I am considering the equation for simple harmonic motion, which is $\ddot x +\omega ^2x=0$ To solve this, I have seen three approaches. This is confusing as I do not know which approach is physically correct or, if there is no correct approach, what is the physical significance of the three different approaches. My guess is that the first approach is slightly redundant as it is only using the well known result (of cosine and sin solutions of this form of a second order differential equation) which was derived from using methods of the full solution using $e^{i\phi}$, as in the the second and third approaches. However I do not know which of these is correct or why we can view it in two different ways.
1. Assume the sin and cosine results
Cosine and Sine are both solutions of the above equation, so the full solution is a linear combination of the two $x=A\cos(\omega t)+B\sin(\omega t)$ which is equivalent to $x=A\cos t(\omega t +\phi)$, giving the common form.
2. Solve more generally for a complex x.
I have seen the equation also be solved by solving the homogeneous differential equation for $x=Ae^{i\omega t}+Be^{-i\omega t}$. Then the constants are used to give the real solution, so the boundary condition (that $x$ is real) yields that $A=B^*$. So here we have complex constants, and we get the result $x=(A+B)\cos(\omega t)+i(A-B)\sin(\omega t)=2\Re(A)\cos(\omega t) -2\Im(A)\sin(\omega t)$ which, as before with constants multiplying sin and cos terms, reduces to $x=D\cos(\omega t +\phi)$
3. The final approach I have seen is by solving the differential equation for $z$ and just setting $x=\Re(z)$.
Solving the equation for z gives $x=Ae^{i\omega t}+Be^{-i\omega t}$ where I believe (if I got the method correctly), the constants are here REAL. As this still reduces to $z=(A+B)\cos(\omega t)+i(A-B)\sin(\omega t)$, but this time for real A and B and therefore $z$ is a complex number, you can see that this is equivalent to $z=De^{i\phi}e^{i\omega t}$ where the complex amplitude component $e^{i\phi}$ rotates the complex number $e^{i\omega t}$ in the complex plane such that the ratio of the constants of the cosine and sin terms of $z$ is correct and as given in the $z=(A+B)\cos(\omega t)+i(A-B)\sin(\omega t)$ form of $z$. This reduces to $z=e^{i(\omega t+\phi)}$ yielding $x=\Re(z)=D\cos(\omega t+\phi)$ as before.
As mentioned, I am not sure which is the 'correct' way to approach this, if any. But there must at least be a physical significance to tackling this in the different methods which I would be grateful if someone could explain.
Some of the thoughts I have had so far are:
- The first sin and cos solution is just a shortcut using the known result derived from the more formal solution using $e^{i\phi}$. However the sin/cos form ONLY applies when the variable in the differential equation is REAL. For example, it gives the correct solution for real $x$, but if had a complex $z$ in there which I knew was meant to be complex, then assuming the sin/cos form would be wrong and would not give the complete solution/picture- it would only give the real component of $z$.
- The difference between the second and third approaches seems to be that in one case we know we have a complex variable $z$. In that case, we actually find that the constants can be real (although I think they may also be not real. Provided that $A\neq B^*$, this still yields a complex $z$ as the complex parts do not cancel out). And in the second case we find that the constants must be real and that $A=B^*$ such that we get a real $x$, as mandated by the boundary conditions. So considering this I can't quite see how approaches 2 and 3 above would yield the same x form through these two different ways: using boundary conditions for a complex A and B so that complex parts cancel, or allowing z to be complex and thus $A\neq B^*$ and making x the real part of this complex solution.
Apologies for the long post. I hope I have made my confusion and thoughts sufficiently clear!