Simple Pendulum Damping: Finding Amplitude and Energy Loss Rate

[toothpaste666]

Homework Statement

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?

Homework Equations

ω=sqrt(g/L) D = xcosωt

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 doesn't use Q. this problem has me stuck

 

[Orodruin]

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.

 

[haruspex]

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.

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

 

[Orodruin]

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

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.

 

[olivermsun]

but that is wrong and it doesn't use Q.

Do your notes contain the definition of the Q factor?

 

[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?

 

[olivermsun]

You might be able to go simpler. Can you find anything in your textbook that talks about the physical interpretation of Q?

 

[Orodruin]

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.

 

[toothpaste666]

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.

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? I am 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})
?
I am very confused about this problem, particularly about how Q ties in

 

[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.

 

[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 can't 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

 

[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.

 

[olivermsun]

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.

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.

 

[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

 

[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

 

[olivermsun]

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

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

 

[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 doesn't have anything about this. Sorry that I am getting stuck at every step . I really want to understand this

 

[Orodruin]

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?

 

[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?

 

[Orodruin]

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.

 

[toothpaste666]

the answer says |dE/dt|t=0| = (blank where the answer should go)

so if my differentiation is correct
dE/dt = d/dt(1.95(.029e^{-.0056t))^2)
dE/dt = (1.95*2)(.029e^{-.0056t})(.029*-.0056)e^{-.0056t}
thats if i am doing the chain rule right
but then i would plug 0 into the t's and get
3.9 * .029 * .029 * -.0056
=-1.8x10^-5
i guess it makes sense that its a negative number because it is losing energy but that seems very small. perhaps i made a mistake when using the chain rule?

 

[Orodruin]

Yes you are correct (although I did not check the actual numbers, I find the conceptual understanding more important), it is negative due to energy dissipating out of the system. That it is small should not a priori bother you if you do not expect it to be large. It may also depend on the units you use (you really should put some). With regards to the derivative, it will be easier if you perform the square before doing the derivative. Keep in mind that $(a^b)^c = a^{bc}$.

That is about as much as there is to this problem I would say.

 

[toothpaste666]

Thank you so much for your help. I did the derivative again after doing the square first and got the same answer so I am pretty sure that is right but its still being marked wrong. could that be because they want the answer in watts? I am going to recheck my other calculations that i plugged into it

 

[toothpaste666]

nevermind it was a rounding error although it did make me switch the sign. I guess that's what the || meant in the answer label

 

[Orodruin]

nevermind it was a rounding error although it did make me switch the sign. I guess that's what the || meant in the answer label

Yes, the |x| would signify the absolute value of x.