Transcript Forced Oscillations and Magnetic Resonance

Forced Oscillations and Magnetic
Resonance
A Quick Lesson in Rotational Physics:
TORQUE is a measure of how much a force acting on an object
causes that object to rotate.
Moment of Inertia, I, is the rotational analogue to mass.
Angular acceleration,  , is the second derivative of angular position.
Newton’s Second Law of Motion:
 F  ma  mx
Rotational Equivalent of Newton’s Second law:



I


Where:
 is the torque
I is the moment of inertia
 is the angular acceleration

Binnercoil

BHelmholtz

Diagram of the magnetic field due to both coils, along with the
angle associated with the angular displacement from equilibrium.
There are three forces acting on the compass
needle:
1) magnetic force due to Helmholtz coils
2) magnetic force due to inner coils
3) damping force due to friction between the
compass needle and the holding pin
Since we are dealing with rotational motion,
these forces are actually torques.
Constants and Variables
• B = magnetic
induction due to
Helmholtz coils
 b = damping constant
• F = amplitude of the
driving field
 w  the driving
frequency
 m  dipole moment of
the compass needle
• I = rotational inertia of
the compass needle
   angular
displacement from
equilibrium
Newton’s Second Law of Motion
(rotational)
net = driving field + restoring + damping
The torque due to the driving field is:
 drivingfield  F coswt
The torque due to the restoring field is:
 restoring  mB
The torque due to the damping force is:
 damping  mb
The Second Order Differential
Equation



I  F coswt  mB  mb
Which corresponds to Newton’s Second Law of Motion:
net = driving field + restoring + damping
Dividing through by I, the rotational inertia, and rearranging
gives:
mb
m
B
F
 
 
  coswt
I
I
I
then we let...
w1 
2
mb
I
w0 
2
mB
I
Substituting w02 and w12 into the differential
equation yields:
F



  w1   w 0   cos wt
I
2
2
We assume a particular solution:
 p  c1 coswt  c2 sin wt
Since we are dealing with a oscillatory function it makes sense
to assume the most general oscillatory solution involving the
two oscillatory functions, sin and cos.
Solving for c1 and c2 we get:
F
(w 0  w )( )
I
c1 
2
4
2 2
2
(w 0  w )  (w w 1 )
2
c2 
(w o
2
2
F
2
(ww1 )
I
4
2 2
2
 w )  (w w 1 )
Letting:
D  (w0  w )  w w1
2
2 2
This is the denominator of the c1 and c2 solutions
2
4
Further calculations lead us to the following equation:
p 
(w 0
2
F
F
2
 w )( )
(ww1 )
I coswt  I
sin wt
D
D
2
We can rewrite the particular equation, by
using some simple trigonometry:
 p  R cos(wt   )
where
2
2
F
F 2 4
(w o  w ) 2  2 w w 1
I
I
R
2
4 2
2 2
2
[w 0  w ]  w w 1
2

2 2

Through further calculations we can rewrite our
particular solution in terms of amplitude of the
driving field:
p 
F
z
cos(wt   )
where z  I (w0  w )  (mb ) w
2
2
2 2
2
2
We want to do this, in order to use our experimental data
directly in the equation. The ratio in front of the cosine function
is the amplitude of the compass needle.
Let’s introduce a new relationship,
E osc





z 
F
2
With any kind of wave motion the relationship of the oscillatory
energy is directly proportional to the square of the amplitude of
the motion.
F which is the amplitude of the driving field, remains constant,
thus the relationship
E osc
 F 


 z
2
can lead us to the conclusion:
Emax  z min
In other words, when z is at a minimum, the oscillatory energy
of the compass needle is at a maximum
-When we take the derivative of z and set it equal to
zero, that is when z is at a minimum.
-When z is at a minimum, E, oscillatory energy, is at
a maximum.
-When E is at a maximum, the deflection of the
compass needle is at a maximum, and we get
resonance.
dz
2
2
2
2
 0  2 I (w  w 0 )  ( mb )  0
dw
When solved for w2 will yield:
w  w0
2
2
( mb )

2
2I
2
Using the fact that
w  2f
We can make a substitution and come up with:
4 f 
2
2
mB
I

m b
2
2I
2
2
Rearranging and making the substitution
B
8m 0 Ni
125R
Gives us the final equation
f
2

2m 0 N
b
m m
*
i

*




2
2
8
125 R  I   I 
2
2
The following graph was generated from experimental data:
A plot of frequency squared vs. current, from experimental data
350
300
250
f2
200
150
100
50
0
-50
0.5
1
1.5
2
2.5
3
I
3.5
4
4.5
5
The graph is a line with the equation in the form of
f 2  Ai  B
Where f2 is the frequency measured, i is the current measured,
A is the slope of the graph which is directly proportional to the
ratio m / I , and B is the y-intercept which is directly proportional
to b 2
Thus, we now can determine from our experimental data that the
magnetic dipole moment-rotational inertia ratio to be
3.66 * 10^6, and the damping constant 2.03 * 10^-5 .
A resonant solution to our second order differential equation
8000
6000
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theta
2000
0
-2000
-4000
-6000
-8000
0
0.1
0.2
0.3
0.4
0.5
t
0.6
(Current used 1A)
0.7
0.8
0.9
1
A non-resonant solution to our second order differential equation
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2000
theta
1000
0
-1000
-2000
-3000
-4000
-5000
0
0.1
0.2
0.3
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t
0.6
(Current used 1A)
0.7
0.8
0.9
1
Another resonant solution to our second order D.E.
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2000
theta
1000
0
-1000
-2000
-3000
0
0.1
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0.4
0.5
t
0.6
(Current used 5A)
0.7
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1
Another non-resonant solution to our second order D.E.
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theta
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0
-500
-1000
-1500
-2000
0
0.1
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0.4
0.5
t
0.6
(Current used 5A)
0.7
0.8
0.9
1
The NMR for ethyl acetate, C4H8O2: