How to interpret complex solutions to simple harmonic oscillator?

  • #1
schniefen
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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.
 

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  • #2
Orodruin
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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^*##).
 
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  • #3
PhDeezNutz
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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.
 
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  • #4
hutchphd
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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 .
 
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  • #5
vanhees71
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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}##.
 
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