Download Waves & Oscillations Physics 42200 Spring 2014 Semester

Physics 42200
Waves & Oscillations
Lecture 6 – French, Chapter 3
Spring 2014 Semester
Matthew Jones
Damped Harmonic Motion
• Viscous damping: = −
• Differential equation:
(force proportional to velocity)
+
=
− ±
+
=
+
+
+
+
2
−4
=−
=0
=0
=0
2
±
4
−
• Three possible forms of the solution depend on the sign of
the discriminant.
Overdamped Case
( )
=
4
−
=
4
−
+!
>0
Underdamped Case
=
( )
4
−
=
sin
4
−
+!
<0
* = *+
cos
,
−
4
Critically Damped Case
( )
4
−
=0
=( +! )
RLC Circuits
Review of electricity and magnetism:
• Capacitors store energy in an electric field
• Resistors dissipate energy by heating
• Inductors store energy in a magnetic field
• Voltage
– Energy per unit charge
– SI units: Volt = J/C
• Current
– Charge passing a point in a circuit per unit time
– SI units: Ampere = C/s
Resistors
1( )
A
3
B
• Current flows from A to B in the direction indicated
by the arrow.
• Charges at point B have less energy than at point A
because some of their energy was dissipated as heat.
• Potential difference:
∆. = ./ − .0 = 1( )3
• Resistance, 3, is measured in ohms in SI units
Capacitors
A
• Potential difference:
+4
−4
9
B
4
∆. = ./ − .0 =
5
• No current can flow across the capacitor, so any charge that
flows onto the plates accumulates there:
4
=4 + 6 1
7
• Potential difference:
1
∆. = 6 1
5
• SI units for capacitance: farad
7
Inductors
1( )
A
<
B
• The inductor will establish a potential difference that
opposes any change in current.
– If 71 ⁄7 < 0 then .0 > ./
• magnetic field is being converted to energy
– If 71 ⁄7 > 0 then .0 < ./
• energy is being stored in the magnetic field
71
∆. = ./ − .0 = ;
7
• SI units for inductance: henry
Kirchhoff’s Loop Rule
• The sum of the potential differences around a loop in
a circuit must equal zero:
d
?
a
?
∆ABC
?
c
∆.=> + ∆.>? +
∆.?@ + ∆.@= = 0
?
b
Kirchhoff’s Loop Rule
9
D(E)
<
F
Sum of potential differences:
71
1
−; − 1 3 − 6 1
7
5
7 =0
Kirchhoff’s Loop Rule
9
D(E)
<
F
Sum of potential differences:
71
1
−; − 1 3 − 6 1
7
5
7 =0
Kirchhoff’s Loop Rule
9
D(E)
<
F
Sum of potential differences:
71
1
−; − 1 3 −
4 +6 1
7
5
7
=0
Initial charge, 4 , defines the initial conditions.
Kirchhoff’s Loop Rule
71
1
; +1 3+
4 +6 1 7 =0
7
5
Differentiate once with respect to time:
7 1
71 1
;
+3 + 1
=0
7
7
5
This is of the same form as the equation for a
damped harmonic oscillator:
7
7
+
+ ( )=0
7
7
Solutions
71 1
7 1
+3 + 1
;
7
7
5
Suppose 1
=
Then
@G
@
=
and
@ G
@
=0
=
Substitute into the differential equation:
1
;+ 3+
=0
5
True for any only if
H
I
; + 3 + = 0.
Solutions
Roots of the polynomial:
1
;+ 3+ =0
5
3
3
1
=− ±
−
2;
4;
;5
Define some new symbols:
1K
;5
= 3/;
=
Then the roots can be written
=− ±
2
4
−
Example
9
D(E)
<
F
• If ; = 2.2MN and 5 = 10O what is the
frequency of oscillations when 3 = 0?
• What is the largest value of 3 that will still allow
the circuit to oscillate?
Example
=
1K =
;5
1
2.2MN 0.01M
= 6.74 × 10T U
= 3/;
=− ±
2
4
−
• Critical damping:
4
−
=0
3
1
= =2
;
;5
;
2.2MN
3=2
=2
= 29.7Ω
5
0.01M
H
= 2.15WNX
Example
9
D(E)
<
F
; = 2.2MN
5 = 10O
3 = 2Ω
• Suppose the initial charge on the capacitor
was 10 nC… What voltage is measured across
3 as a function of time?
Example
• Calculate the discriminant:
2Ω
3
1
−
=
4;
;5 4 2.2MN
1
−
2.2MN 0.01M
<0
• The circuit will oscillate with frequency:
1
1
3
[=
−
= 1.07WNX
2\ ;5 4;
• Time constant:
3
2Ω
=
=
= 4.55 × 10] U
2 2; 2 2.2MN
• Current will be:
/ cos
1
=1
+^
H
Example
• Initial conditions:
1
=1
/
cos
+^
1 0 = 0 because the inductor produces a potential difference that
opposes the change in current.
Therefore, ^ = \/2
Initial potential across capacitor:
4 10O5
∆. = =
= 1._`
5 10O
Initial voltage across inductor:
71
∆. = ; = 1._`
7
71
1.
=
= 4.55 × 10] ⁄U = 1
7
2.2MN
1 =
.]]×H a //b
T.cd×H e /b
= 68
Example
g
(._` U)
• Current in circuit:
/
1
=1
sin
where ⁄2 = 4.55 × 10] U H and 1 = 68
• Potential difference across the resistor:
∆. = 1 3
/ sin
g
=g
where g = 136 ..
Time (seconds)