Seperation constant giving a harmonic dependence. (Seperation of variables)

Click For Summary
SUMMARY

The discussion centers on the separation constant in the context of harmonic time dependence in differential equations. The user deduced the equation \(\frac{d^{2}T}{dt^{2}}\frac{1}{T} = k^{2}\) and questioned the implications of choosing a positive versus a negative constant. It was established that a positive constant leads to exponential solutions, while a negative constant results in sinusoidal solutions. The consensus is that only the sinusoidal solutions, represented by \(A \cos(kt) + B \sin(kt)\), exhibit harmonic time dependence due to their periodic nature.

PREREQUISITES
  • Understanding of differential equations and separation of variables
  • Familiarity with harmonic functions and periodicity
  • Knowledge of exponential and trigonometric functions
  • Basic grasp of boundary value problems in physics or engineering
NEXT STEPS
  • Study the properties of harmonic functions in differential equations
  • Explore the implications of boundary conditions on solutions
  • Learn about the applications of Fourier series in solving differential equations
  • Investigate the role of eigenvalues in determining the nature of solutions
USEFUL FOR

Students and professionals in mathematics, physics, and engineering who are dealing with differential equations and seeking to understand the implications of separation constants on harmonic solutions.

xago
Messages
60
Reaction score
0

Homework Statement


http://img18.imageshack.us/img18/8970/bose.png

The Attempt at a Solution


I'm on part b) where it asks which separation constation gives a harmonic time dependence. From part a) I deduced the equation [itex]\frac{d^{2}T}{dt^{2}}[/itex][itex]\frac{1}{T}[/itex] = a constant. I'm choosing the constant [itex]k^{2}[/itex] and my question is does it matter if the constant is negative or positive? I have seen in textbooks that a positive constant gives the solution T(t) = Aexp(-kt) + Bexp(kt) whereas a negative one would be Acos(kt) + B sin(kt). Are both solutions equivalent or does only one of them give a harmonic time dependence (My guess would be the sin/cos one is the proper answer for this question.)
 
Last edited by a moderator:
Physics news on Phys.org
hi xago! :smile:
xago said:
… I have seen in textbooks that a positive constant gives the solution T(t) = Aexp(-kt) + Bexp(kt) whereas a negative one would be Acos(kt) + B sin(kt). Are both solutions equivalent or does only one of them give a harmonic time dependence (My guess would be the sin/cos one is the proper answer for this question.)

yes, you're right … harmonic has to be periodic, and only sin/cos will be periodic :smile:

(exp will be either runaway or decay :redface:)
 

Similar threads

Replies
16
Views
2K
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 5 ·
Replies
5
Views
10K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 5 ·
Replies
5
Views
1K