Where does the equation C=2*sqrt(km) for Critical Damping come from?

AI Thread Summary
The equation C=2*sqrt(km) for critical damping is derived from the analysis of damped harmonic oscillators, specifically in the context of a ruler cantilever experiment. Critical damping occurs when the damping force is precisely balanced to prevent oscillation, allowing the system to return to equilibrium in the shortest time possible. The discussion highlights the mathematical approach to solving the equation of motion for a damped spring system and the significance of the parameter values in determining the system's behavior. Resources such as college-level physics textbooks provide foundational insights into critical damping and its derivation. Understanding this equation is essential for accurately analyzing damping in mechanical systems.
physics51
Messages
1
Reaction score
0
Homework Statement
what's the background on this equation, I can't find anything about I, what can you guys tell me about it related to damping, when to use it, how to derive it, what it means, etc.? Anything is useful!! Thank you!!
Relevant Equations
C= 2 sqrt km
Im using this equation to find the damping from a ruler cantilever experiment. Any information about what critical damping really means and how it reflects in a ruler cantilever is also really helpful. Thank you again.
 
Last edited:
Physics news on Phys.org
From the context I guess that you are discussing a damped spring pendulum with the equation of motion
$$m \ddot{x}=-C \dot{x} - k x.$$
To solve this equation a nice trick is to make the substitution
$$x(t)=\exp(\lambda t) y(t).$$
Plug this into the equation and check that then ##y## fulfills
$$m \ddot{y}+(C+2 \lambda m) \dot{y} + [k+\lambda(C+\lambda m)]y=0.$$
Now make ##\lambda=-C/(2m)## to get rid of the term with ##\dot{y}##. Then the equation of motion for ##y## simplifies to
$$m\ddot{y} + \left (k-\frac{C^2}{4m} \right) y=0.$$
Now discuss what happens for the different values of the expression in the bracket,
$$k-\frac{C^2}{4m}<0, \quad k-\frac{C^2}{4m}=0, \quad k-\frac{C^2}{4m}>0.$$
 
For an ODE with constant coefficients I’d just make the ansatz ##x(t) = A \exp(\lambda t)## and solve the characteristic equation, but any way that works works.
 
Well, then you've the trouble with exactly the case of critical damping ;-).
 
physics51 said:
what's the background on this equation, I can't find anything about I,
Where did you look? Most college-level introductory physics textbooks will have a discussion of critical damping and will derive this formula.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Back
Top