Is w Constant in Damped Oscillations?

  • Thread starter Thread starter Shreya
  • Start date Start date
  • Tags Tags
    Damping Forces
Click For Summary
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.
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}##.
 
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.
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

Similar threads

  • · Replies 1 ·
Replies
1
Views
736
  • · Replies 2 ·
Replies
2
Views
2K
Replies
9
Views
2K
Replies
1
Views
1K
  • · 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
809
Replies
3
Views
729