Undergrad How to interpret complex solutions to simple harmonic oscillator?

[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.

 

[Orodruin]

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 $B = A^*$).

 

[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.

 

[hutchphd]

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 .

 

[vanhees71]

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 $x(0)=x_0 \in \mathbb{R}$, $\dot{x}(0)=v_0 \in \mathbb{R}$.