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#24 | |
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Romulan Interpreter
Jun 2011
Thailand
226138 Posts |
Quote:
Last fiddled with by LaurV on 2012-09-15 at 06:10 |
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#25 | |
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Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88
722110 Posts |
Quote:
A problem I turned in last week and got back yesterday: Code:
2. As a Spokane police officer tools down Interstate 90 after confiscating Mrs. Rumpdock's XK8, he realizes that the vehicle needs new shock absorbers. When the Jag hits a pothole, it oscillates up and down for the rest of the trip, apparently moving under the influence of the potential |
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#26 | |
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"Forget I exist"
Jul 2009
Dumbassville
26×131 Posts |
Quote:
Last fiddled with by science_man_88 on 2012-09-15 at 13:18 |
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#27 |
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"Lucan"
Dec 2006
England
2×3×13×83 Posts |
Assuming the TEX means U(y) = U0(y-2 - y-1) we get
U'(y)/U0 = -2y-3 + y-2 U"(y)/U0 = 6y-4 - 2y-3 U'(y) = 0 when y = 2 U"(2) = U0/8 From the Taylor expansion, U(x) ~ U(2) + U0x2/16 where x = y-2 Conserving energy, v2 = U0(a2 - x2)/8 This is a standard SHM equation, with omega2 = U0/8 David |
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#28 |
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Aug 2010
Kansas
10438 Posts |
Hey Dubs-
How many of your MA453 classmates have you referred to this board and/or mersenne.org (or other affiliated sites)? Think of all the fancy computers that parents may have bought them!!! Best Wishes! |
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#29 | |
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Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88
1C3516 Posts |
Quote:
It's just a lecture. |
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#30 |
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"Lucan"
Dec 2006
England
647410 Posts |
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#31 |
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Sep 2012
116 Posts |
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#32 |
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"Mike"
Aug 2002
25·257 Posts |
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#33 | |
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Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88
3·29·83 Posts |
Quote:
(How exactly did you find this? To be clear, I did have the solution before turning it in last week. Edit: That parabolic coordinates stuff looks pretty tough... but I've got a few days yet to work on it.)
Last fiddled with by Dubslow on 2012-09-16 at 05:51 |
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