Function for (damped) SHM as a linear combination of two exponentials

Click For Summary
SUMMARY

The discussion centers on the derivation of the damped simple harmonic motion (SHM) equation from the differential equation m\ddot{x}=-\frac{s}{m}x-r\dot{x}. The solution is expressed as x=Ae^{-pt}(C_{1}e^{qt}+C_{2}e^{-qt}), where p=\frac{r}{2m} and q=\sqrt{p^{2}-\frac{s}{m}}. Participants clarify that the transition to the final form involves recognizing that any linear combination of distinct solutions to a linear equation remains a valid solution. This principle applies universally to linear equations, not just differential ones.

PREREQUISITES
  • Understanding of differential equations, specifically second-order linear differential equations.
  • Familiarity with concepts of simple harmonic motion (SHM) and damping.
  • Knowledge of exponential functions and their properties in mathematical solutions.
  • Basic grasp of linear combinations in the context of linear algebra.
NEXT STEPS
  • Study the derivation of solutions for second-order linear differential equations.
  • Explore the implications of damping in simple harmonic motion through practical examples.
  • Learn about the role of arbitrary constants in the general solutions of differential equations.
  • Investigate the properties of linear combinations of solutions in linear algebra.
USEFUL FOR

Students of physics and mathematics, particularly those studying mechanics and differential equations, as well as educators seeking to clarify concepts of damped SHM and linear combinations of solutions.

quozzy
Messages
15
Reaction score
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.)
 
Last edited:
Physics news on Phys.org
It is true that if f(x) and g(x) are distinct solutions of a linear equation then their sum is also a solution as is any arbitrary linear combination of them.

Note this is true of any linear equation, not only differential ones.

In your penultimate line you have two distinct functions obtained by taking either the = or the - sign in

[tex]{e^{ \pm qx}}[/tex]

Does this help?

You also thought that there was a problem whilst composing and using the preview. This is a problem that PF has with Latex where the preview does not rflect changes. You can cure it by refreshing the whole page or quitting the thread and reacquiring it.

go well
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 131 ·
5
Replies
131
Views
9K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 8 ·
Replies
8
Views
17K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 14 ·
Replies
14
Views
3K
Replies
3
Views
1K