Is w Constant in Damped Oscillations?

  • Thread starter Thread starter Shreya
  • Start date Start date
  • Tags Tags
    Damping Forces
Click For Summary
SUMMARY

The discussion centers on the behavior of the angular frequency (ω) in damped oscillations, specifically under conditions of small damping where the damping ratio b/√(k*m) is significantly less than 1. Participants clarify that the correct formulation for angular frequency is ω = √(k/m - b²/4m²), emphasizing that while the displacement x(t) decreases exponentially due to damping, the angular frequency remains constant at its adjusted value. This indicates that the effect of damping on angular frequency is minimal when the damping is small.

PREREQUISITES
  • Understanding of basic mechanics and oscillatory motion.
  • Familiarity with the concepts of damping and restoring forces.
  • Knowledge of the equations of motion for oscillating systems.
  • Ability to manipulate and interpret mathematical expressions involving square roots and ratios.
NEXT STEPS
  • Study the effects of varying damping ratios on oscillatory systems.
  • Explore the derivation and implications of the equation ω = √(k/m - b²/4m²).
  • Learn about the physical significance of damping in real-world oscillatory systems.
  • Investigate the relationship between damping and energy loss in oscillations.
USEFUL FOR

Students and professionals in physics, particularly those focusing on mechanics and oscillatory motion, as well as engineers dealing with systems subject to damping effects.

Shreya
Messages
187
Reaction score
64
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
 
Physics news on Phys.org
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}##.
 
  • Like
Likes   Reactions: Shreya
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.
 
  • Like
Likes   Reactions: Shreya

Similar threads

  • · Replies 1 ·
Replies
1
Views
831
  • · Replies 2 ·
Replies
2
Views
3K
Replies
1
Views
1K
Replies
9
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
13
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
5
Views
905
Replies
3
Views
847