How to find damping ratio, revision questions.

In summary, it appears that the mass is oscillating on a spring and it is observed that amplitude reduces to 5% of its initial value after 2 oscillations. The damping ratio is found to be 0.43 and the natural frequency is found to be 3.51Hz. The spring stiffness is found to be 2915N/m and the critical damping coefficient is found to be 264.5Ns/m. Finally, it is found that the actual frequency is 3.17Hz and the actual damping coefficient is 113.7Ns/m.
  • #1
rockhead
13
0
Hi guys,
Revision for an exam tomorrow, I hope someone can help?

I've been given the answers but I can't work out how to find the damping ratio without being given a velocity or distance for the mass to travel?A mass of 6kg is suspended on a spring and set oscillating. it is observed that amplitude reduces to 5% of its initial value after 2 oscillations, which takes 0.57 seconds.

Determine the following.

The damping ratio (0.43)
The natural frequency (3.51)
The actual frequency (3.17)
The spring stiffness (2915N/m)
The critical damping coefficient (264.5Ns/m)
The actual damping coefficient (113.7Ns/m)

I can use the damping ratio to help solve some of the other values once I know how to work it out.

Any help appreciated.
 
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  • #2
OK this is what I have,
Natural frequency = 2/0.57 = 3.51Hz
Spring stiffness 3.51Hz = 1/2pi x square root of K/m (k works out to be about 2915)
Critical damping square root of 4mk = 264.5

I need the damping ratio to find actual damping and actual frequency, I don't think I can do it without a velocity?
The 5% of amplitude must come into it somehow?
 
  • #3
We start with the diff. eq. F = ma:
mx'' = -kx - cx'
where k = spring constant
Assume an initial x = x0 and solve the equation for x(t).
Then, specify that at t = 0.57s two full oscillations have occurred, and that x(0.57) = 0.05x(0). This will enable you to solve for c, k, and the actual ω. You should already know the formula for the natural frequency as a function of m and k.

The damping coefficient ζ = c/cm where cm is the damping coefficient for critical damping. You can look up that formula on Wikipedia (it's a function of k and m also), or derive it as being the value of c for which the roots of your auxiliary equation are both just barely real (any lower c would give you complex-conjugate roots). OK, I'll give it to you: cm = 2√(km).

(I haven't actually solved the problem but I can if necessary).
 
  • #4
Thanks!

Could you solve please so I can see how it's done?
 
  • #5
rockhead said:
Thanks!

Could you solve please so I can see how it's done?

We are not allowed to do that. But if you post your work we can tell you where, if at all, you went wrong. The numerical answers are given to you already, apparently.

Start with the expression for x(t) as I suggested.
 
  • #6
I mentioned it in the second post.
I can work out Natural frequency,Spring stiffness and Critical damping.
And when given the one of the other 3, (damping ratio, actual frequency, actual damping coefficient), I can solve for the other 2.

I now need to differentiate to find x, once I have x I can find velocity and solve the rest from there.

I've just been on Khan Academy to study differentiation but there is a lot to learn and I have to be up early.

Cheers for the push in the right direction.
 
  • #7
OK, you probably solved this equation in class, and it's a chore to do it again so I will give you the sol'n:
x = x0exp(-ct/2m)cos(ωt)

where
ω = √(ω02 - c2/4m2)
and the natural frequency ω0 = √(k/m).

Now you should be able to use the givens in your problem to arrive at the missing parameters ω, ω0, c, cm, k, and ζ.

P.S. you don't need to solve for velocity.
 
  • #8
Thanks, but is there a simpler way to explain what is going on?

It's all getting very complicated.

I'm going to have to dedicate my weekend to it I think.
 
  • #9
It's really just algebra at this point.

For example: what's the simple relation connecting ω and 0.57 sec.?
What about the 5% and exp(-ct/2m)?
Then ω0 pops right out, as does k.
And so do c and ζ.
 
  • #10
what's the simple relation connecting ω and 0.57 sec.?
Natural Frequency is square root of k/m = 22.04 (radians) divide by 2pi to get 3.51Hz
0.57 for 2 revs = 0.285 secs per revolution. 1/0.285 = 3.51Hz

What about the 5% and exp(-ct/2m)?
I can't for the life of me find a value for c (unless you mean critical?)

x = x0exp(-ct/2m)cos(ωt)
I don't understand where cos(ωt) comes into it.

This is the best I can do so far
2915 x e1 = 7923.79
2915 x e0.95 = 7537.34
7923.79 - 7537.34 = 386.45
386.45/3.51 = 110.1 (nearest I can get to 113)

Any chance you can solve it or solve a similar problem so I can see what you did?
Then I have a chance of working out what's going on.

I'm running out of time, the exam is on Monday and I've already dodged it twice.
 
  • #11
rockhead said:
what's the simple relation connecting ω and 0.57 sec.?
Natural Frequency is square root of k/m = 22.04 (radians) divide by 2pi to get 3.51Hz
0.57 for 2 revs = 0.285 secs per revolution. 1/0.285 = 3.51Hz

Where did you get k?
Anyway, ω is not the natural frequency. ω is the actual frequency = 2πf. So period T = 1/f and 0.57s = 2T according to what you're given.

The natural radian frequency is √(k/m).

What about the 5% and exp(-ct/2m)?
I can't for the life of me find a value for c (unless you mean critical?)
Solve exp(-ct/2m) = 5% = 0.05 for c. No, c is the actual damping coefficient, not the critical one.
x = x0exp(-ct/2m)cos(ωt)
I don't understand where cos(ωt) comes into it.

As I said, it's the solution to the differential equation based on F = ma. You can understand that only if you've followed its derivation, which is lengthy & you should have covered it in class or in your text.

You need to understand this stuff rather than trying to memorize solutions to "similar" problems. Is the test going to have this very problem? Anyway, I'm not allowed to go any further than I have. Try to understand what I've written so far. I've given you every single formula you need to get all six parameters asked for.
 
  • #12
i got k from the equation
Natural freq = 1/2pi x √k/m
(3.51 x2pi)2 x 6= k
=2915

"You need to understand this stuff rather than trying to memorize solutions"
I agree totally, but unfortunately we are given little more than equations to memorise.
I find myself having to spend hours online trying to teach myself something that would have taken just a few minutes to explain.

"You can understand that only if you've followed its derivation, which is lengthy & you should have covered it in class or in your text."
No the hand outs we were given only cover questions that have force and velocity to find the damping ratio.

Could you help me out with the calculus part?
 
  • #13
I think I've stumbled onto something.

Critical damping = 264.5Ns/m

But that's per second, so 264.5 x 0.57 = 150.765
leaving 113.735 actual damping?

I'm not sure if it makes sense but its the right number that I need?
 
  • #14
rockhead said:
i got k from the equation
Natural freq = 1/2pi x √k/m
(3.51 x2pi)2 x 6= k
=2915
And where did you get the natural frequency from?
"You need to understand this stuff rather than trying to memorize solutions"
I agree totally, but unfortunately we are given little more than equations to memorise.
I find myself having to spend hours online trying to teach myself something that would have taken just a few minutes to explain.

"You can understand that only if you've followed its derivation, which is lengthy & you should have covered it in class or in your text."
No the hand outs we were given only cover questions that have force and velocity to find the damping ratio.

Could you help me out with the calculus part?

I have. You don't need any calculus now. Just use the formulas I gave you and try to understand the equation for x(t): it's a damped sinusoid. Sorry I can't draw it for you. Lots of people on this forum could.

If you want a derivation of x(t) look up damped harmonic oscillations in Wikipedia.
 
  • #15
I have natural frequency given on the revision sheet as 3.51Hz
Are you hinting that this is wrong and 3.51 is the actual frequency?

Makes more sense to me, and it wouldn't surprise me if they were giving us wrong answers.
 
  • #16
rockhead said:
I have natural frequency given on the revision sheet as 3.51Hz
Are you hinting that this is wrong and 3.51 is the actual frequency?

Makes more sense to me, and it wouldn't surprise me if they were giving us wrong answers.

I haven't computed any actual numeical answers.
 

Related to How to find damping ratio, revision questions.

1. What is damping ratio?

The damping ratio is a measure of how quickly a system will return to its equilibrium position after being disturbed. It is a dimensionless quantity and is usually denoted by the symbol ζ (zeta).

2. How do you calculate damping ratio?

To calculate the damping ratio, you first need to determine the natural frequency (ωn) and the damped natural frequency (ωd) of the system. The damping ratio can then be calculated using the equation ζ = (c/2m) * (√(m/k)), where c is the damping coefficient, m is the mass of the system, and k is the spring constant.

3. What is the significance of damping ratio in a system?

The damping ratio is an important parameter in a system as it determines the type of response the system will have: overdamped, critically damped, or underdamped. It also affects the stability and responsiveness of the system.

4. How does the damping ratio affect the response of a system?

The damping ratio affects the response of a system by controlling the rate at which the system returns to its equilibrium position after being disturbed. A higher damping ratio results in a slower response, while a lower damping ratio results in a faster response.

5. How can I improve the damping ratio of a system?

There are several ways to improve the damping ratio of a system, such as increasing the damping coefficient, reducing the mass of the system, or using a different type of damping mechanism. It is also important to properly tune and calibrate the system to achieve the desired damping ratio.

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