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

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

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

Width of Power Resonance Curve

(Full Width at Half Maximum (FWHM))

0

: FWHM

)(P

2Pm

mP

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

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

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

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

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

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

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

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

For an atom :

Hz10~ 150810~Q;

Hz10~ 7

8

0

10~

At 0A5000

05 A105

8

0

10~

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