How to solve 2nd order diff. equation for simple harmonic motion

Click For Summary

Discussion Overview

The discussion revolves around solving second-order differential equations related to simple harmonic motion (SHM), particularly in the context of LC and LRC circuits. Participants explore the methods of finding solutions to these equations and the reasoning behind the proposed solutions.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about how the solution x=Acos(ωt+φ) satisfies the differential equation m(d²x/dt²)+(k/m)x=0, questioning the method used in their textbook.
  • Another participant suggests that one should manipulate the assumed solution to verify if it satisfies the ordinary differential equation (ODE) and mentions the use of initial conditions to determine unknown constants.
  • A different viewpoint emphasizes that solving differential equations is an art that requires practice, highlighting the structure of linear differential equations with constant coefficients as particularly nice for this context.
  • This participant provides a specific solution approach using an exponential ansatz and discusses the resulting solutions and their linear combinations.
  • Another participant challenges the notion that the book's author was merely guessing, suggesting that the book likely demonstrates why certain functions do not satisfy the ODE.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the methods of solving the differential equation or the characterization of the textbook's approach, indicating multiple competing views and unresolved questions regarding the solution process.

Contextual Notes

Participants mention the need for initial conditions and the structure of linear differential equations, but there are unresolved assumptions about the methods of solution and the validity of different approaches.

cjurban
Messages
7
Reaction score
0
In my physics class we're talking about LC and LRC circuits, and the equations are analogous to those for SHM. However, I don't see how x=Acos(ωt+[itex]\varphi[/itex]) satisfies m(d^2x/dt^2)+(k/m)x=0. I've never done differential equations and in the book it seemed like the author just guessed and checked until he found the right solution, and this doesn't seem like a satisfactory answer. How did he solve this equation?
 
Physics news on Phys.org
Take the assumed solution, do the requisite manipulation, and see if it satisfies the ODE. Any unknown constants can be determined using initial conditions. You have taken derivatives of trig functions, haven't you?
 
Well, solving differential equations is an art. You just have to practice it to find solutions. There is of course a lot of work done by the mathematicians to provide general theorems about their structure and that of their solutions.

In your case of circuit theory or the harmonic oscillator you have a particularly nice class of differential equations, namely a linear differential equation. Here, it's even the kind with constant coefficients. For the undamped harmonic oscillator it reads
[tex]\ddot{x}+\omega^2 x=0.[/tex]

Here the standard ansatz to find a solution is
[tex]x(t)=A \exp(\lambda t).[/tex]
Plugging this into the equation you find
[tex]\lambda^2+\omega^2=0.[/tex]
There are two solutions for [itex]\lambda[/itex], namely
[tex]\lambda_{1/2}=\pm \mathrm{i} \omega.[/tex]
Now the mathematicians have proven that any solution is given as the superposition of two linearly independent solutions, and these we just have found! So the general solution is
[tex]x(t)=A_1 \exp(\mathrm{i} \omega t) + A_2 \exp(-\mathrm{i} \omega t).[/tex]
The constants [itex]A_1[/itex] and [itex]A_2[/itex] are determined by giving initial values, e.g., position and velocity at [itex]t=0[/itex].
 
I have a book that does something similar and I'm not sure the book was guessing and checking. I think they were showing why other functions don't satisfy the given ODE.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 11 ·
Replies
11
Views
16K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
6K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
8K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 51 ·
2
Replies
51
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K