It's About Time C

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boomvroomshroom
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Re: It's About TIme

Postby boomvroomshroom » February 25th, 2015, 4:20 pm

Hey Ionizer, its considered bad forum etiquette to double post. No hard feelings, but in the future keep in mind that you can edit your post.
That was fine since Ionizer was addressing two different topics. Double posting is when two consecutive posts basically say the same thing (except for grammar issues).

Anyway, question: Gary, Barry, Harry, Darry, Larry, Kerry, and Mary have a pendulum each. Gary's is 1m long, Barry's is 2m...etc. up until Mary, who has a 7m long pendulum. They each let go of their pendulums in 1s intervals (so Gary lets go of his at 0s, Barry at 1s, Darry at 2s...up until Mary, at 6s after Gary). When will the periods of the pendulums first coincide?

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Re: It's About TIme

Postby finagle29 » February 26th, 2015, 11:10 am

Double posting is when two consecutive posts basically say the same thing (except for grammar issues).
I guess I took double posting a little too literally :oops: Sorry Ionizer.

About the question at hand, when you say the first time the periods coincide do you mean the first time all of the pendulums are making the same (signed) angle to the vertical?
I don't have a rigorous proof for this, but it has its roots in modular arithmetic and number theory as a type of resonance problem.

The periods of the pendulums will never coincide.  In order for the periods of any grouping of more than two pendulums that were not released simultaneously to coincide, two requirements must be met.  Only the first is sufficient to prove that Gary et al.'s pendulums will never coincide.  The ratio of all possible pairs of two pendulums' angular frequencies must be a rational number.  Since the angular frequencies of Gary et al.'s pendulums are [math]\sqrt{g}, \sqrt \frac{g}{2}, \sqrt \frac{g}{3}, \sqrt \frac{g}{4}, \sqrt \frac{g}{5}, \sqrt \frac{g}{6}, \sqrt \frac{g}{7}[/math] we can examine Gary and Barry's pendulums and find that the ratio of their pendulums' angular frequencies is irrational and therefore their pendulums will not coincide regularly and thus all of the pendulums will never all coincide.
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Re: It's About TIme

Postby boomvroomshroom » February 26th, 2015, 8:33 pm

yep. kind of a trick question, but nice explanation.

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Re: It's About TIme

Postby finagle29 » February 26th, 2015, 8:48 pm

Thank you!

Gary, Barry, Harry, Darry, Larry, Kerry, and Mary return, but their pendulums have changed. Gary's is 1m, Barry's is 4m, Harry's is 9m, Darry's is 16m, Larry's is 25m, Kerry's is 36m, and Mary's is 49m. Their pendulums are on individual stands in a row such that they can all swing parallel to each other. They let go of their pendulums from the same angle at the same time. How long does it take for their pendulums to all come into alignment?
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Re: It's About TIme

Postby JonB » February 27th, 2015, 3:50 pm

Thank you!
Gary, Barry, Harry, Darry, Larry, Kerry, and Mary return, but their pendulums have changed. Gary's is 1m, Barry's is 4m, Harry's is 9m, Darry's is 16m, Larry's is 25m, Kerry's is 36m, and Mary's is 49m. Their pendulums are on individual stands in a row such that they can all swing parallel to each other. They let go of their pendulums from the same angle at the same time. How long does it take for their pendulums to all come into alignment?
Is it anywhere around 842.5 seconds? 
I took the LCM of the relative lengths of the periods (1 through 7), 
found that it was 420 (indicating that after 420 swings of the shortest pendulum all the pendulums will be in their same start positions), 
and multiplied by the period of the shortest pendulum.

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Re: It's About TIme

Postby boomvroomshroom » February 27th, 2015, 5:02 pm

Thank you!
Gary, Barry, Harry, Darry, Larry, Kerry, and Mary return, but their pendulums have changed. Gary's is 1m, Barry's is 4m, Harry's is 9m, Darry's is 16m, Larry's is 25m, Kerry's is 36m, and Mary's is 49m. Their pendulums are on individual stands in a row such that they can all swing parallel to each other. They let go of their pendulums from the same angle at the same time. How long does it take for their pendulums to all come into alignment?
Is it anywhere around 842.5 seconds? 
I took the LCM of the relative lengths of the periods (1 through 7), 
found that it was 420 (indicating that after 420 swings of the shortest pendulum all the pendulums will be in their same start positions), 
and multiplied by the period of the shortest pendulum.
That totally reminds me of this video:
https://www.youtube.com/watch?v=7_AiV12XBbI

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Re: It's About TIme

Postby finagle29 » March 1st, 2015, 4:40 pm

Thank you!
Gary, Barry, Harry, Darry, Larry, Kerry, and Mary return, but their pendulums have changed. Gary's is 1m, Barry's is 4m, Harry's is 9m, Darry's is 16m, Larry's is 25m, Kerry's is 36m, and Mary's is 49m. Their pendulums are on individual stands in a row such that they can all swing parallel to each other. They let go of their pendulums from the same angle at the same time. How long does it take for their pendulums to all come into alignment?
Is it anywhere around 842.5 seconds? 
I took the LCM of the relative lengths of the periods (1 through 7), 
found that it was 420 (indicating that after 420 swings of the shortest pendulum all the pendulums will be in their same start positions), 
and multiplied by the period of the shortest pendulum.
Correct!

your turn
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Re: It's About TIme

Postby boomvroomshroom » March 5th, 2015, 4:41 pm

Reviving this thread...

Ron and Don both have pendulums of equal length. Ron is freezing his butt off at the North Pole, and Don is sweating his head off at the equator. Whose pendulum swings faster, and by approximately how much, taking the radius of the earth to be 6.38E6 meters?

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Re: It's About TIme

Postby elephantower » March 12th, 2015, 6:31 pm

The period at the poles should be around
99.6%
of that at the equator. I don't understand what you do with the radius though (I just used the International Gravity Formula).
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Re: It's About TIme

Postby boomvroomshroom » March 13th, 2015, 9:40 am

The period at the poles should be around
99.6%
of that at the equator. I don't understand what you do with the radius though (I just used the International Gravity Formula).
That works too. I wasn't thinking of that when I came up with this question (something to do with adding centripetal acceleration to normal acceleration) - but that's definitely a cleaner way to do this.


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