Is w Constant in Damped Oscillations?

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Homework Help Overview

The discussion revolves around the behavior of angular frequency (ω) in the context of damped oscillations, particularly focusing on the effects of damping on ω and the conditions under which it remains constant.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between damping ratio (b/√(k*m)) and its impact on angular frequency (ω). Questions arise regarding the choice of this particular ratio and its significance in determining the behavior of ω in damped oscillations.

Discussion Status

Some participants have offered clarifications regarding the equations involved and the effects of damping on ω. There is an acknowledgment of the relationship between damping and the constancy of ω, although the discussion does not reach a definitive conclusion.

Contextual Notes

Participants reference specific equations and ratios related to damping, indicating a focus on theoretical understanding. The discussion includes a mix of assumptions about the behavior of the system under small damping conditions.

Shreya
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Homework Statement
In my text, it is written that small damping means that b/√(k*m) is much less than 1


My question is , Why is this particular ratio chosen?
Please be kind to answer.
Relevant Equations
F (damping)=-bv
F(restoring)= -kx
w =√(k/m - b^2/4m^2)
My first intuition is that for small damping Fd<<Fr
 
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Shreya said:
Homework Statement:: In my text, it is written that small damping means that b/√(k*m) is much less than 1My question is , Why is this particular ratio chosen?
Please be kind to answer.
Relevant Equations:: F (damping)=-bv
F(restoring)= -kx
w =√(k/m - b/4m^2)

My first intuition is that for small damping Fd<<Fr
You mean w =√(k/m - b^2/4m^2), right?

The interesting question in regards to the behaviour is whether b much affects w.
Writing ##\omega=\sqrt{\frac km(1-\frac {b^2}{4km}})##, we can see that the strength of b's effect depends on ##\frac {b^2}{4km}##.
 
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haruspex said:
You mean w =√(k/m - b^2/4m^2), right?
Yes, that was a typo, I have edited it.
haruspex said:
The interesting question in regards to the behaviour is whether b much affects w.
Writing ω=km(1−b24km), we can see that the strength of b's effect depends on
I understand now, Thanks a lot! 🙏
By the way, even though x(t) decreases exponentially, w is constant (with its new value) even in a damped oscillation, right?
 
Shreya said:
By the way, even though x(t) decreases exponentially, w is constant (with its new value) even in a damped oscillation, right?
Yes.
 
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