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Math Help - Physics and Complex numbers

  1. #1
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    Physics and Complex numbers

    Hey guys,

    I was looking through one of the past Advanced Physics exam papers and I came across a question which is a little confusing at the end (the answer that is):


    Consider a forced, damped simple harmonic oscillator, which obeys the equation of
    motion:
     m\frac{d^2x}{dt^2} = -kx-bv+Fcos(\Omega t)

    a) First consider the un-driven case ( F=0)
    Using the substitution
     x(t) = Ae^{\lambda t}
    where A is an arbitrary constant, together with de Moivre’s theorem

    e^{i \omega t}=cos(\omega t)+isin(\omega t)

    derive the expression giving the damped SHM:

     x(t)= Ae^{\frac{-bt}{2m}}cos(\omega t)

    Answer:
    And the answer is too long so I'll just post the last few lines. Basically you let x(t)= Ae^(lambda)t dx/dt = A(lambda)e^(lambda)t and
    d^2x/dt^2= A(lambda)^2 e^(lambda)t
    sub d^2x/dt^2 into original equation (given above).
    Divide by Ae^(lambda)t to elimintae on both sides.
    Take everything over to the left hand side to get a quadratic in (lambda)^2.
    Quadratic formula.
    You get:
    x(t) = A e ^(-bt/2m) e^ (+-i (omega) t) where omega = the discriminant of the quadratic= i((k/m - b^2/4m^2)^1/2)

    I apologise for the lack of LaTex but it would have taken like 30 minutes. The last 2 lines of the proof/answer are in LaTex:

    =  Ae^{\frac{-bt}{2m}}(cos(\omega t) + i sin (\omega t))
    (I know i have neglected the -iomega t case [as did the answers], but I believe this just gives cos(\omega t) - isin(\omega t))
    Now here is where I get confused.
    The next line is:
     x(t)=Ae^{\frac{-bt}{2m}}(cos(\omega t)

    Why/how did they eliminate the sin? Is it because we are dealing with a real oscillator and an expression for oscillation cannot have an imaginary part? Does this hold with any example (that is if your dealing with something physical in the real world and you use complex numbers to prove, can you just eliminate the imaginary part at the end of the proof)?

    Also if we end up eliminating the imaginary part in the end, why did we use complex numbers to prove in the first place? I am unsure why complex nmbers are used so rigorously to prove mathematical equations (especially if part of it is just deleted at the end of the proof).

    Much thanks for any responses.
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  2. #2
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    Quote Originally Posted by behemoth100 View Post
     m\frac{d^2x}{dt^2} = -kx-bv+Fcos(\Omega t)

    a) First consider the un-driven case ( F=0)
    The solution that you give: x(t) = Ae^{\frac{-bt}{2m}}\cos (\omega t) is wrong!

    Do you know how to solve this differencial equation?
    Since F=0 rewrite this equation as:
    m x'' + kx = - bv

    This is a Non-homogenous differencial equation.
    We begin by finding the charachteristic equation:
    r^2 + k =0 \Rightarrow r^2 = -k.
    Note that -k<0 since k the spring coefficient is positive.
    Thus, r = \pm i \sqrt{k}

    But that is still not the solution to the Non-homogenous differcinal equation, we need to find a particular solution which is x = -\frac{bv}{k}.

    So the general solution is given by:
    x(t) = C_1 \sin (\sqrt{k}) + C_2 \cos (\sqrt{k}) - \frac{bv}{k}
    -----
    This looks nothing what you posted. And I cannot help you out with their derivation because Mathematical Physics derivations' never make sense to me. They make you feel like an idiot.
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  3. #3
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    That is true, but it can also be said that for many equations (not all but some) there are is more than one way to prove it. However in the question it states that you MUST use the substitution x(t) =Ae^ (lambda t)
    and you MUST use de moivres theorem, and your resulting answer MUST be the answer THEY give, in terms of A, e, b, t, m, and omega, as the equation is used in later questions.

    Edit: Also its not the proof I am having trouble with really its the principles they employed in the end (i.e. eliminating isin (omega t)). I like to learn and understand the principles behind the maths, rather then just the method of doing a qs. Id prefer to know WHY you can get rid of isin(omega t), if you can do it in any and every case, why we used complex numbers (in this qs and in other mathematic proofs which any student will inevitably do in a maths lecture) to prove equation, rather than just remembering or being told that you CAN eliminate isin(omega t) (with no explanation as to why).
    P.s. I do know that complex numbers can be useful, I'm just wondering how they are useful in this case, and other cases involving the proof of mathematical equations that are physically applicable (e.g. period of a pendulum, or driven oscillation, Newtonian physics, etc)
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  4. #4
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    Quote Originally Posted by behemoth100 View Post
    That is true, but it can also be said that for many equations (not all but some) there are is more than one way to prove it. However in the question it states that you MUST use the substitution x(t) =Ae^ (lambda t)
    and you MUST use de moivres theorem, and your resulting answer MUST be the answer THEY give, in terms of A, e, b, t, m, and omega, as the equation is used in later questions.
    My point is that the solution that you posted for x(t) is totally wrong. Just substitute it into the equation and see for yourself. I am not saying that I have a different way of deriving this I am saying that what you did is wrong. Unless I am missing something here because of my limited physics skills.
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    The solution that you give: x(t) = Ae^{\frac{-bt}{2m}}\cos (\omega t) is wrong!

    Do you know how to solve this differencial equation?
    Since F=0 rewrite this equation as:
    m x'' + kx = - bv

    This is a Non-homogenous differencial equation.
    We begin by finding the charachteristic equation:
    r^2 + k =0 \Rightarrow r^2 = -k.
    Note that -k<0 since k the spring coefficient is positive.
    Thus, r = \pm i \sqrt{k}

    But that is still not the solution to the Non-homogenous differcinal equation, we need to find a particular solution which is x = -\frac{bv}{k}.

    So the general solution is given by:
    x(t) = C_1 \sin (\sqrt{k}) + C_2 \cos (\sqrt{k}) - \frac{bv}{k}
    -----
    This looks nothing what you posted. And I cannot help you out with their derivation because Mathematical Physics derivations' never make sense to me. They make you feel like an idiot.
    v=\frac{dx}{dt}

    RonL
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    Quote Originally Posted by ThePerfectHacker View Post
    My point is that the solution that you posted for x(t) is totally wrong. Just substitute it into the equation and see for yourself. I am not saying that I have a different way of deriving this I am saying that what you did is wrong. Unless I am missing something here because of my limited physics skills.
    Oh right I see what your saying. Actually this is not my working, this is the worked answer given for the End of Semester exam for Physics advanced 2003 paper.
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  7. #7
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    And can you believe this question was only worth 2 marks? Far out
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  8. #8
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    Quote Originally Posted by behemoth100 View Post
    Hey guys,

    I was looking through one of the past Advanced Physics exam papers and I came across a question which is a little confusing at the end (the answer that is):


    Consider a forced, damped simple harmonic oscillator, which obeys the equation of
    motion:
     m\frac{d^2x}{dt^2} = -kx-bv+Fcos(\Omega t)

    a) First consider the un-driven case ( F=0)
    Using the substitution
     x(t) = Ae^{\lambda t}
    where A is an arbitrary constant, together with de Moivre’s theorem

    e^{i \omega t}=cos(\omega t)+isin(\omega t)

    derive the expression giving the damped SHM:

     x(t)= Ae^{\frac{-bt}{2m}}cos(\omega t)

    Answer:
    And the answer is too long so I'll just post the last few lines. Basically you let x(t)= Ae^(lambda)t dx/dt = A(lambda)e^(lambda)t and
    d^2x/dt^2= A(lambda)^2 e^(lambda)t
    sub d^2x/dt^2 into original equation (given above).
    Divide by Ae^(lambda)t to elimintae on both sides.
    Take everything over to the left hand side to get a quadratic in (lambda)^2.
    Quadratic formula.
    You get:
    x(t) = A e ^(-bt/2m) e^ (+-i (omega) t) where omega = the discriminant of the quadratic= i((k/m - b^2/4m^2)^1/2)

    I apologise for the lack of LaTex but it would have taken like 30 minutes. The last 2 lines of the proof/answer are in LaTex:

    =  Ae^{\frac{-bt}{2m}}(cos(\omega t) + i sin (\omega t))
    (I know i have neglected the -iomega t case [as did the answers], but I believe this just gives cos(\omega t) - isin(\omega t))
    Now here is where I get confused.
    The next line is:
     x(t)=Ae^{\frac{-bt}{2m}}(cos(\omega t)

    Why/how did they eliminate the sin? Is it because we are dealing with a real oscillator and an expression for oscillation cannot have an imaginary part? Does this hold with any example (that is if your dealing with something physical in the real world and you use complex numbers to prove, can you just eliminate the imaginary part at the end of the proof)?

    Also if we end up eliminating the imaginary part in the end, why did we use complex numbers to prove in the first place? I am unsure why complex nmbers are used so rigorously to prove mathematical equations (especially if part of it is just deleted at the end of the proof).

    Much thanks for any responses.
    The general solution of the homogeneous equation is of the form:

    <br />
x=e^{-bt/2m}\left[Ae^{i\omega t}+Be^{-i \omega t}\right]<br />

    (that is the indical equation has two roots -bt/2m \pm i \omega t)

    Now if we impose real initial conditions on x and x' we find A=\bar{B} and x is always
    real, but what you have for the solution is:

    <br />
x=e^{-bt/2m}\left[a \cos(\omega t)+b \sin(\omega t)\right]<br />
.

    What you give for the general real solution for the homogeneous equation cannot
    be right as it has only one arbitary constant.

    RonL
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  9. #9
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    Quote Originally Posted by behemoth100 View Post
    And can you believe this question was only worth 2 marks? Far out
    the question itself is easy enough, its the worked solution that is crap.

    RonL
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  10. #10
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    Uhh ok you guys are getting me a little confused .
    I havent actually heard the term homogenous equation before (though looking it up on google, I realise what your basically saying) but there is no mention of general forms of HE, the lectures (remember its physics not maths) haven't mentioned it, so its safe to say (trust me on this) that the general formulas you guys are talking about are not need nor used.

    I do greatly appreciate the sppedy responses though .

    If I may, I have posted the link to the worked answers for that question in the 2003 exam. It is question 9 a)
    http://www.physics.usyd.edu.au/ugrad...tions_2003.pdf
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  11. #11
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    Quote Originally Posted by CaptainBlack View Post
    the question itself is easy enough, its the worked solution that is crap.

    RonL
    I agree totally, I got down to the end easily enough, just got stuck with the sudden deletion of the sin.

    NOTE THAT FOR THE LINK I PROVIDED THERE IS A TYPO:
    Second line of the equation they have -b(lambda)e ^(lambda t) where as there should be an A there. However their eqns do take this into account, they just skipped the A key when typing it up, as can be seen on the next line once they have divided through by Ae^(lambda t).

    Also remember that this is the first semester of first year undergraduate physics .
    Last edited by CaptainBlack; June 10th 2007 at 09:09 PM.
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  12. #12
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    Quote Originally Posted by behemoth100 View Post
    Uhh ok you guys are getting me a little confused .
    I havent actually heard the term homogenous equation before (though looking it up on google, I realise what your basically saying) but there is no mention of general forms of HE, the lectures (remember its physics not maths) haven't mentioned it, so its safe to say (trust me on this) that the general formulas you guys are talking about are not need nor used.

    I do greatly appreciate the sppedy responses though .

    If I may, I have posted the link to the worked answers for that question in the 2003 exam. It is question 9 a)
    http://www.physics.usyd.edu.au/ugrad...tions_2003.pdf
    The model answer is wrong, it takes the step:

    x(t) = A e^{-bt/2m}e^{\pm i \omega t}

    ....... = A e^{-bt/2m}\left(\cos(\omega t) + i \sin(\omega t) \right)<br />

    which is invalid.

    In fact even the first equality above is wrong it should be:

    x(t) =  e^{-bt/2m}\left(A e^{+ i \omega t}+B e^{- i \omega t}\right)

    RonL
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  13. #13
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    Quote Originally Posted by behemoth100 View Post
    I agree totally, I got down to the end easily enough, just got stuck with the sudden deletion of the sin.

    NOTE THAT FOR THE LINK I PROVIDED THERE IS A TYPO:
    Second line of the equation they have -b(lambda)e ^(lambda t) where as there should be an A there. However their eqns do take this into account, they just skipped the A key when typing it up, as can be seen on the next line once they have divided through by Ae^(lambda t).

    Also remember that this is the first semester of first year undergraduate physics .
    But it still involves the solution of second order linear constant coefficient ordinary
    differential equations, which are essential for an awfull lot of Physics, and the model
    answer should still give the correct solution.

    Suppose the initial conditions are x(0)=0,\ x'(0)=1 the first of these forces
    A=0, and so the second cannot be satisfied by the model solution.

    RonL
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  14. #14
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    Quote Originally Posted by CaptainBlack View Post
    The model answer is wrong, it takes the step:

    x(t) = A e^{-bt/2m}e^{\pm i \omega t}

    ....... = A e^{-bt/2m}\left(\cos(\omega t) + i \sin(\omega t) \right)<br />

    RonL
    May I ask why this is wrong?
    By euler's formula:
    e^{i\theta}= cos(\theta) + isin (\theta)
    Let  \theta = \omega t
    Then:
     e^{i\omega t} = cos (\omega t) + i sin (\omega t)
    for e^{-i\omega t}
    we rewrite  e^{i(-\omega) t}
     e^{-i\omega t}= cos(-\omega t) + i sin (-\omega t)
    e^{-i\omega t}= cos (\omega t) - isin (\omega t)
    as  cos(-\theta)=cos(\theta)
    and  sin(-\theta)= - sin(\theta)
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  15. #15
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    I should also add that the general formula for damped oscillation IS given by
     x(t)=Ae^{-bt/2m}cos(\omega t) in physics textbooks, so the answer should be correct.
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