Physics and Complex numbers

**behemoth100** · June 10th 2007, 07:26 PM

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.

**ThePerfectHacker** · June 10th 2007, 07:40 PM

Originally Posted by behemoth100

$m\frac{d^2x}{dt^2} = -kx-bv+Fcos(\Omega t)$

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.

**behemoth100** · June 10th 2007, 07:56 PM

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)

**ThePerfectHacker** · June 10th 2007, 07:59 PM

Originally Posted by behemoth100

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.

**CaptainBlack** · June 10th 2007, 08:02 PM

Originally Posted by ThePerfectHacker

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

**behemoth100** · June 10th 2007, 08:03 PM

Originally Posted by ThePerfectHacker

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.

**behemoth100** · June 10th 2007, 08:06 PM

And can you believe this question was only worth 2 marks? Far out

**CaptainBlack** · June 10th 2007, 08:15 PM

Originally Posted by behemoth100

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:

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

(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:

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

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

RonL

**CaptainBlack** · June 10th 2007, 08:19 PM

Originally Posted by behemoth100

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

**behemoth100** · June 10th 2007, 08:23 PM

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

**behemoth100** · June 10th 2007, 08:26 PM

Originally Posted by CaptainBlack

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

.

**CaptainBlack** · June 10th 2007, 09:01 PM

Originally Posted by behemoth100

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) $

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

**CaptainBlack** · June 10th 2007, 09:09 PM

Originally Posted by behemoth100

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

**behemoth100** · June 10th 2007, 09:27 PM

Originally Posted by CaptainBlack

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) $

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 $sin(-\theta)= - sin(\theta)$

**behemoth100** · June 10th 2007, 09:35 PM

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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