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Physics 42200
Waves & Oscillations
Lecture 9 – French, Chapter 4
Spring 2014 Semester
Matthew Jones
Comments on the Assignment
• Question 1 – Primarily concerned with the
case when ≪ (weak damping).
– In this case ≈
and the total energy changes
much more slowly than the position,
.
• Question 2 – A graph will help a lot…
( )
Forced Oscillators and Resonance
+
+
=
• Natural oscillation frequency:
=
• Amplitude of steady-state oscillations:
⁄
=
−
cos
=
+
• Phase difference:
= tan$%
−
−
2
Resonance Phenomena
• Change of variables:
=
•
Natural oscillation frequency:
=
•
=
Amplitude of steady-state oscillations:
=
•
Strong damping force
Strong damping force
& =
−
/
−
Phase difference:
= tan
$%
1/&
−
2
=
1
+
&
1−
%/
1
2&
large
small &
Resonance Phenomena
• Steady state amplitude:
( ⁄ )
Q=5
Q=4
Q=3
Q=2
Q=1
Peak is near ⁄
/
≈ 1. The peak occurs at exactly ⁄
= 1.
Average Power
• The rate at which the oscillator absorbs energy is:
)*( ) =
)*( )
/2
2 &
−
1
1
+
&
& = 10
Full-Width-at-Half-Max:
&=5
./0 =
&=3
&=1
/
&
=
Examples
4
• Resonant RLC circuit:
2(3)
1
5
ℰ cos
89
7 +9
8
1 =
:+ < 9
;
8
The transformer does not play a role in the analysis of the circuit. It is just
a convenient way to isolate the driving voltage source from the part of the
circuit that oscillates.
Resonant Circuit
89
1 =
7 + 9 : + < 9 8 = ℰ cos
8
;
• Differentiate once:
8 9
89 1
7
+: + 9
= −ℰ sin
8
8
;
• Redefine what we man by “ = 0”:
8 9
89 1
7
+: + 9
= ℰ cos
8
;
8
• Change of variables:
8 9
89
ℰ
+
+
9
=
cos
8
8
7
Resonant Circuit
8 9
89
ℰ
+
+
9
=
cos
8
8
7
• Amplitude of steady state current oscillations:
/
=ℰ ;
%/
1
−
+
&
• Voltage across the capacitor:
9
= 9 ?@ =
;
• Amplitude of voltage oscillations measured across C:
/
=ℰ
%/
1
−
+
&
Actual Data
Voltage
• If we know that 7 = 235A/, can we estimate : and ;?
7 = 235A/
Peak voltage is at
BD EF = 290 /H
D EF = 2C BD EF
B = /2C
[Mhz]
Energy
%
• Energy stored on a capacitor is I = ;
• The graph of the stored energy is proportional to the
square of the voltage graph.
• If we defined 2∆ as the FWHM on the graph of
power vs frequency, then it will correspond to 1/ 2
of the peak voltage.
Voltage
Resonant Circuit
2∆B = 110 /H
= 2C ∆B = 6.91 × 10M N $%
7 = 235A/
But
B=
/2C
Peak position is
[Mhz]
D EF
=
= :/7 so we can find ::
: = 7 = 6.91 × 10M N $% 235A/
= 162Ω
%
P@
−
QR
; = 1.20nF
;=
P
%
R YR
STUVW X R
Lifetime of Oscillations
• Amplitude of a damped harmonic oscillator:
= [ $Q=/ cos
• Maximum potential energy:
1
\=
∝ [ $Q=
2
• After time = 1/ , the energy is reduced by the
factor 1/[.
• We call ^ = 1/ the “lifetime” of the oscillator.
Other Resonant Systems
Other Resonant Systems
Other Resonant Systems
Wire Bond Resonance
• Wire bonds in a magnetic field:
_
f
`
Lorentz force is a = b c 8ℓ × e
The tiny wire is like a spring.
A periodic current produces
the driving force.
Wire Bond Resonance
Resonance in Nuclear Physics
• A proton accelerated through a potential
difference gains kinetic energy g = [ :
i
Phys. Rev. 75, 246 (1949).
Resonance in Nuclear Physics
• In quantum mechanics, energy and frequency
are proportional:
I=ℏ
• A given energy corresponds to a driving force
with frequency .
• When a nucleus resonates at this frequency,
the proton energy is easily absorbed.
Nuclear Resonance
“Lifetime” is
defined in terms of
the width of the
resonance.
Resonance
• Resonances are the main way we observe fundamental
particles.