Forced Oscillations and Magnetic Resonance

Preview:

DESCRIPTION

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. - PowerPoint PPT Presentation

Citation preview

Forced Oscillations and Magnetic Resonance

A Quick Lesson in Rotational Physics:

TORQUE is a measure of how much a force acting on an objectcauses 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:

Rotational Equivalent of Newton’s Second law:

xmmaF

I

Where: is the torque I is the moment of inertia is the angular acceleration

innercoilB

Diagram of the magnetic field due to both coils, along with the angle associated with the angular displacement from equilibrium.

HelmholtzB

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

= damping constant• F = amplitude of the

driving field the driving

frequency

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:

tFlddrivingfie cos

The torque due to the restoring field is:

Brestoring

The torque due to the damping force is:

damping

The Second Order Differential Equation

BtFI cos

net = driving field + restoring + damping

Which corresponds to Newton’s Second Law of Motion:

Dividing through by I, the rotational inertia, and rearranginggives:

tI

F

I

B

I cos

I

21 I

B 20

then we let...

Substituting and 1

2 into the differential equation yields:

tI

F cos20

21

We assume a particular solution:

tctcp sincos 21

Since we are dealing with a oscillatory function it makes senseto assume the most general oscillatory solution involving the two oscillatory functions, sin and cos.

Solving for c1 and c2 we get:

)()(

))((

41

22220

220

1

I

F

c

)()(

)(

41

2222

21

2

o

I

F

c

Letting:

41

22220 )( D

This is the denominator of the c1 and c2 solutions

Further calculations lead us to the following equation:

tD

I

F

tD

I

F

p

sin)(

cos))(( 2

122

0

We can rewrite the particular equation, byusing some simple trigonometry:

)cos( tRp

where

241

22220

41

22

2

2

2222

][

)(

I

F

I

F

Ro

Through further calculations we can rewrite our particular solution in terms of amplitude of the driving field:

)cos( tz

Fp

We want to do this, in order to use our experimental datadirectly in the equation. The ratio in front of the cosine functionis the amplitude of the compass needle.

222220

2 )()( Izwhere

Let’s introduce a new relationship,

2

z

FEosc

With any kind of wave motion the relationship of the oscillatoryenergy 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

2

z

FEosc

can lead us to the conclusion:

minmax zE

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.

0)()(20 220

22

Id

dz

When solved for will yield:

2

22

02

2

)(

I

Using the fact that

f 2

We can make a substitution and come up with:

2

2222

24

II

Bf

Rearranging and making the substitution

R

NiB

125

8 0

Gives us the final equation

2

22

2

02

8**

125

2

Ii

IR

Nf

The following graph was generated from experimental data:

0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-50

0

50

100

150

200

250

300

350A plot of frequency squared vs. current, from experimental data

I

f2

The graph is a line with the equation in the form of

BAif 2

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 , and B is the y-intercept which is directly proportionalto

I/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 .

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-8000

-6000

-4000

-2000

0

2000

4000

6000

8000A resonant solution to our second order differential equation

t

theta

(Current used 1A)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-5000

-4000

-3000

-2000

-1000

0

1000

2000

3000

4000

5000A non-resonant solution to our second order differential equation

t

theta

(Current used 1A)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-3000

-2000

-1000

0

1000

2000

3000Another resonant solution to our second order D.E.

t

theta

(Current used 5A)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-2000

-1500

-1000

-500

0

500

1000

1500

2000

Another non-resonant solution to our second order D.E.

t

theta

(Current used 5A)

The NMR for ethyl acetate, C4H8O2:

Recommended