Yes I set phi = 0 for simplicity.
Well thanks for all the help on this problem I thought it was a bit dodgy .
But just to bottom line it, lets forget the steps up to the 2nd last line. If we pretend that the rest of the qs is correct (even though it isn't) if you have a problem such as
$\displaystyle x(t)= Ae^{\frac{-bt}{2m}}(\cos(\omega t)+i \sin(\omega t))$
Are we allowed to eliminate the imaginary part if we are assuming its applicable in real physical situations? For example its not possible for a dampened oscillator to have 'imaginary dampening' or an 'imaginary restoring force'. Or are
$\displaystyle x(t)= Ae^{\frac{-bt}{2m}}(\cos(\omega t)+i \sin(\omega t))$
and
$\displaystyle x(t)= Ae^{\frac{-bt}{2m}}\cos(\omega t)$
completely unrelated? I just need to know for if we are asked a question that requires complex numbers. Also I am quite interested to find out what mathematicians can do to a complex number. For example does applying it to the 'real world' automatically make the assumption that the imaginary part of all complex numbers is equal to zero? If your not to busy: If my previous statement is incorrect could you give me an example where imaginary numbers can exist in real (world), physical applications?
Thanks alot!
What is happening when you just take the real part of the complex function
is that you lose the phase information. This is often OK in signal proccesing,
but will get you into difficulties eventualy if you do habitualy.
In the case like yours you get the real signal (a mixture of sin's and cos's,
or with a phase term like phi in either the cos or sin) by applying real initial
conditions to the complex general solution.
No, everything that we measure is real, but that does not mean that we are just taking the real partcompletely unrelated? I just need to know for if we are asked a question that requires complex numbers. Also I am quite interested to find out what mathematicians can do to a complex number. For example does applying it to the 'real world' automatically make the assumption that the imaginary part of all complex numbers is equal to zero? If your not to busy: If my previous statement is incorrect could you give me an example where imaginary numbers can exist in real (world), physical applications?
Thanks alot!
of a complex entity. When you come to do Qantum Mechanics you will find that you are dealing with
complex wave functions all of the time, but experimental observations are always real numbers, but
not simply the real part of the wave function. How they are related to the wave functions depends on
what the measurement is.
RonL
RonL
I've taken a look at these posts and it's too bad I couldn't have cut in on the conversation sooner.
The whole problem here is due to the solution method and a little detail that never got mentioned in the first place.
The equation for forced, damped harmonic motion is
$\displaystyle m \frac{d^2x}{dt^2} + c \frac{dx}{dt} + kx = Fcos( \omega t)$
(This is obviously not the most general equation, but it is the one that the original poster started with. For the record, this kind of driving force produces resonance.)
What the problem did to make the solution "easier" was to introduce
$\displaystyle Fcos( \omega t) = Re (Fe^{i \omega t})$
The solution method should carry through the Re operator but I don't think anyone does it because it "gunks up" the solution. This produces the a solution that contains a complex part. But since we are only interested in the "real" nature of the driving force, we take only the real part of the solution. If you do keep the Re operator then the solution comes out without any confusion.
(This problem is probably part of the reason why TPH seems to dislike Mathematical Physics so much: notation gets dropped for a while out of convenience, then put back in when its needed.)
-Dan
The major reason is how they manipulate equations. They just say "make this substitution and then take partial derivaties ....." It in the end it produces the answer. But when you want to "make a substitution and take partial derivatives ....." on your own problem it almost never works. Because they overlook serious mathematical necessities, and make it look so simple. In reality it is much more complicated then the way they do it. So in the end it is not about skill, it is about memorization. Memorization of what "make this substitution .... " techinques actually work and which do not.