Is w Constant in Damped Oscillations?

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In damped oscillations, the ratio b/√(k*m) is significant as it indicates the level of damping, with small damping defined when this ratio is much less than 1. The discussion clarifies that the angular frequency w is affected by damping, expressed as w = √(k/m - b^2/4m^2). The impact of the damping coefficient b on w is determined by the term b^2/4km. Despite the exponential decay of the displacement x(t), the modified angular frequency w remains constant in damped oscillations. Understanding this relationship is crucial for analyzing the behavior of damped systems.
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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}##.
 
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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