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Offline syhprum

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« Reply #100 on: 25/06/2009 16:19:44 »
I take it you mean the shape of a path of length h*2^.5 a solution although not very elegant mathematically would be an animation program where the shape of the curve could be pushed about similar to gamma control animation.
When I was at Siemens Hell thirty years ago we had a Basic program where we had to land a Moon landing vehicle with limited fuel with blasts of power from its engines I see certain similarities, the solution was to expend half the fuel in the first burn, half of the remainder in the second and so on.
congratulations on your solution.
 

Offline syhprum

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« Reply #101 on: 25/06/2009 17:30:53 »
I calculate that if the object takes an "L" shaped path to traverse the same distance it takes 85.4% of the time the object on the inclined plane takes but to reach the same horizontal position takes 105.9% of the time.
The path length of the circular route is only 78.54 (pi/4) the length of the "L" shaped route that opens up the possibility that it could be quicker although it is 111.1% the length of the inclined plane route but the initial acceleration will be greater. 
« Last Edit: 25/06/2009 20:00:26 by syhprum »
 

Offline JP

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« Reply #102 on: 25/06/2009 20:24:13 »
That's a novel way of looking at it.
But does it show that the third option I suggested takes longer? It would suggest that, as the height is lost quicker, it would get there faster so the inequality wouldn't always always apply.
So what is the shape of fastest path from top to bottom?

I had posted a response, but then realized my math was off, so here goes again. 

The constraints on the problem are that total length traveled = R*Pi/2, and that the mass drop a distance of R total.  I think that your numbers are off in the last case you posted.  Here's what I get:
Step 1: The mass falls vertically the distance R.  The time for this is
tv=Sqrt[2]*Sqrt[R/g],
and the final velocity is
v=tv*g=Sqrt[2gR]

Step 2: The mass maintains its velocity and finishes the total distance horizontally.  The remaining distance is R*Pi/2-R=R*(Pi-2)/2.  The time to travel this distance is just
th=R*(Pi-2)/(2*tv)=R*(Pi-2)/(2*Sqrt[2*g*R])=(Pi-2)/[2*Sqrt(2)]*Sqrt[R/g]

The total time is then

t=tv+th={(Pi-2)/[2*Sqrt(2)]+Sqrt(2)}*Sqrt[R/g]~1.82*Sqrt[R/g].

This is the fastest of the three times, as it should be.
« Last Edit: 25/06/2009 23:19:37 by jpetruccelli »
 

Offline syhprum

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« Reply #103 on: 25/06/2009 20:39:15 »
But What is the problem, does it have to travel the same distance in its chosen path or does it have to reach the same position horizontaly as the body moving down the incined plane ?.
 

Offline syhprum

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« Reply #104 on: 25/06/2009 20:57:49 »
I have just stumbled upon the circuit diagram problem, was it deliberately intended that the diode was polarized in the the non conductive direction or was this an error?.
Was the purpose of the question just to test ones understanding of circuit drawing conventions as the arithmetic involved was negligible ?. 
 

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« Reply #105 on: 25/06/2009 22:23:52 »
Jp
I don't see how pi can come into an expression for a vertically falling body.
« Last Edit: 25/06/2009 22:46:42 by sophiecentaur »
 

Offline JP

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« Reply #106 on: 25/06/2009 23:18:39 »
From a height R falls (from zero speed) a mass point which follows a curved trajectory of 1/4 circumference (so the total lenght is π/2*R).
From the same height R falls, at the same instant of time, from zero speed, a mass point along a stright inclined plane which is long π/2*R as well.

I read the original problem to mean that all tracks were required to be R*Pi/2 in length, hence the Pi entering.
« Last Edit: 25/06/2009 23:20:38 by jpetruccelli »
 

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« Reply #107 on: 25/06/2009 23:48:01 »
I thought that the starts and destinations were the same.
No wonder we disagree!!
Ah well.
 

Offline erickejah

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« Reply #108 on: 26/06/2009 00:09:34 »
Is there any program like Multisim for physics?
 

Offline lightarrow

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« Reply #109 on: 26/06/2009 09:50:34 »

I had posted a response, but then realized my math was off, so here goes again. 

The constraints on the problem are that total length traveled = R*Pi/2, and that the mass drop a distance of R total.  I think that your numbers are off in the last case you posted.  Here's what I get:
Step 1: The mass falls vertically the distance R.  The time for this is
tv=Sqrt[2]*Sqrt[R/g],
and the final velocity is
v=tv*g=Sqrt[2gR]

Step 2: The mass maintains its velocity and finishes the total distance horizontally.  The remaining distance is R*Pi/2-R=R*(Pi-2)/2.  The time to travel this distance is just
th=R*(Pi-2)/(2*tv)=R*(Pi-2)/(2*Sqrt[2*g*R])=(Pi-2)/[2*Sqrt(2)]*Sqrt[R/g]

The total time is then

t=tv+th={(Pi-2)/[2*Sqrt(2)]+Sqrt(2)}*Sqrt[R/g]~1.82*Sqrt[R/g].

This is the fastest of the three times, as it should be.
Yes, this is the fastest path, you are correct.
 

Offline lightarrow

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« Reply #110 on: 26/06/2009 09:53:44 »
That's a novel way of looking at it.
But does it show that the third option I suggested takes longer? It would suggest that, as the height is lost quicker, it would get there faster so the inequality wouldn't always always apply.
So what is the shape of fastest path from top to bottom?
It is just the one you guessed, your third option; the inequality holds for that path, furthermore, h(s) is minimum in that case.
 

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« Reply #111 on: 26/06/2009 12:26:44 »
You are right, of course and it solves the problem, as presented.
I read it wrong but the fastest journey from A to B (i.e. displacement) is a more likely scenario to deal with than how quickly you can travel a certain distance.
I guess, if you had to drop a ball down a fixed length of tubing from a given height above ground, the original question would deal with that scenario.
 

Offline lightarrow

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« Reply #112 on: 26/06/2009 16:13:05 »
You are right, of course and it solves the problem, as presented.
I read it wrong but the fastest journey from A to B (i.e. displacement) is a more likely scenario to deal with than how quickly you can travel a certain distance.
I guess, if you had to drop a ball down a fixed length of tubing from a given height above ground, the original question would deal with that scenario.
Yes. Note (but probably you already know it) that if, instead, you have to find the quicker path between *two fixed points* A and B with the second at lower height than the first, it would be a different problem, known as the 'Brachistochrone' problem:
http://en.wikipedia.org/wiki/Brachistochrone_curve
which solution is an hyperbolic cosine a cycloid.

To solve this last problem we can use variation calculus and the resulting Lagrange equation. Only one function y(x) solve the equation (the cycloid).
I say this because I tried to use the same method for the problem I posted and I found that there is no solution to the resulting equation, that is no function h(s) for which the Functional F(h): h(s)-->T   (T = total time) is stationary.

For a normal function f(x), if it's never stationary, you can say that its maximum and minimum values (if they exist) must be at the border of the domain. But which is the border of the domain for the functional F(h), if we can use the same reasoning (not sure of it)?

Intuitively it seems that the function h(s) for which the functional is minimum is that one which graph is horizontal for a lenght (π/2 - 1)R and then vertical for a lenght R (kind of opposite of the quicker path). Does it mean that those two functions h(s) represent the 'border' of the domain of F(h)?
« Last Edit: 26/06/2009 19:59:45 by lightarrow »
 

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« Reply #113 on: 26/06/2009 18:25:13 »
You've gone and spoiled my evening now. I shall have to start thinking!
 

Offline Chemistry4me

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« Reply #114 on: 03/07/2009 05:21:40 »
An LCR series circuit is connected to an AC voltage source:




The voltages across all the components were measured and this data was obtained:




After looking at the data it was argued that the voltmeter was faulty because the inductor voltage was larger than the source voltage. Is the voltmeter faulty? Why? Why not?
 

Offline syhprum

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« Reply #115 on: 03/07/2009 08:21:17 »
Insufficient data is given to calculate frequency etc but this is quite normal behavior for a AC circuit incorporating capacitors and inductors, at the resonant frequency the impedance of the capacitor and inductor will cancel each other out and the current will be determined by the resistor while the voltage across the reactive components will be determined by their reactance times the current. 
It is assumed that the source impedence of the AC supply is zero.
« Last Edit: 03/07/2009 08:27:55 by syhprum »
 

Offline syhprum

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« Reply #116 on: 06/07/2009 11:40:32 »
Does no one want to calculate the phase angle or the "Q" ?.
 

Offline lightarrow

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« Reply #117 on: 06/07/2009 11:45:24 »
Does no one want to calculate the phase angle or the "Q" ?.
I have computed the quantities ωL/R = 75/36, ω2LC = 31/15, ωRC = (4/45)√481, but I made some error in the computations because the numbers are not consistent.

Edit: Found and fixed the error. The correct values are:
ωL/R = 25/12
ω2LC = 25/9
ωRC = 4/3
--> Q = (1/R)√(L/C) = 5/4
  (Exact values).
« Last Edit: 07/07/2009 10:24:46 by lightarrow »
 

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« Reply #118 on: 06/07/2009 12:23:24 »
Using phasor notation:
  The triangle of voltages has 7.2 real. 9.6, resultant of Xl - Xc. Square and add gives you (12)squared, which you'd expect.
The angle between input voltage and voltage across resistor is 53degees.

Say there's1A flowing, to make it easier.
I reckon that the resonant frequency will be 0.6 of the test frequency (that would be when 15ω = 5.4/ω, when Xl = Xc).
So we would know the impedance at resonance (7.2) and Modulus(impedance) at the test frequency (7.2+j9.6) and also the ratio of frequencies is 1:1.67.
That should allow you to work out the Q but I need to think about how.
Anyone else, please?

[Edit for clarity of maths]
« Last Edit: 06/07/2009 14:55:44 by sophiecentaur »
 

Offline syhprum

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« Reply #119 on: 07/07/2009 05:53:47 »
Given that the resonant frequency f1 equals the original freqency f0 times 0.6 (Sophiecentaur post 5643) the impedance of the inductor is now 0.6 of its previous value.
At f0 the ratio of resistance to impedance was 15 to 7.2 now it is 1.25 which is the "Q" value at this frequency.
Can we go further do we have sufficient data to work out the value of the resistor ?

Arithmetical error corrected!
« Last Edit: 07/07/2009 11:59:51 by syhprum »
 

Offline Chemistry4me

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« Reply #120 on: 07/07/2009 06:22:51 »
I think my head just exploded after all of that.
 

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« Reply #121 on: 07/07/2009 08:09:19 »
As we've assumed the current yo be 1A, the value of R is 7.2 ohms.
 

Offline Chemistry4me

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« Reply #122 on: 07/07/2009 08:23:17 »
Didn't we already know that R was 7.2 ohms?
 

Offline lightarrow

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« Reply #123 on: 07/07/2009 10:23:51 »
The exact value of Q = (1/R)√(L/C) is 5/4 (see post up).
 

Offline syhprum

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« Reply #124 on: 07/07/2009 11:54:45 »
How can we include √(L/C) in our calculations when we are told nothing about them ?.
We are told nothing about the value of R we can only assume a convinient value.
In my opinion the only information we can derive from the data given is the Q factor (1.25) and the phase angle of the current (53.13).
« Last Edit: 07/07/2009 19:49:40 by syhprum »
 

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