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#1
05-07-2004, 01:14 PM
 CC Charter Member Join Date: Apr 2000 Location: not elsewhere Posts: 3,521
what's wrong with my basic math and/or physics?

Group - I may be more ignorant than I usually assume, but I can't for the life of me see what I'm missing here.. I'm playing with the basic formula for computing the period, in seconds, (T) of a pendulum when you know its length (L).
The formula I use is that T = 2pi times the square root of (the length divided by acceleration due to gravity(g). If I plug in, say, 1 meter for length, and use 9.8 m/sec/sec as (g), I find that T comes out to about 2 seconds. Reasonable enough. Then, I say, suppose I know the time and I want to know the length. With a little basic algebra, I think I find that L = g times (time divided by 2 pi)squared. Then, for the fun of it, I try to figure out how long a pendulum would have to be to have a period of 60 seconds. I get 893 meters - maybe a half a mile. But that surely is too small. It doesn't seem at all reasonable. Is it? A pendulum that hung down 893 meters would take a minute to get back and forth just once? Seems wrong. Any help?

(excuse my additional ignorance of how to use mathematical symbols in these posts. That would certainly make things more clear)
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#2
05-07-2004, 01:36 PM
 Jinx Member Join Date: Dec 1999 Location: Lost In Space Posts: 6,837
Seems ok to me

First, I verified your method, and it seems correct. Then, I made a ratio of the two equations which simplified to SQRT(L2/L1) = T2/T1 which became:
SQRT(893m/1m) = 30:1 ...or, 60s/2s

At first, it seems a little surprising how long a pendulum would be needed for a period of one minute, but your numbers seem correct to me. Now, go build such a pendulum and prove it!
- Jinx
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#3
05-07-2004, 01:37 PM
 Bytegeist Guest Join Date: Jul 2003
I don't think there is anything wrong with your approach. This page confirms the formula you're using. So the inverse is indeed L = g (T / 2 pi )2. Finally, checking your arithmetic, I get the same numerical result.
#4
05-07-2004, 01:40 PM
 Q.E.D. Charter Member Join Date: Jan 2003 Location: Richmond, VA Posts: 22,536
Yep, that's correct. Remember the relationship isn't linear; the time of the period increases with the square of the length, not in proportion to it.. Graph a square funtion and you'll see.
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#5
05-07-2004, 02:31 PM
 Jurph Guest Join Date: Oct 2003
Try to imagine the pendulum: eight hundred and ninety-three meters! There aren't even any buildings that tall right now. That is really long; the world record for running 800m is just under two minutes, and that's damn-near a sprint. Now imagine you draw your pendulum back and release it... ignoring the air resistance over the 893-meter-long massless cable ( ), the difference in deflection at the top and bottom of the cable, and coriolis effects ( ) and yeah, that sounds about right. A minute is a long damn time, but that is one hella long cable you got there.
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#6
05-07-2004, 04:34 PM
 wisernow Charter Member Join Date: Apr 2004 Location: at my desk Posts: 663
Quote:
 Originally Posted by CC I get 893 meters - maybe a half a mile. But that surely is too small. It doesn't seem at all reasonable. Is it? A pendulum that hung down 893 meters would take a minute to get back and forth just once?
Maybe this will help it look not so "wrong". For us to realize the reality of the above, we have to imagine and look at it in the way that our mind is used to which is to say our mind grasps linear motions better. Therefore let us translate the movement of this pendulum into the linear distance that the bottom of the pendulum would have to cover if it were to travel an angle of say 30 degrees at the point of suspension. The angle could be anything since it does not matter to the Time Period which is constant for this pendulum. With a length of 893 meters and an angel of traverse = 30 degrees, the length of the arc traversed comes to 142212 meters approx. That would be a total travel of approx 142.2 kms(88miles). One minute to travel 88 miles does not look too slow now does it?
#7
05-07-2004, 04:49 PM
 biqu Guest Join Date: Jun 2003
Quote:
 Originally Posted by wisernow The angle could be anything since it does not matter to the Time Period which is constant for this pendulum.
For a simple pendulum, the equation of motion is nonlinear, involving the sine of the displacement angle. If the small angle approximation is valid, then the linearization of sine reduces the equation of motion to the equation of simple harmonic motion, whose period is independent of amplitude. For larger angles, the period is actually a more complicated beast, which does depend on amplitude. I haven't done the calculation recently, so I can't recall whether 30 degrees is a large enough angle to demonstrate the amplitude dependence.
#8
05-07-2004, 05:01 PM
 wisernow Charter Member Join Date: Apr 2004 Location: at my desk Posts: 663
I think we are talking in terms of Simple Harmonic Motion here. The OP was wondering how it would take 1 minute(so...long...) for a 893 m long pendulum. Without going into the inticate physics and attempting to keep it simple I was only trying to show that it is not unimaginable.
As an addition to what I said, it would also make sense to realize that the 893 meter long pendulum would start with an initial velocity of zero at the begining of the 30 degree swing and reach a highest speed of approx 87 meters/sec at its lowest position. So now with that speed, travelling 82 miles in one minute does not seem too "slow" either.
#9
05-07-2004, 05:09 PM
 Xema Guest Join Date: Mar 2002
Quote:
 Originally Posted by wisernow With a length of 893 meters and an angel of traverse = 30 degrees, the length of the arc traversed comes to 142212 meters approx. That would be a total travel of approx 142.2 kms(88miles).
I'm having some trouble with this. A circle whose radius is 893 meters has a circumference of (2 * pi * 893), or 5610 meters. If the pendulum is oscillating through 30 degrees, that's one-twelfth of the full circumference -- a path length of 467.5 meters. I don't see how 142 km comes into this.
#10
05-07-2004, 05:17 PM
 zut Charter Member Join Date: Apr 2000 Location: Detroit, MI Posts: 3,569
Quote:
 Originally Posted by wisernow With a length of 893 meters and an angel of traverse = 30 degrees, the length of the arc traversed comes to 142212 meters approx. That would be a total travel of approx 142.2 kms(88miles). One minute to travel 88 miles does not look too slow now does it?
No way! Arc length for 30 degrees ought to be same order of magnitude as the radius.

[on preview] What Xema said. 467 meters.
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#11
05-07-2004, 05:24 PM
 ftg Guest Join Date: Feb 2001
Note that Foucault pendulums are the most common lengthy pendumlums and have periods that aren't all that long. I Googled for some info: one listed a length of 18m and a period of 8.5 seconds. Etc.
#12
05-07-2004, 05:25 PM
 CC Charter Member Join Date: Apr 2000 Location: not elsewhere Posts: 3,521
yes and no

boys and girls, thank you so much for verifying what was counterintuitive to me. However, I, too, question wisernow's analysis. A pendulum that is less than a mile long is certainly not going to describe an arc that is 88 miles long. Something went wrong somewhere. Knowing precious little physics, I'm guessing that a decimal was misplaced somewhere. If a pendulum had a length of one mile, the circumference of its transit would be 6+ miles (pi times the diameter). Thirty degrees away from the vertical and back and then thirty degrees the other way would circumscribe 60 degrees of this circle - one sixth of the distance. Traveling this entire path back and forth would be about a two mile trip, not 88. Still, I'll try to accept the wisdom of the masses. Even though my pendulum is shorter by half almost. Any other ideas would be most welcome. xo C.
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#13
05-07-2004, 05:44 PM
 John Mace Guest Join Date: Dec 2002
Also keep in mind that the simple equations ignore the mass of the string and air friction. As the pendulum length gets to be really long, these things are going to come into play. You'd probably have to have some serious cable and a very massive wieght to keep it taught. Not that the mass matters, but just picture how huge this whole setup would have to be in order to work properly.
#14
05-07-2004, 06:07 PM
 CC Charter Member Join Date: Apr 2000 Location: not elsewhere Posts: 3,521
well, yeah

also keep in mind that I'm dealing with the purely theoretical. I mean, for starters, what are ya going to hang the thing from? No - my question had to do with the time of the period, and I guess that I'll accept what the numbers say. Counterintuitive, as I say.
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#15
05-07-2004, 08:02 PM
 Jinx Member Join Date: Dec 1999 Location: Lost In Space Posts: 6,837
Refresh my memory: For a given pendulum, why doesn't amplitude matter? Wouldn't you think a greater amplitude would have a greater period? Woudn't it have a greater angle through which to swing? (I'd have to dust off the ole physics book and refresh myself).

I guess it is the fact that, for any given pendulum where obviously "length" is fixed, they are both driven by gravity...and it sorta parallels the problem of dropping two objects of different mass simultaneously.

Mr. Galileo, please come refresh my memory, and mange, eh? And, call your mamma this Mamma's day! - Jinx
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#16
05-07-2004, 08:34 PM
 biqu Guest Join Date: Jun 2003
Quote:
 Originally Posted by Jinx Refresh my memory: For a given pendulum, why doesn't amplitude matter? Wouldn't you think a greater amplitude would have a greater period? Woudn't it have a greater angle through which to swing? (I'd have to dust off the ole physics book and refresh myself).
For any system that can be modeled by the differential equation of simple harmonic oscillation,

Code:
`x''(t) + w^2 x(t) = 0,`
the frequency of oscillation is precisely w. Depending on the problem that is being modeled, w might involve various parameters like masses, spring constants, lengths, and local acceleration due to gravity. The general solution is a linear combination of sines and cosines of the angle wt, and it's the linearity of the differential equation that allows sinusoidal oscillation of arbitrary amplitude.

From a less mathematical perspective, consider that the restoring force is stronger for larger displacements, and recall from Newton's second law that the acceleration is correspondingly larger when the mass is displaced farther from equilibrium. Therefore, when the oscillation has a greater amplitude, the mass is moving faster and can cover more distance in the same amount of time.

For the case of the simple pendulum, though, the equation of motion is only approximately linear (when the displacement is not large and the small angle approximation is valid). For larger amplitudes, the equation of motion must be treated more delicately, leading to a period that does depend on the amplitude.
#17
05-07-2004, 08:44 PM
 CC Charter Member Join Date: Apr 2000 Location: not elsewhere Posts: 3,521
in other words...

Yeah - the amplitude may have a slight effect when it's quite high, compared to quite low, but by and large, the only factor affecting frequency is length. Most people also feel that mass should have an effect, but it doesn't. And in those cases where measurements are pretty gross, say, in a middle school classroom, where just about everything is gross, you can just about ignore all other variables besides length. xo C.
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#18
05-07-2004, 08:59 PM
 Jinx Member Join Date: Dec 1999 Location: Lost In Space Posts: 6,837
Yes, this once surprised a younger me!

Quote:
 Originally Posted by CC Yeah - the amplitude may have a slight effect when it's quite high, compared to quite low, but by and large, the only factor affecting frequency is length. Most people also feel that mass should have an effect, but it doesn't. And in those cases where measurements are pretty gross, say, in a middle school classroom, where just about everything is gross, you can just about ignore all other variables besides length. xo C.
Yeah, one would think mass would affect the period, but then you let the kids play with this to reach the right conclusion in the lab. Speaking of which, I recall doing freefall experiments in high school where different weights had a strip of ticker-tape trailing behind it. The tape fed through a "ticker" which made an impression every 1/60s. The tape hindered our experiment so badly, our results showed acceleration due to gravity=0! Instinctively, I knew this didn't sound right. Some kids got better results having better luck with feeding the ticker tape through more freely. We almost walked away with wrong conclusion! - Jinx
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#19
05-07-2004, 10:50 PM
 David Simmons Charter Member Join Date: Nov 2001 Posts: 12,684

60 seconds is a long time for the period of a pendulum. When I was attending the University of Iowa there was a Foucault Pendulum in an elevator shaft of the Physics Building. The thing was about 70 ft long and it's period as I remember it was at most 10 seconds.

In fact I just ran it on Mathcad and it comes out 9.2 sec., assuming my guess as to the length is close.
#20
05-08-2004, 08:38 AM
 CC Charter Member Join Date: Apr 2000 Location: not elsewhere Posts: 3,521
yep - I think so

The more I play with this, the more I think 893 M is about right. If one dropped a weight from that height, it would take about 13.5 sec to fall (d=1/2 a t(squared)). And if it were swinging from that height, it would take a bit longer to swing from the release point to the center point - probably about 15 seconds. So, if it takes about 15 seconds to travel 1/4 of the entire cycle, then I guess one minute would be about right. Interesting that in the absence of any relevant experience on that scale, it's about impossible to intuit how this would come out. By the way, Jinx, I used to do that tape-through-the-timer bit with my kiddos. With a heavy enough weight to pull the tape, we got pretty respectable numbers. But I wasn't after numbers. I just wanted them to see that the longer the thing fell, the farther it was travelling every second (or tick or whatever). Observational science - that's what I tried to do. Thanks again, boys and girls.
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#21
05-08-2004, 01:18 PM
 MonkeyMensch Charter Member Join Date: Jul 2001 Location: Encinitas Posts: 2,012
I'm going to build this thing!

I remember as a child watching the Foucault pendulum at the San Diego Museum of Natural History swing back and forth and having to wait forever to see it knock down the next wooden peg. The wooden pegs, of course, marked the pendulum's progress as its plane of rotation turned underneath the Earth's motion. So with this bad boy they're will be no waiting!

With a period of 60 seconds and an swing amplitude of say 10 degrees, not that it matters, we get to see pegs knocked over on every swing that are spaced 1 meter apart! Granted they might have to be tall pegs, but I'm going to Home Depot right now! Who's with me?
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#22
05-08-2004, 01:43 PM
 Q.E.D. Charter Member Join Date: Jan 2003 Location: Richmond, VA Posts: 22,536
Quote:
 Originally Posted by MonkeyMensch I'm going to build this thing!... Who's with me?
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#23
05-08-2004, 03:55 PM
 Omphaloskeptic Guest Join Date: Oct 2001
If you want a long-period pendulum without building a kilometer-high building you can build a compound pendulum: a rod with weights on both ends, suspended from a point near the center of mass. These have much longer periods than a simple pendulum of the same length.
#24
05-08-2004, 05:05 PM
 wisernow Charter Member Join Date: Apr 2004 Location: at my desk Posts: 663
Quote:
 Originally Posted by Xema I don't see how 142 km comes into this.
My mistake guys! I must have been drinking. I used the formula for area pi r sq instead of 2 pi r! Sorry!
Quote:
 Originally Posted by zut What Xema said. 467 meters.
My apologies!
#25
05-09-2004, 05:11 AM
 MonkeyMensch Charter Member Join Date: Jul 2001 Location: Encinitas Posts: 2,012
Quote:
 Originally Posted by Omphaloskeptic If you want a long-period pendulum without building a kilometer-high building you can build a compound pendulum: a rod with weights on both ends, suspended from a point near the center of mass. These have much longer periods than a simple pendulum of the same length.

Are you describing a single rigid rod with the CM near the pivot point or a set of rods with connected pivots? Do tell...
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#26
05-09-2004, 08:53 AM
 Q.E.D. Charter Member Join Date: Jan 2003 Location: Richmond, VA Posts: 22,536
Quote:
 Originally Posted by MonkeyMensch IAre you describing a single rigid rod with the CM near the pivot point or a set of rods with connected pivots? Do tell...
The former. If you've ever used a triple-beam balance and observed its motion as it settles down on a reading, you know how short the period is relative to the beam length compared to a pendulum of similar dimensions. Same idea.
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#27
05-09-2004, 05:24 PM
 moriah Member Join Date: Jun 1999 Location: NJ, USA ♂ Posts: 3,747
Pulling back this hypothetical mass to a thirty degree declension and letting go would create a one minute period. Traveling speed would be an average of two miles/min (120 ml/hr). OK calc majors, what would it's maximum speed actually be at the nadir of the arc?

But it would also take a full minute period if you pulled the mass back just 3 inches. Imagine how painfully slowly the mass would move then (one foot/min, actually).

Boggling.

Peace.

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