a1=-k1x1+k2(x2-x1)
    and
    m2a2=-k3x2-k2(x2-x1)

    Sub in values and use laplace transforms and rearrange partial fraction and I found that

    By doing this I am assuming that the xII and xIII will equal 0 when t=0 because this is not stated at all in the question. do you believe this is correct?


    x1=-(1/3)cos2t-(2/3)sin2t+(4/3)cost+(8/3)sint
    x2=-(2/3)cos2t+(1/3)sin2t+(8/3)cost-(4/3)sint

    Finding the initial displacements by subbing in t=0 for both x_1 and x_2 comes out with what is written in the question 1 and 2, respectively.

    However, when I try to sub t=0 into the differentials of the 2 above equations. I believe I should receive the initial velocities stated in the question. however I do not receive these results.

    I receive 4/3 for x1 and -2/3 for x2
    the difference between these values and the actual values appears to differ once I differentiate the sin2t and it is multiplied by 2

    Does anyone know if I should receive the values listed in the question using this methods and have just made a calculation error in my working earlier on, or should I have done something differently.

    Also, the next part of the questions asks to use a matrix and eigenvalues/eigenvectors to solve it. any pointers to help me get started

    Thanks very much
     
  2. jcsd
  3. May 3, 2015 #2
    It doesn't seem like you used Laplace Transforms to solve this. What is the Laplace Transform of x1''(t) in terms of s, x1(0), and x1'(0)?

    Chet
     
  4. May 3, 2015 #3
    I got (s3+2s2+5s+10)/((s2+4)(s2+1))

    I found this by rearranging my equation earlier to find x2 and xII2 in terms of x1

    This came out as xIV1+5xII1+4x=0, then used laplace transform to find what I had above.

    Does this seem correct?
     
  5. May 3, 2015 #4
    I then split the equation into partial fractions and it worked out as the sin and cos equations earlier after I used the inverse laplace
     
  6. May 3, 2015 #5
    Try it by first taking the Laplace Transform of each equation, and then doing the algebra.

    Chet
     
  7. May 3, 2015 #6
    ok, ill post how I go soon
     
  8. May 3, 2015 #7
    ok, I just did it for x2 quickly and got (2s3-s2+7s-1)/((s2+4)(s2+1))
    which resulted in (1/3)cos2t-sin2t+(5/3)cost

    this also gives me the displacement as 2 which is correct, however the velocity is -2 instead of -1

    ill check with x1 now
     
  9. May 3, 2015 #8
    again with x1, displacement came out as the same (1) but velocity was doubled at 4 instead of 2.
    I've noticed the displacement always equals what is infront of the s3. looks like this is because the x(0)= whatever is the only factor to affect the s3 value causing the displacement to be equal each time
     
  10. May 3, 2015 #9
    It seems to me that this is just a matter of getting the algebra correct, and then properly inverting the transforms.

    Chet
     
  11. May 3, 2015 #10

    vela

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    It would help if you at least showed the initial equations you get from taking the Laplace transform. For all we know, you made an error there and propagated it the rest of the way.
     
  12. May 3, 2015 #11
    I started with x1II+3x1-2x2=0
    and x2II+2x2-x1=0
     
  13. May 3, 2015 #12
    I rearranged those to get x2IV+5x2''+4x2=0

    from laplace I got this for x2
    s4X2-s3x2(0)-s2x2'(0)-sx2''(0)-x2'''+5(s2X2-sx2(0)-x2'(0))+4X2

    From there I subbed in x2(0)=2 and x2'(0)=-1, and assumed that the second and third derivatives equal 0

    Then simplified that to X2= (2s3-s2+10s-5)/((s2+4)(s2+1))

    Used partial fractions and inverse laplace to get what I had for x2 in the original question
     
  14. May 3, 2015 #13
    Good. Now let's see the laplace transform of each equation.

    Chet
     
  15. May 3, 2015 #14
    Then doing the laplace before algebra I got
    s2X1-s-2+3X1-2X2=0
    s2X2-2s+1+2X2-X1=0

    these included the inclusion of the initial x and x' values

    then I simplified both equations to find X2 which I found as
    (2s3-s2+7s-1)/((s2+4)*(s2+1))
     
  16. May 3, 2015 #15
    that gave me x2=(1/3)cos2t-sin2t+(5/3)cost
     
  17. May 3, 2015 #16

    vela

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    I got a factor of 1/2 on the sin 2t term.
     
  18. May 3, 2015 #17
    I got a factor of 1/3 on the sin 2t term.

    Chet
     
  19. May 3, 2015 #18
    Did you get similar factors for the other terms?
    Do you know where our work differs to result in this?
     
  20. May 3, 2015 #19

    vela

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    I cranked it out in Mathematica. There was a factor of 1/3 on the cosine term, but on the sine term, it was 1/2.

    I got the same ##X_2(s)## as you did, so the problem is in your inversion.
     
  21. May 3, 2015 #20
    From here
    (2s3-s2+7s-1)/((s2+4)*(s2+1))

    I used partial fraction expransions and found that 2s3-s2+7s-1=(A+C)s3+(B+D)s2+(A+4C)s+(B+4D)

    that gave me A+C=2, B+D=-1, A+4C=7 and B+4D=-1

    therefore A=(1/3), B=-1, C=(5/3) and D=0
     
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