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Old 2012-09-15, 01:06   #23
Dubslow
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So you all, go look at the PHYS 325 description; this is my professor.
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Old 2012-09-15, 06:07   #24
LaurV
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Quote:
On the door to his university office, Gollin has taped the Randocks’ prison inmate numbers
Huh! Genial! Taking down my hat and bowing to the floor! I would like to nominate that guy for something like Nobel award in education! This reminds me about the huge collection of idiots in our (Romanian) parliament who bought diplomas, translated foreign papers and published as their own, etc. There was (and still is) a big scandal about it. We need more guys like that Gollin. Again, my respect! You are very fortunate to be there and learn from such people (no joke intended!).

Last fiddled with by LaurV on 2012-09-15 at 06:10
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Old 2012-09-15, 06:27   #25
Dubslow
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Quote:
Originally Posted by LaurV View Post
Huh! Genial! Taking down my hat and bowing to the floor! I would like to nominate that guy for something like Nobel award in education! This reminds me about the huge collection of idiots in our (Romanian) parliament who bought diplomas, translated foreign papers and published as their own, etc. There was (and still is) a big scandal about it. We need more guys like that Gollin. Again, my respect! You are very fortunate to be there and learn from such people (no joke intended!).
Mrs. Rumpdock is one of the main villains in our homework problems.

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

U(y)=U_0\left[\frac{1}{y^2}-\frac{1}{y}\right]

(a) For what value of y is the potential a minimum?
(b) Assuming the XK8 has mass m, calculate the (angular) frequency \omega for 
small oscillations about the minimum of the potential.

Copyright © 2012 George Gollin
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Old 2012-09-15, 13:12   #26
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Quote:
Originally Posted by Dubslow View Post
Mrs. Rumpdock is one of the main villains in our homework problems.

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

U(y)=U_0\left[\frac{1}{y^2}-\frac{1}{y}\right]

(a) For what value of y is the potential a minimum?
(b) Assuming the XK8 has mass m, calculate the (angular) frequency \omega for 
small oscillations about the minimum of the potential.

Copyright © 2012 George Gollin
if I knew what the brackets mean't it doesn't seem too hard, \frac{1}{y^2}-\frac{1}{y} goes to \frac{1}{y^2}-\frac{y}{y^2} or \frac{1-y}{y^2} wither this is mean't to be high or low depends on what the brackets are meant to be used for since it's that value times U_0 that needs to be the minimum if we assume the brackets are just there for for appearance the value of y^2 grows as y grows and 1-y becomes negative so you end up in negative fractions as y grows since fractions decrease in negative nature as y grows y=2 is the minimum I believe if we say the brackets are absolute values then y=1 should be the minimum since 1-y = 0 and the result would be 0.

Last fiddled with by science_man_88 on 2012-09-15 at 13:18
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Old 2012-09-15, 23:17   #27
davieddy
 
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Default Physics lesson for kids

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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Old 2012-09-16, 02:46   #28
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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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Old 2012-09-16, 03:05   #29
Dubslow
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Quote:
Originally Posted by davieddy View Post
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
Yes of course, but I didn't post the answers for a reason.
Quote:
Originally Posted by c10ck3r View Post
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!
It's just a lecture.
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Old 2012-09-16, 03:42   #30
davieddy
 
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Quote:
Originally Posted by Dubslow View Post
Yes of course, but I didn't post the answers for a reason.
I was trying to tell sm88 to stop playing fast and loose with division by zero
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Old 2012-09-16, 04:15   #31
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Quote:
Originally Posted by Dubslow View Post
Yes of course, but I didn't post the answers for a reason.

It's just a lecture.
You could always come see me to ask about an approach to the problem-- GG
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Old 2012-09-16, 05:08   #32
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Old 2012-09-16, 05:46   #33
Dubslow
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Quote:
Originally Posted by gollin View Post
You could always come see me to ask about an approach to the problem-- GG
Why hello professor (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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