# How to interpret complex solutions to simple harmonic oscillator?

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schniefen
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.

Staff Emeritus
Homework Helper
Gold Member
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^*##).

• schniefen, vanhees71, PeroK and 1 other person
PhDeezNutz
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.

• schniefen and vanhees71
Homework Helper
2022 Award
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 .

• schniefen, vanhees71, VVS2000 and 1 other person
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