How Does Hooke's Law Relate to Angular Frequency?

  • Context: Undergrad 
  • Thread starter Thread starter misogynisticfeminist
  • Start date Start date
  • Tags Tags
    Angular Hooke's law Law
Click For Summary

Discussion Overview

The discussion revolves around the relationship between Hooke's Law and angular frequency, particularly in the context of the differential equation governing simple harmonic motion. Participants explore how to derive the angular frequency from the equation and the implications of substituting variables in the solution.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant references a video that states the solution to the differential equation \( x'' + \frac{k}{m} x = 0 \) is \( x = x_0 \cos(\omega t + \phi) \) if and only if \( \omega = \sqrt{\frac{k}{m}} \).
  • Another participant suggests differentiating the solution twice and substituting it back into the differential equation to verify the relationship.
  • A participant expresses doubt about how to derive \( \omega = \sqrt{\frac{k}{m}} \) from the original differential equation without prior knowledge of the solution.
  • Another response indicates that substituting \( x = x_0 \cos(\omega t - \phi) \) into the differential equation will reveal the required value of \( \omega \).
  • One participant clarifies that \( \omega \) refers to the angular frequency and suggests it is a standard substitution that may not have been explicitly defined.
  • A later reply proposes considering \( \omega^2 \) as a mathematically sound representation of the positivity of \( \frac{k}{m} \), implying that this choice is both useful and meaningful.

Areas of Agreement / Disagreement

Participants express varying levels of understanding regarding the derivation of \( \omega \) from the differential equation. While some suggest methods to verify the solution, others remain uncertain about the initial assumptions and substitutions needed to arrive at \( \omega = \sqrt{\frac{k}{m}} \.

Contextual Notes

There is an ongoing discussion about the assumptions made in the derivation process, particularly regarding the introduction of \( \omega \) and its physical significance. The relationship between the variables in the differential equation and the solution is not fully resolved, leaving room for further exploration.

misogynisticfeminist
Messages
370
Reaction score
0
hmm ok, i was watching the MIT opencourseware video on oscillations and there was a part where it was mentioned that,

the diff. eq. [tex]x''+ \frac {k}{m} x = 0[/tex] has solution [tex]x= x_0 cos (\omega t + \phi)[/tex] if and only if [tex]\omega= \sqrt{\frac {k}{m}}[/tex]

how do i show that omega is the sqaureoot of k over m? thanks a lot.
 
Physics news on Phys.org
Differentiate the solution twice with respect to time and substitute it back into the differential equation - then tell us what you discovered! :)
 
ohhh, i see, it can be gotten by verifying the solution. I thought we needed to do something to hooke's law, but verifying the solution works great too.

: )

edit: a tinge of doubt crosses my mind though. When we solve the original differential equation, we get the solution in terms of k, m and x and no omega. While verifying the solution works when the solution is given, and we see that [tex]\omega= \sqrt{\frac {k}{m}}[/tex]. How do we know that [tex]\omega= \sqrt{\frac {k}{m}}[/tex] when we are solving it?
 
Last edited:
If you just substitute [itex]x = x_0 \cos(\omega t - \phi)[/itex] into the differential equation the required value of [itex]\omega[/itex] will jump out at you!
 
misogynisticfeminist said:
edit: a tinge of doubt crosses my mind though. When we solve the original differential equation, we get the solution in terms of k, m and x and no omega. While verifying the solution works when the solution is given, and we see that [tex]\omega= \sqrt{\frac {k}{m}}[/tex]. How do we know that [tex]\omega= \sqrt{\frac {k}{m}}[/tex] when we are solving it?

In this case, [tex]\omega[/tex] is referring to the angular frequency, which is the multiplicative factor in front of the independent variable in the sine or cosine function. It's just a standard substitution that they probably just didn't bother to define.
 
try the following:

in your original ODE, consider [itex]\omega ^2[\itex] just as a mathematically sound way to express the positiveness of the factor [itex]\frac{k}{m} [\itex]. After all the calculations you will realize that this choice has proven itself useful and physically meaningful.[/itex][/itex]
 

Similar threads

  • · Replies 30 ·
2
Replies
30
Views
4K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 20 ·
Replies
20
Views
5K
  • · Replies 21 ·
Replies
21
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 87 ·
3
Replies
87
Views
8K