Find the value of the damping coefficient

[fruitl00p]

Homework Statement

A 95.0 cm pendulum is released from a small angle. After 115 oscillations, the amplitude is one half of its original value. The damping is proportional to the speed of the pendulum bob. Find the value of the damping coefficient.

Homework Equations

b=sqrt k*m
A= Ao*factor^N

The Attempt at a Solution

I can't seem to figure out how to approach this problem. According to the equation, I need to know what k is and what the mass is...but all they provide in the problem is the length of the pendulum.

I think it is a simple pendulum, so I used the equation
T= 2pi*sqrtL/g
knowing T, I can find the frequency and the angular frequency...but I'm stuck after that. What am I overlooking?

 

[Mindscrape]

Your equations are either confusing or wrong. What is k? Are you sure you don't mean A = Ao e^-kt? Have you been given anything other than those?

 

[denverdoc]

Since its a hard problem, I assume its a physics class that assumes calculus knowledge, especially since you're required to find damping coefficient. Usually such a problem leads to damped exponential SHO. Do you know solutions for differential eqns?

 

[fruitl00p]

k is supposed to be the spring constant. In my textbook, they have the equation b= sqrt k*m, where b is the damping coefficient.

Mindscrape, I wasn't offered any equations for this problem, I am just trying to use the info from my textbook to solve for this problem. But I have seen the equation you are wrote, its just that when I attempted to solve a problem simliar to this, I was given a hint that suggested to solve A after N oscillations try A= Ao*factor of^N...but the equation you mentioned is probably the proper one.

denverdoc, I don't know the solutions for differential equations.

 

[fruitl00p]

I found an equation that might be helpful.
it is A(t)=Ao*e^((-b/2m)*t)
However, I am not given what the mass is and the time is in the problem ...so I don't know if that is the correct equation to use at all.

 

[denverdoc]

Oh ok, then let's cheat a bit and use the approximation

where (1/2pi)*ln(amplitude ratio after n cycles)=damping constant

here the ratio would be 1/2. For full details you might look here:

http://meweb.ecn.purdue.edu/~me563/Lectures/Free/Single_DOF/Example_02/example_02.pdf

 

[fruitl00p]

denverdoc, I was looking at the info you attached and I am wondering
about the n in the eqaution. In the attached info, the formula is 1/(2*pi*n)*ln(amplitude ratio) = damping constant. Is the n the number of oscillations?
(In the equation you typed it is 1/(2pi)" " )

 

[fruitl00p]

well I did
damping coefficient= [1/(2*pi*115)] *ln(1/2) = -9.59e^-4
115 is the number of oscillations. Should the coefficient, or rather can the coefficient be negative? Also, my answer is considered wrong, so that is why I am wondering if its the sign or if I did something wrong.

 

[denverdoc]

maybe it should be ln of 2, i didn't look at the formula carefully but that would take care if the minus sign, should have noticed that would be negative but was rushed this am.

 

[fruitl00p]

Yes, I got a positive value. However, I forgot to mention that the problem wants the damping coefficient in Hz. Aren't the units for the damping coefficient kg/s?

 

[Mindscrape]

Yeah, the "half-life" of an exponential decay is $$t_{1/2} = \frac{ln2}{\lambda}$$ where lambda is the eigenvalue. Now I put in these terms so that fruitl00p doesn't just use my formula (though I suppose if he spent as much time researching as it would take to actually do the problem he could).

Do some dimensional analysis to figure out the units of the damping coefficient. You probably used angular velocity. How does angular velocity relate to frequency?

Btw, it's pretty easy to just derive a half life from given information rather than get mixed up in confusing plug-and-chug equations.

 

[fruitl00p]

Actually mindscrape I have been trying to read more information on damped oscillations. Unfortunately for me my textbook only has 2 pages dedicated to the information. I am trying to understand how to do the problem, and quite frankly, rude remarks are not helpful. I have more problems that are due that I have solved by doing some research. This one has left me stumped.

 

[denverdoc]

Yeah, the "half-life" of an exponential decay is $$t_{1/2} = \frac{ln2}{\lambda}$$ where lambda is the eigenvalue. Now I put in these terms so that fruitl00p doesn't just use my formula (though I suppose if he spent as much time researching as it would take to actually do the problem he could).

Do some dimensional analysis to figure out the units of the damping coefficient. You probably used angular velocity. How does angular velocity relate to frequency?

Btw, it's pretty easy to just derive a half life from given information rather than get mixed up in confusing plug-and-chug equations.

I'm curious and maybe even a little offended as well that you suggest a guy with no knowledge of differential eqns, should be able to intuitively find a soln from first principles. Yes the period is invariant, and multiplying it by N gives a t to use in a e^-kt problem and its fairly simple to figure a half life expression. To the initiated at least. And I agree wholeheartedly that better to teach to fish than give a fish...

 

[Mindscrape]

I'm not trying being rude, but I don't want to give you the answer. I can tell you are very confused, and probably frustrated, so it will be a lot more helpful for your learning if you go back to square one and figure out how to derive a half life for a general equation (from there it will be merely applying it to your situation). I will give you an example. Consider a differential equation that has a solution

$$A(t) = A_0 e^{-\lambda t}$$

then you want when A(t) = 0.5*Ao

$$A(t_{1/2}) = \frac{1}{2}A_0$$

put the half-life formula into the general formula

$$\frac{1}{2}A_0 = A_0 e^{-\lambda t_{1/2}}$$

so the Ao will cancel, and then take ln of both sides

$$ln(\frac{1}{2}) = -\lambda t_{1/2}$$

$$-ln(\frac{1}{2}) = \lambda t_{1/2}$$

$$ln2 = \lambda t_{1/2}$$

$$t_{1/2} = \frac{ln2}{\lambda}$$

Now, I'm not really sure what kind of pendulum you have, but if you want I can give you the ODE solution for the general pendulum with damping (i.e. in some kind of oil).

Maybe your textbook, or professor, has overlooked some fact with pendulums that you need. Still, all that differential equations will do for you is give you the solution. The general idea of finding a half life, whether it be a spring mass (which generally use k and m, and what you seem to be thinking about) or a pendulum (which generally use g and l) does not need to explicitly use ODEs.

 

[fruitl00p]

Thank you for clarifying your earlier remark mindscrape. If I understand what you are asking, I think the pendulum in the problem is a simple pendulum. And I would very much like to see the ODE solution for the general pendulum damping.

 

[fruitl00p]

ok, this is what I've tried after looking for more info online.

ln (A/Ao)= -lambda*time
since A is just 1/2 Ao, I changed it to
ln(1/2)= -lambda*time

T= 2pi *sqrt L/g = 2pi *sqrt .95/9.8 = 1.956
time, t = T* number of oscillations
t= 1.956*115 = 224.9
[ln(1/2)]/ -time = lambda
[ln(1/2)]/-224.9 = .003082

t1/2 = ln2/lambda
=224.9 s

 

[fruitl00p]

and since it needs to be in Hz, 1/224.9 = .004446 Hz, but when I tried that answer, it was considered incorrect

 

[denverdoc]

fl,

the damping constant is dimensionless, and can't be converted to Hz. The original answer you had looks better, did you try to submit it?

PS: as a footnote, the period is not actually invariant, which gives rise to an error by assuming t=115*T. But I would still give the answer 1/224.9 a try if you haven't already. for more on derivation of ODE, look here:

http://www.ocw.cn/NR/rdonlyres/Mathematics/18-03Spring2004/B76E6F4F-7B05-4DA0-A5A5-03FA4ACCB6B2/0/sup_13.pdf

 

[Mindscrape]

So the general equation for a damped will be the solution to an ODE of the form $$\ddot{\theta} + 2\beta \dot{\theta} + \frac{g}{l}\theta = 0$$ (if you haven't seen the dot notation before it means second derivative with respect to time), where the term with beta is related to damping. The solution will be $$\theta(t) = e^{-\beta t}(A_0 cos(\omega t + \delta)$$ where omega is the angular frequency, and equal to √(g/l).

Essentially what is going on with this equation is that cos function is oscillating how a normal undamped oscillator would, but the exponential is slowly crunching the oscillations' amplitude.

Here is a picture (the blue points represent the damped oscillator, and the black background represents the oscillator if it were not damped):
http://www.physics.usu.edu/classes/4020/soundnotes/Image29.gif

So what is controlling the amplitude is that exponential term.

You have been given the number of oscillations have gone by. So, you know that after 115 periods the amplitude is at half its value. The period relates to the angular frequency by a factor of 2π, so you can figure out how many oscillations happen per second and then figure out how many seconds have elapsed by the period they give you.

Once you know t in seconds then you should be able to figure out the damping coefficient, which would be in 1/s (since the exponential must be dimensionless). 1/s is equivalent to hertz.

So far, it looks like you have:
Figured out the period. (The angular frequency is actually sqrt(g/l))
Made the relation between period and seconds elapsed. (Good!)
Found the supposed time.

I think you went wrong with figuring out the angular frequency.

 

[Mindscrape]

Also, I am not quite sure if you used that half-life formula correctly. Remember that lambda will be the damping constant. So if you want to solve for lambda (or beta in my case) you would want to take ln2/t_(1/2).

*Edit:
Actually you did find lambda, and then you calculated the half-life again?

 

[fruitl00p]

Also, perhaps I got the answer incorrect because I used the equation "twice"
I didn't view lambda as the damping coefficient, so I thought I needed to find lambda to find the time...oh denverdoc, I did try .004446 and that was considered wrong.

 

[fruitl00p]

Oh, I didn't see the last part of your last post. Yes, I redid that portion and got the correct answer. Thanks denverdoc and mindscrape!