# Simple pendulum with damping

1. Sep 19, 2014

### toothpaste666

1. The problem statement, all variables and given/known data
Consider a simple pendulum (point mass bob) 0.55m long with a Q of 370.
a) How long does it take for the amplitude (assumed small) to decrease by two-thirds?
b) If the amplitude is 2.9cm and the bob has mass 0.22kg , what is the initial energy loss rate of the pendulum in watts?

2. Relevant equations
ω=sqrt(g/L) D = xcosωt

3. The attempt at a solution

a) 2/3A = Acos(sqrt(g/L)t)
2/3 = cos(sqrt(g/L)t)
cos^-1(2/3) = sqrt(g/L)t
sqrt(L/g)cos^-1(2/3) = t
t = .20

but that is wrong and it doesnt use Q. this problem has me stuck

2. Sep 19, 2014

### Orodruin

Staff Emeritus
First of all: What is Q and why does it not have units?

Second: You seem to be ignoring the damping in the system and to be solving for when the pendulum reaches a point which is 2/3 of the amplitude, not for when the amplitude itself is reduced to 2/3 of the original value.

3. Sep 19, 2014

### haruspex

Third, it says reduced by 2/3, not reduced to 2/3.

4. Sep 19, 2014

### Orodruin

Staff Emeritus
This is of course correct, I must have been looking at the attempted solution where he put 2A/3 equal to the deviation when I wrote that.

5. Sep 19, 2014

### olivermsun

Do your notes contain the definition of the Q factor?

6. Sep 19, 2014

### toothpaste666

ok i found the formula for damped motion

x = Ae^(-γt)cos(ω't)

where γ=b/2L (b is the damping constant) and ω' = sqrt(g/L - b^2/4L^2)

so for this problem the set up should be

$A-\frac{2}{3}A = Ae^\frac{-bt}{2L}cos(sqrt(\frac{g}{L}-\frac{b^2}{4L^2})t)$

$1-\frac{2}{3} = e^\frac{-bt}{2L}cos(sqrt(\frac{g}{L}-\frac{b^2}{4L^2})t)$

$\frac{1}{3} = e^\frac{-bt}{2L}cos(sqrt(\frac{g}{L}-\frac{b^2}{4L^2})t)$

my book calls Q the quality factor of the resonant peak and says that Q = (ω0)(L)/(b)

how is this so far?

7. Sep 19, 2014

### olivermsun

8. Sep 19, 2014

### Orodruin

Staff Emeritus
You are still making the mistake of just trying to put the position of the pendulum to 1/3 of the initial amplitude. Note that the amplitude is the number that multiplies the sinusoidal function (in this case the cosine). It is this amplitude that should be a third of the initial amplitude.

9. Sep 20, 2014

### toothpaste666

So is this correct: The A on the left side of the equation is the current distance from the equilibrium point and the A on the right is the current max amplititude? Im a little confused about the difference. so the equation should be:

$A = \frac{1}{3}Ae^{\frac{-bt}{2L}}cos(sqrt(\frac{g}{L}-\frac{b^2}{4L^2})t)$

the only other definition for Q i can find in my book is : (deltaOmega/Omega_0) = 1/Q (sorry for not using the symbols i am still getting used to the new layout)
since Q is talking about the peak of a resonant frequency could I say that omega' = omega_0 or in other words

$omega_0 = \frac{Qb}{L} = sqrt(\frac{g}{L}-\frac{b^2}{4L^2})$
?

Last edited by a moderator: Sep 20, 2014
10. Sep 20, 2014

### olivermsun

You have everything you need in post #6.

$x = Ae^{-\gamma t} \cos(\omega^\prime t)$ is an oscillation with exponentially decaying amplitude $Ae^{-\gamma t}$ and slowly varying frequency $\omega^\prime$. Your problem is interested in the amplitude. You can get $\gamma$ using the definitions in terms of $b$ and $Q$ respectively.

11. Sep 20, 2014

### toothpaste666

but b is still an unknown. i think if i knew what either b or omega_0 was I would be able to solve it but i cant figure out what piece I am missing here. Is it correct that I can ignore the cosine since the max it can be is 1 and I'm focusing on the amplitude?
if i can say that
$x = Ae^{-\frac{b}{2L}t}$
and I know that
$Q = \frac{(omega_0)(L)}{b}$
omega_0 and b are both still unknown.

forgive me if this is obvious. my book has only a very small section on Q which did very little to clarify its use for me

12. Sep 20, 2014

### toothpaste666

if i solved the Q definition for b and plugged that into the b in the top equation i would still need to know omega_0 to find t.

13. Sep 20, 2014

### olivermsun

Check your very first post for the natural frequency $\omega$ of the pendulum.

Also, the Wikipedia entry linked above is pretty useful for understanding some of the "whys" about Q and this system.

14. Sep 20, 2014

### toothpaste666

omega_0 = sqrt(g/L) ?
so omega_0 = sqrt(9.8/.55) = 4.2
Q = (omega_0)(L)/(b) so
b = (omega_0)(L)/(Q) = (4.2)(.55)/370 = .0062

$A= (1/3)Ae^{\frac{-(.0062)}{2(.55)}t}$
$A= (1/3)Ae^{-(.0056)t}$
$1= (1/3)e^{-(.0056)t}$
$3= e^{-(.0056)t}$
$ln(3)= -(.0056)t$
$ln(3)/-(.0056)= t$

t = -196 .... i think i did something wrong

15. Sep 20, 2014

### toothpaste666

ahh i didnt see that that was linked before. Ill look it over, I am a lot less confident about my understanding of SHM because of this problem

16. Sep 21, 2014

### olivermsun

You want to do it the other way around. If the original amplitude is $A_0$, then you want to know $t$ such that
$$A(t) = A_0 e^{-\gamma t} = \frac{1}{3} A_0.$$

17. Sep 21, 2014

### toothpaste666

Thank you so much! for part b) to find the energy I would use E=.5kA^2
k = 4pi^2mf^2
where f = Omega_0/2pi = 4.2/2pi = .67
they give us m as .22 so
k = 4pi^2(.22)(.67)^2 = 3.9
they give us A = .029m
so
E = .5(3.9)(.029)^2 = .0016

so i think that would be the initial energy of the system. the initial energy loss rate in watts, however, i am not sure what to do. My textbook doesnt have anything about this. Sorry that I am getting stuck at every step . I really want to understand this

18. Sep 21, 2014

### Orodruin

Staff Emeritus
If you have the energy of the system expressed as a function of A and A as a function of time, what is the time derivative of the energy?

19. Sep 21, 2014

### toothpaste666

E(t) = .5k(Ae^{-yt})^2
= (.5*3.9)(.029e^{-.0056t})^2
= 1.95(.029e^{-.0056t})^2
then find the derivative of that with respect to t?

20. Sep 21, 2014

### Orodruin

Staff Emeritus
That would be my suggestion.

Note that this is the energy loss averaged over the oscillations though, which I would guess is what the problem is after. This is fine as long as the typical time of one oscillation is much shorter than the typical decay time of the amplitude.

In reality, the energy will be lost only when the pendulum is moving and there is drag as a result. In this case the initial energy loss would be zero if the pendulum is released from rest.