I How to interpret complex solutions to simple harmonic oscillator?

AI Thread Summary
The discussion focuses on the equation of motion for a simple harmonic oscillator, highlighting that solutions can be expressed in complex form. It emphasizes that both the real and imaginary parts of the solution are valid and correspond to the differential equation, with real initial conditions leading to specific constraints on the complex coefficients. The relationship between the constants A and B is crucial, as they can be complex but must satisfy the condition B = A*. Ultimately, the solutions must align with physical boundary conditions, ensuring that the final results remain real. This approach allows for a comprehensive understanding of the harmonic oscillator's behavior.
schniefen
Messages
177
Reaction score
4
Consider the equation of motion for a simple harmonic oscillator:
##m\ddot {x}(t)=-kx(t).##​
The solutions are
##x(t)=Ae^{i\omega t}+Be^{-i\omega t},##​
where ##\omega=\sqrt{\frac{k}{m}}##, and constants ##A## and ##B##. Physically, what does it mean for a solution to be complex? Is it only the real part that is of interest or also the imaginary part? Of course, using Euler's formula, the solution can be rewritten as
##(A+B)\cos{(\omega t)}+i(A-B)\sin{(\omega t)},##​
and one can introduce new, complex constants. However, there is still the issue of a real and imaginary part with that representation.
 
Physics news on Phys.org
Both the real and imaginary parts are solutions to the differential equation by construction. If you put in real initial values, your determined constants ##A## and ##B## will result in a real solution (i.e., their sum will be real and their difference imaginary - in other words ##B = A^*##).
 
  • Like
Likes schniefen, vanhees71, PeroK and 1 other person
You can form linear combinations of your basis solutions to get real values solutions in terms of sine and cosine. That can form your new basis if you wish.

Edit: and your constants A and B can be complex in order to make this happen.
 
  • Like
Likes schniefen and vanhees71
Of course at some point your solution to a physical problem must be matched to boundary conditions in space and time. That will constrain the actual result .
 
  • Like
Likes schniefen, vanhees71, VVS2000 and 1 other person
Another way to think about it is to say that you only look for solutions with ##x(t) \in \mathbb{R}##. That constrains the "allowed values" for the complex coefficients of you solution to ##B=A^*##. You get still the complete solutions for the real differential equation, because the complex coefficient ##A=A_r + \mathrm{i} A_i## consists of the two real numbers ##A_r## and ##A_i##, which can be used to satisfy the real (!) initial conditions ##x(0)=x_0 \in \mathbb{R} ##, ##\dot{x}(0)=v_0 \in \mathbb{R}##.
 
  • Like
Likes schniefen and PhDeezNutz
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
This has been discussed many times on PF, and will likely come up again, so the video might come handy. Previous threads: https://www.physicsforums.com/threads/is-a-treadmill-incline-just-a-marketing-gimmick.937725/ https://www.physicsforums.com/threads/work-done-running-on-an-inclined-treadmill.927825/ https://www.physicsforums.com/threads/how-do-we-calculate-the-energy-we-used-to-do-something.1052162/
Back
Top