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 Author Topic: Pendulum sets with different initial angle plus damping  (Read 10517 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
Fu-Kwun Hwang
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 « Embed this message on: December 03, 2010, 09:08:38 am » posted from:,,Taiwan

This simulation show 11 pendulums with different initial angle(from minimum angle to minimum angle+angle range)
It can be run in two different modes:
1. $\frac{d^2\theta}{dt}=-\frac{\ell}{g}\sin\theta-b\frac{d\theta}{dt}$ (equation for real pendulum)
2. $\frac{d^2\theta}{dt}\approx-\frac{\ell}{g}\theta-b\frac{d\theta}{dt}$ (equation for small angle approximation :assume $\sin\theta\approx\theta$)

The damping factor b=0 is the default.
You can change it and find out what will happened.
For the second mode: the period and damping rate are the same for all pendulums. Do you know why?

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Fu-Kwun Hwang
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 « Embed this message Reply #1 on: December 15, 2010, 07:47:40 pm » posted from:,,Taiwan

For a mass m attached to a  spring with spring constant k
The force is proportional to the displacement$x$.
$\vec{F}=-k*x$
Assume there is a damping force which is proportional to the velocity
$F=m\frac{d^2m}{dt^2}=-k*x-b*v$
The above equation can be rewrite as $m\frac{d^2m}{dt^2}+b \frac{dx}{dt}+k*x=0$

Do you notice the similarity between equation for spring and pendulum set?
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lookang
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 « Embed this message Reply #2 on: December 17, 2010, 09:33:04 am » posted from:SINGAPORE,SINGAPORE,SINGAPORE

This simulation show 11 pendulums with different initial angle(from minimum angle to minimum angle+angle range)
1. $\frac{d^2\theta}{dt}=-\frac{\ell}{g}\sin\theta-b\frac{d\theta}{dt}$ (equation for real pendulum)
this mode is demonstrate real pendulum swings at different period T for different magnitude of angle ϑ.
well done!
i like the view of all pendulum all superposition together.

2. $\frac{d^2\theta}{dt}\approx-\frac{\ell}{g}\theta-b\frac{d\theta}{dt}$ (equation for small angle approximation :assume $\sin\theta\approx\theta$)
For the second mode: the period and damping factor are the same for all pendulums. Do you know why?

This is due to the small angle assumption say ϑ = 5 degree, the period T has to be the same for the all ϑ = 5 degree.

i don't understand the question on "damping factor are the same for all pendulums", if b = 0.5, it will be the same of all models of the pendulums because that is the way the model is made to obey this mathematical equation $\frac{d^2\theta}{dt}\approx-\frac{\ell}{g}\theta-b\frac{d\theta}{dt}$ (equation for small angle approximation :assume $\sin\theta\approx\theta$).
is my interpretation of the question correct?
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Fu-Kwun Hwang
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 « Embed this message Reply #3 on: December 17, 2010, 10:38:49 am » posted from:Taipei,T\'ai-pei,Taiwan

Why the angle $\theta$ has to be smaller than 5 degree to be able to satisfy the small angle approximation?
Why 5.5 can not be a small angle? Does 5 a magic number???

Quote
i don't understand the question on "damping factor are the same for all pendulums"
What I mean is "The period and damping rate are the same for all pendulums."
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lookang
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 « Embed this message Reply #4 on: December 17, 2010, 11:02:55 am » posted from:SINGAPORE,SINGAPORE,SINGAPORE

Why the angle $\theta$ has to be smaller than 5 degree to be able to satisfy the small angle approximation?
Why 5.5 can not be a small angle? Does 5 a magic number???

students can key in 5/180*PI = 0.08726
in radian mode, students can key in sin(5/180*PI) = 0.08715
the students can realise the approximation is accurate for 2 decimal place only.

students can key in 5.5/180*PI = 0.09599
in radian mode, students can key in sin(5.5/180*PI) = 0.09584
the students can realize the approximation is accurate for 2 decimal place only.

it is a balance between practically observably angle of oscillation and accuracy in approximation of the assumption of $\frac{d^2\theta}{dt}\approx-\frac{\ell}{g}\theta-b\frac{d\theta}{dt}$

i don't understand the question on "damping factor are the same for all pendulums"
What I mean is "The period and damping rate are the same for all pendulums."

ic damping rate.
then my earlier answer is acceptable? made to obey this mathematical equation?
 « Last Edit: December 17, 2010, 11:04:59 am by lookang » Logged
Fu-Kwun Hwang
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 « Embed this message Reply #5 on: December 17, 2010, 11:49:42 am » posted from:Taipei,T\'ai-pei,Taiwan

$\sin\theta=\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\frac{\theta^7}{7!}+\frac{\theta^9}{9!}-\frac{\theta^{11}}{11!}+...$
The error due to approximation can be estimated as $\frac{\theta^3/3!}{\theta}=\frac{\theta^2}{6}$
For 5 degree, the error is about $\frac{(5\pi/180)^2}{6}=0.00127= 0.127%$ which is a very small error.
For 10 degree, the error is about 4*0.127%= 0.50%
For 20 degree, the error is about 16*0.127=2.0%

What I want to point out is:
1. 5 degree is not a magic number: user should provide the precission required in order to determined the maximum angle for good approximation.
2. Even 20 degree only produce 2% of error which is normally smaller than experiment error when student perform real experiment.

The solution for RLC circuit is: $Q(t)=Q_0 e^{-\alpha t}e^{i\omega t}$
where $\alpha=\frac{R}{2L}, \omega=\sqrt{\frac{1}{LC}}$
Because $\alpha$ is independent of charge or current: i.e. damping rate is the same for the same R/L value.
You should be able to find out solution for pendulum(compare equations for RLC and pendumum set)
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Knowledge and practice are one. ..."Wang Yang Ming (1472-1529, Chinese Philosopher) "