Small Damping Forces: Fd << Fr

[Shreya]

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

 

[haruspex]

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 $\frac {b^2}{4km}$.

 

[Shreya]

You mean w =√(k/m - b^2/4m^2), right?

Yes, that was a typo, I have edited it.

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?

 

[haruspex]

By the way, even though x(t) decreases exponentially, w is constant (with its new value) even in a damped oscillation, right?

Yes.