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Forced Oscillations and Nuclear Magnetic Resonance Home Page
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Forced Oscillations and Nuclear Magnetic Resonance
Dominika Pcion and Tallie Noble
May 19, 2000
Abstract
When analyzing forced oscillations and nuclear magnetic resonance,
both math and physics play a signicant role. Thus, during this analysis
we will be referring to an experiment that is easily done in the labora-
tory. A pair of Helmholtz coils is set up. By denition, when current is
run through the coils, a magnetic eld is produced. This magnetic eld is
uniform because the coils are separated by a distance that is equal to the
radius of the coils. Within the Helmholtz coils, another coil is placed in
such a fashion as to produce a magnetic eld that is perpendicular to that
of the Helmholtz coils. Inside the inner coil a compass rests, acting like a
magnetic dipole, and it attempts to adjust itself to the two perpendicular
elds, as shown in
Figure 1
. This attempt causes an oscillatory deection
of the compass needle. By varying the frequency, the resonant state for the
compass needle can be found. Home Page
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BHelmholtz
BInnercoil Figure 1: Diagram of the magnetic eld due to both coils, along with the angle
associated with the angular displacement of the compass needle from equilibrium Home Page
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1.
The Physics of Forced Oscillations and Magnetic Res-
onance
Before starting the analysis, it is most helpful to review a few basic physics
relationships. We begin with Newtons second law for rotational motion,
= I
(1)
where the sum of the torques is equal to the product of I, the rotational inertia
of the compass needle, and the angular acceleration. We dene to be the
angular displacement of the compass needle from equilibrium.
There are three torques acting on the compass needle. One torque is mag-
netic, due to the Helmholtz coils. There is another magnetic torque due to the
inner coils. Lastly, there is the damping torque due to the friction between the
compass needle and the holding pin. Now, the left hand side of Newtons second
law can be rewritten as: net
=
driving eld
+
restoring
+
damping
(2)
The torque due to the driving eld is driving eld
= F cos t,
(3)
where F is the amplitude of the driving eld and is the driving eld frequency,
and t represents time.
Equation 3
holds true because the driving eld torque
is caused by a sinusoidal wave input, which in the experiment was generated
by a power amplifying device. This sinusoidal input is modeled by the cosine
function. Home Page
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The restoring torque can be expressed as restoring
= µB.
(4)
where µ is the magnetic dipole moment of the compass needle, and B is the
magnetic induction due to the Helmholtz coils. By denition the restoring torque
is the cross product of µ and B, which can be rewritten as restoring
= µ × B = |µ||B| sin .
(5)
Making use of the small angle approximation, sin ,
Equation 5
becomes restoring
= µB.
(6)
The damping torque is dened to be damping
= µ .
(7)
Based on previous research, the assumption that damping
µ
(8)
can be made. Thus, the damping torque is dependent on the angular veloc-
ity and magnetic dipole moment. In
Equation 7
, represents the constant of
proportionality, which is also known as the damping constant.
Now Newtons second law can be rewritten in the form of a second order
dierential equation.
I = F cos t µB µ
(9) Home Page
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In the above equation, the quantity µB is negative because the restoring torque
is opposite the displacement. In other words, it tries to compensate for the
positive change caused by the driving eld torque, F cos t. The term µ is
negative because it is a damping torque, which always opposes motion. A clear
correspondance can be seen between Newtons second law and the second order
dierential equation. Home Page
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2.
Mathematical Analysis
The rst thing is to divide the dierential
Equation 9
through by I, the rotational
inertia. This yields
= F
I cos t
µB
I
µ
I .
(10)
Next, to simplify the equation, we will make the following two substitutions: 2
1
= µ
I
(11) 2
0
= µB
I
(12)
Substituting
Equation 11
and
Equation 12
into
Equation 10
and rearranging
gives,
+
2
1
+
2
0
= F
I cos t.
(13)
The above equation can be solved by rst assuming a particular solution, p
= A cos t + B sin t
(14)
Since we are dealing with an oscillatory function, it makes sense to assume the
most general solution involving the two oscillatory functions, sine and cosine.
Taking the rst and second derivatives gives, p
= A sin t + B cos t,
(15) p
= A
2
cos t B
2
sin t.
(16) Home Page
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Next we substitute the above derivatives along with the assumed particular
solution into
Equation 13
and this yields:
A
2
cos t B
2
sin t +
2
1
(A sin t + B cos t)
+
2
0
(A cos t + B sin t) = F
I cos t
(17)
Combining like terms
cos t(A
2
+ B
2
1
+
2
0
A)
+ sin t(B
2
A
2
1
+ B
2
0
) = F
I cos t + 0 sin t,
(18)
allows us to equate the coecients of
Equation 18
in order to solve for A and
B.
A
2
+ B
2
1
+
2
0
A = F
I
(19)
B
2
A
2
1
+ B
2
0
= 0
(20)
Cramers rule can be used to solve for A,
A =
F/I 2
1
0 2
0

2 2
0

2 2
1 2
1 2
0

2
=
(
2
0

2
)F/I
(
2
0

2
)
2
+
2 4
1
,
and B,
B = 2
0

2
F/I 2
1
0 2
0

2 2
1 2
1 2
0

2
=
(F/I)
2
1
(
2
0

2
)
2
+
2 4
1 Home Page
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To simplify things, let D equal the denominator of A and B.
D = (
2
0

2
)
2
+
2 4
1
(21)
Substituting A and B back into our assumed particular solution,
Equation 14
,
yields, p
= (
2
0

2
)
D
F
I cos t + 2
1
D
F
I sin t
(22)
Next we use the following trigonometric identity,
y sin + x cos =
y
2
+ x
2
cos( )
(23)
to simplify our particular solution. Between
Equation 22
and
Equation 23
we
see the following correspondences.
= t
y =
F
I 2
1
(
2
0

2
)
2
+
2 4
1
x =
F
I 2
0

2 2
0

22
+
2 4
1
Referring to
Figure 2
we can see that the following are true,
= tan
1
y
x
R =
x
2
+ y
2 Home Page
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x
y
x
2
+ y
2 Figure 2: Using right triangle trigonometry.
Using the above conclusions we can rewrite our particular solution from
Equa-
tion 22
in terms of cosine only, p
= R cos(t )
(24)
where R is equal to:
R =
F
2
1
/I
(
2
0

2
)
2
+
2 4
1
2
+
F (
2
0

2
)/I
(
2
0

2
)
2
+
2 4
1
2
(25)
We can rewrite this complex looking equation, by performing some simple alge-
bra, not shown here.
R =
F
I
2
(
2
0

2
) + I
2 2 4
1
(26) Home Page
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Furthermore, we want to substitute µ/I back in for
2
1
, in order to have an
equation containing the damping constant, .
I
2 2 4
1
= I
2 2
µ
I
2
=
2
µ
2 2
(27)
Next we want to simplify what is under the radical in
Equation 26
by setting it
equal to z,
z = I
2 2
0

22
+
2
µ
2 2
(28)
We now want to substitute z into
Equation 26
,
R = F
z
(29)
When we substitute R into our particular solution,
Equation 24
, we get the
following result, p
= F
z cos(t )
(30)
This is a useful form of the particular solution because it allows for the experi-
mental data to be used directly in the equation. The ratio in front of the cosine
function in
Equation 30
corresponds to the amplitude of the deection of the
compass needle.
Our analysis deals with oscillatory functions, and with any kind of wave mo-
tion, the oscillatory energy is directly proportional to the square of the amplitude
of the motion. Thus the following is true,
E
osc F
z
2
(31) Home Page
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Therefore, the maximum energy of the oscillations will correspond to a minimum
value of z, due to the fact that F , the amplitude of the driving eld, is xed,
E
max
z
min
(32)
In order to get a minimum value of z, the derivative must be taken and set equal
to zero,
dz
d = 0
(33)
Thus
2I
2
(
2
0

2
)(2) + 2µ
2 2
= 0
4I
2
(
2
0

2
) + 2µ
2 2
= 0
4I
2 2
0
+ 4I
2 3
+ 2µ
2 2
= 0
2(2I
2 2
0
+ 2I
2 2
+ µ
2 2
) = 0
2I
2
(
2
0

2
) + µ
2 2
= 0
2I
2
(
2

2
0
) + µ
2 2
= 0
2I
2
(
2

2
0
) + (µ)
2
= 0
Solve the last equation for
2
. 2
=
2
0
(µ)
2
2I
2
(34)
By denition, we know that the magnetic eld due to the Helmholtz coils is
given by
B =
8µ
0
N i
125
2
r ,
(35)