Chapt 4 French (2) (2)
description
Transcript of Chapt 4 French (2) (2)
-
Power Input to a Driven
Oscillator in the Steady
State
vFP
tcosFF 0
)tsin()(Adt
dxv
Instantaneous power input to the oscillator
by the driving force :
)tsin(v0
-
22
0
0
000
Q
1
A)(Av
Resonance for velocity amplitude occurs
exactly at the natural frequency
)tsin(tcosvF)t(P 00
)tcossintsintcos(cosvF 200
T
0
dt)t(PT
1)(P sinvF
2
100
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Prob. 4.10 The power required to maintain
forced vibration must be equal to the power
loss due to damping.
a) Find the instantaneous rate of doing work
against the damping force.
Ans vF
dt
dWdamp
2vb
)t(sinAb 222
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Ans :
dt)t(sinT
1Ab
dt
WdT
0
222
22 Ab2
1
b) Find the mean rate of doing work
against damping
c) Substitute the value of A at any
arbitrary frequency and hence obtain the
expression for average P.
-
sinAF2
1sinvF
2
1P 000driveAns:
Since 220
tan
2tan1
tansin
22222
0 )(m
b
0F
Ab
22
drive Ab2
1P
2
2
0
0
2
0
22
0
1
)2(
Q
mbF
-
22Ab2
1)(P
2
2
0
0
2
0
22
0
1
)2(
Q
mbF
0 at maximum is P
Power Resonance Curve
2
2
0
0
2
m
Q
1
1
Q
P)(P
b
F
m
QbFPm
22
2
0
2
0
2
22
0
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Width of Power Resonance Curve
(Full Width at Half Maximum (FWHM))
0
: FWHM
)(P
2Pm
mP
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Equating 2
Pto)(P m
2
2
0
0
2
Q
1
1
Q
1
2
1
1Q
2
0
02
Q
1
0
0
Finding FWHM
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Putting where 1,10
Q
1)1(
1
1
Q2
1
Q
0
0
)(P
20 20
0Q
Q2
11
0
-
Q = 1
Q = 3
Q = 5
Q = 30
)(P
0
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Prob. 4.12 A mass of 2 kg is hung from a
spring that is extended by 2.5 cm. The
top end of the spring is oscillated up and
down with an amplitude 1 mm. The Q of
the system is 15.
a) What is for this system? 0
1
0 s20;m/N800k
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b) What is the amplitude of the oscillations
at ?
tydt
dy
dt
yd cos20
2
02
2
0
mm10
y
t cos0
Equation of motion is
2
00 mF
002
0
00 for1
mm
m
FA
mmQAA 15)( 00
-
22
0
0
2
2
0
2
2
0
0
2
1
1
21
1)(
Q
bQ
F
Q
Q
PP m
0
2
000
2
2
0
0
3
0
2
0 &1
1
2
mbQmAF
Q
Q
mA
Putting )02.0(10
2
2
3
0
2
0
Q
14
1
Q2
mAP
w11.0
c) What is the mean power input to maintain
an oscillation at 2% higher than ? 0
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Prob. 4.17 The graph
shows the mean power
absorbed by an
oscillator when driven
by a force of constant
magnitude but variable
frequency.
P, w
atts
5
10
)MHz(
1 0.995 1.005
a) At exact resonance, how much work per
cycle is being done against the resistive force?
s102Ts10 6016
0
0max/ TPcycleWork J5102
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b) At exact resonance, what is the total
mechanical energy of the oscillator? 0E
0
22
0m EAb2
1P
From the power resonance curve :
161001.0 s
mJ0.1P
E m0
14 s10
22
002
1 where AmE
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c) If the driving force is turned off, how long
does it take for the energy of the oscillator
to drop to ? 10 eE
t
0 eE)t(E
s101t 4
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Line Width of Atomic Spectral Lines
0
)(P
Q
0
Out of the broad spectrum of the incident
light, the atom absorbs appreciably, only
within the frequency range around 0
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For an atom :
Hz10~ 150810~Q;
Hz10~ 7
8
0
10~
At 0A5000
05 A105
8
0
10~
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Electrical Resonance Free Oscillations of LC Circuit
0VV LC
0C
q
dt
idL
Or, 01
2
2
qCtd
qdL
Comparing this with the equation of motion
of a SHO we get the correspondence :
C
1k;Lm
C
L
i
q -q
-
LC
10
Angular frequency of free oscillations :
Free Oscillations LCR Circuit
0VVV CRL
0C
qiR
dt
idL
)(cos 00 tqq
C
L
i
R
q -q
-
New correspondence :
L
R
02
2
C
q
dt
qdR
td
qdLOr,
)cos(20 teqq LtR
-
L
i
R ~ tcosVV 0
tcosVC
q
dt
qdR
td
qdL 02
2
Steady State Solution :
)tcos()(q)t(q 0
Forced Oscillations of LCR Circuit