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#1
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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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"And it's just...that...easy!" - The Flying Karamazov Brothers |
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#2
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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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Reality TV gives new meaning to virtual reality! ![]() Reality TV is an oxymoron to the nth degree!
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#4
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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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SnUgGLypuPpY -- TakE BaCk tHe PiT! |
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#5
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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
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#7
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#8
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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.
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#9
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#10
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[on preview] What Xema said. 467 meters.
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Glaring directly down towards her was the stoney, cycloptic face of the bloated diety. Gaping from its single obling socket was scintillating, many fauceted scarlet emerald, a brilliant gem seeming to possess a life all of its own. A priceless gleaming stone, capable of domineering the wealth of conquering empires...the eye of Argon. |
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#11
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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.
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#12
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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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"And it's just...that...easy!" - The Flying Karamazov Brothers |
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#13
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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.
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#14
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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.
__________________
"And it's just...that...easy!" - The Flying Karamazov Brothers |
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#15
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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
__________________
Reality TV gives new meaning to virtual reality! ![]() Reality TV is an oxymoron to the nth degree!
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#16
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x''(t) + w^2 x(t) = 0, 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. |
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#17
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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.
__________________
"And it's just...that...easy!" - The Flying Karamazov Brothers |
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#18
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Yes, this once surprised a younger me!
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- Jinx
__________________
Reality TV gives new meaning to virtual reality! ![]() Reality TV is an oxymoron to the nth degree!
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#19
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Your formula is correct.
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. |
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#20
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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.
__________________
"And it's just...that...easy!" - The Flying Karamazov Brothers |
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#21
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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
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#23
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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.
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#24
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My mistake guys! I must have been drinking. I used the formula for area pi r sq instead of 2 pi r! Sorry! Quote:
My apologies!
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#25
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Quote:
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
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Quote:
__________________
SnUgGLypuPpY -- TakE BaCk tHe PiT! |
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#27
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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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