- #1
quozzy
- 15
- 0
So, in lectures we derived the equation for damped SHM by solving the differential equation relating position (x), mass (m), spring constant (s), and damping coefficient (r):
[tex]m\ddot{x}=-\frac{s}{m}x-r\dot{x}[/tex]
Using a solution of the form [tex]Ae^{\alpha t}[/tex], we find that:
[tex]x=Ae^{-pt}e^{\pm qt}[/tex],
where [tex]p=\frac{r}{2m}[/tex], and [tex]q=\sqrt{p^{2}-\frac{s}{m}}[/tex].
Everything until and including this I understand. However, the final step, with no explanation, turns the solution into the following:
[tex]e^{-pt}(C_{1}e^{qt}+C_{2}e^{-qt})[/tex],
where C1 and C2 are some arbitrary constants. (i.e. a linear combination of the two distinct solutions.) In trying to research this online, I found an article that mentions the solution holds true for all complex values of C1 and C2. I don't understand how, algebraically, you can go from the previous step to the last one. Somebody help me out?
Thanks in advance.
P.S. I don't know why the closing bracket doesn't show up in the last equation, but it should be there. (EDIT: Nevermind, it works now.)
[tex]m\ddot{x}=-\frac{s}{m}x-r\dot{x}[/tex]
Using a solution of the form [tex]Ae^{\alpha t}[/tex], we find that:
[tex]x=Ae^{-pt}e^{\pm qt}[/tex],
where [tex]p=\frac{r}{2m}[/tex], and [tex]q=\sqrt{p^{2}-\frac{s}{m}}[/tex].
Everything until and including this I understand. However, the final step, with no explanation, turns the solution into the following:
[tex]e^{-pt}(C_{1}e^{qt}+C_{2}e^{-qt})[/tex],
where C1 and C2 are some arbitrary constants. (i.e. a linear combination of the two distinct solutions.) In trying to research this online, I found an article that mentions the solution holds true for all complex values of C1 and C2. I don't understand how, algebraically, you can go from the previous step to the last one. Somebody help me out?
Thanks in advance.
P.S. I don't know why the closing bracket doesn't show up in the last equation, but it should be there. (EDIT: Nevermind, it works now.)
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