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Old 2009-02-01, 09:52   #1
davieddy
 
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"Lucan"
Dec 2006
England

145128 Posts
Default Waves on a taut string

At the second attempt, I succeeded in registering
with "Physics Forums" (where I probably belong)

After two homework questions about waves on a string,
and searching for the easiest way of explaining the formula
for their speed, I recalled this (original from my teaching days
although it must have been familiar long before):

Consider the flexible string under tension T passing through a
fixed wiggly tube at speed v.
At any point, the tube's wiggly curve has a well-defined centre of curvature,
and associated radius r.
The acceleration of the string is v^2/r, and the force due to tension T is
T*dl/r where dl is the length of an element of string. Both directed towards the
centre of curvature.
If T*dl/r = m*dl*v^2/r (m being mass/unit length) then there is no need for
the tube!
v^2 = T/m and there is no restriction on the waveform.
*******************************************
And just for good measure, someone was asked to derive
the moment of inertia of a rectangle about a corner by integration
so I volunteered (after some previous help about double integration
and the || axis theorem):

Consider any plane shape and scale it up by two, keeping the mass
per unit area the same. Each "element" has quadrupled in mass and
doubled its distance from the C of M. It follows that the moment of
inertia about an axis through the C of M normal to the plane has gone up
by a factor of 16.
For the a x b rectangle, let I be the moment of inertia about the C of M
and J that about a corner.
The || axis theorem gives J = I + m(a^2 + b^2)/4
Now consider the double sized rectangle formed from placing 4 of the
originals together. We deduce that
16I = 4J = 4I + m(a^2 + b^2)
*****************************

Anything to get rid of that inane "Beam up a Scottie" jest.

David

Last fiddled with by davieddy on 2009-02-01 at 10:01
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Old 2009-02-01, 11:21   #2
Flatlander
I quite division it
 
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"Chris"
Feb 2005
England

31·67 Posts
Default

Quote:
Originally Posted by davieddy View Post
At the second attempt, I succeeded in registering
with "Physics Forums" (where I probably belong)

After two homework questions about waves on a string,
and searching for the easiest way of explaining the formula
for their speed, I recalled this (original from my teaching days
although it must have been familiar long before):

Consider the flexible string under tension T passing through a
fixed wiggly tube at speed v.
At any point, the tube's wiggly curve has a well-defined centre of curvature,
and associated radius r.
The acceleration of the string is v^2/r, and the force due to tension T is
T*dl/r where dl is the length of an element of string. Both directed towards the
centre of curvature.
If T*dl/r = m*dl*v^2/r (m being mass/unit length) then there is no need for
the tube!
v^2 = T/m and there is no restriction on the waveform.
*******************************************
And just for good measure, someone was asked to derive
the moment of inertia of a rectangle about a corner by integration
so I volunteered (after some previous help about double integration
and the || axis theorem):

Consider any plane shape and scale it up by two, keeping the mass
per unit area the same. Each "element" has quadrupled in mass and
doubled its distance from the C of M. It follows that the moment of
inertia about an axis through the C of M normal to the plane has gone up
by a factor of 16.
For the a x b rectangle, let I be the moment of inertia about the C of M
and J that about a corner.
The || axis theorem gives J = I + m(a^2 + b^2)/4
Now consider the double sized rectangle formed from placing 4 of the
originals together. We deduce that
16I = 4J = 4I + m(a^2 + b^2)
*****************************

Anything to get rid of that inane "Beam up a Scottie" jest.

David
You're just stating the obvious.
Any more Krankies' jokes?

Last fiddled with by Flatlander on 2009-02-01 at 11:33
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Old 2009-02-01, 18:19   #3
davieddy
 
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"Lucan"
Dec 2006
England

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Quote:
Originally Posted by Flatlander View Post
You're just stating the obvious.
Any more Krankies' jokes?
:)

I thought making things obvious was a goal of teaching,
so I shall optimistically treat this as a compliment.

Although "any" waveform can travel in one direction on
an ideal taut string, I would guess that the principle of
superposition goes to pot, and if it met another wiggle
coming in the opposite direction, they would not pass
through each other unscathed like ideal waves do.

Last fiddled with by davieddy on 2009-02-01 at 18:20
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Old 2009-02-01, 18:39   #4
cheesehead
 
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"Richard B. Woods"
Aug 2002
Wisconsin USA

22·3·641 Posts
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Quote:
Originally Posted by davieddy View Post
Although "any" waveform can travel in one direction on an ideal taut string, I would guess that the principle of superposition goes to pot, and if it met another wiggle coming in the opposite direction, they would not pass through each other unscathed like ideal waves do.
Hmm ... In my frosh physics lab, we used a wire strung between the lab and another building. I seem to recall that one thing we did was to pluck it once to start a wave, then pluck it again to make a second wave cross the first one coming back (easy to see as we sighted along the wire's length). I think they canceled (approximately) when they met.

Last fiddled with by cheesehead on 2009-02-01 at 18:57
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Old 2009-02-01, 19:32   #5
davieddy
 
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"Lucan"
Dec 2006
England

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Quote:
Originally Posted by cheesehead View Post
Hmm ... In my frosh physics lab, we used a wire strung between the lab and another building. I seem to recall that one thing we did was to pluck it once to start a wave, then pluck it again to make a second wave cross the first one coming back (easy to see as we sighted along the wire's length). I think they canceled (approximately) when they met.
Cancelled displacement-wise, but the velocity was there, and so
did you also witness their "miraculous" reincarnation after they had
passed each other?

PS the number of hours I spent animating f(x-ct) - f(-(x+ct)) (or something like it)
on a blackboard doesn't really bear thinking about!

Last fiddled with by davieddy on 2009-02-01 at 20:23
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Old 2009-02-01, 20:37   #6
davieddy
 
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"Lucan"
Dec 2006
England

2·3·13·83 Posts
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Quote:
Originally Posted by Flatlander View Post
You're just stating the obvious.
Any more Krankies' jokes?
The "homework" section of Physics Forums is ludicrously busy,
and depressingly "formula" oriented.
The recipient of my wisdom was having difficulty reconciling
v = frequency*wavelength with v = SQR(T/m).

Last fiddled with by davieddy on 2009-02-01 at 20:44
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Old 2009-02-01, 22:19   #7
Flatlander
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"Chris"
Feb 2005
England

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I realised some time after posting that my post looked ambiguous, almost aggresive. I apologize; I was just teasing you about your joke.
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Old 2009-02-02, 01:22   #8
cheesehead
 
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"Richard B. Woods"
Aug 2002
Wisconsin USA

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Quote:
Originally Posted by davieddy View Post
Cancelled displacement-wise, but the velocity was there, and so did you also witness their "miraculous" reincarnation after they had passed each other?
Don't remember.
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Old 2009-02-05, 05:10   #9
davieddy
 
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"Lucan"
Dec 2006
England

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Quote:
Originally Posted by Flatlander View Post
I realised some time after posting that my post looked ambiguous, almost aggresive. I apologize; I was just teasing you about your joke.
I prefaced my post with a smily too.
Ever since our real time corresespondence about that
spectacular Lunar Eclipse I have regarded you as a "friend".

My post possibly deserved som putting down, and I can take
it from you.

I'd just like to add that a fellow "mentor" referred to my
C of M derivation as "elegant reasoning".

David
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