• Loud-soft-loud modulations of intensity are produced when waves of slightly different frequencies are superimposed.

• The beat frequency is equal to the difference frequency fbeat = | f1 - f2| 1 beat

Used to tune musical instruments to same pitch

LECTURE 9 Ch 16.7 BEATS Ch 115.7 Doppler Effect

BEATS

CP 52

Beats two interfering sound waves can make beat

Two waves with different frequency create a beat because of interference between them. The beat frequency is the differenceof the two frequencies.

2 1beatf f f

1 2

1 2 1 2

1 2 1 2

sin(2 ) sin(2 )

( ) ( )2 sin(2 )cos(2 )

2 2( ) ( )

2 cos(2 ) sin(2 )2 2

A f t A f t

f f f fA t t

f f f fA t t

Superimpose oscillations of equal amplitude, but different frequencies

Modulation of amplitude Oscillation at the average frequency

frequency of pulses is | f1-f2 |

BEATS

CP 527

1 2

2

f f

1 2

2

f f

beat 1 2f f f

BEATS – interference in time

Consider two sound sources producing audible sinusoidal waves at slightly different frequencies f1 and f2. What will a person hear? How can a piano tuner use beats in

tuning a piano? If the two waves at first are in phase they will interfere constructively and a large amplitude resultant wave occurs which will give a loud sound. As time passes, the two waves become progressively out of phase until they interfere destructively and it will be very quite. The waves then gradually become in phase again and the pattern repeats itself. The resultant waveform shows rapid fluctuations but with an envelope that various slowly.

The frequency of the rapid fluctuations is the average frequencies =

The frequency of the slowly varying envelope =

Since the envelope has two extreme values in a cycle, we hear a loud sound twice in one cycle since the ear is sensitive to the square of the wave amplitude. The beat frequency is

CP 527

0

10

20

30

40

50

60

0 0.05 0.1 0.15 0.2 0.25

time

f =100

f = 104

beats

CP 527

f1 = 100 Hz f2 = 104 Hz frapid = 102 Hz Trapid = 9.8 msfbeat = 4 Hz Tbeat = 0.25 s (loud pulsation every 0.25 s)

0

10

20

30

40

50

60

0 0.05 0.1 0.15 0.2 0.25

time

CP 527

f1 = 100 Hz f2 = 110 Hz frapid = 105 Hz Trapid = 9.5 msfbeat = 10 Hz Tbeat = 0.1 s (loud pulsation every 0.1 s)

0

10

20

30

40

50

60

0 0.05 0.1 0.15 0.2 0.25

time

f =100

f = 120

beats

CP 527

f1 = 100 Hz f2 = 120 Hz frapid = 110 Hz Trapid = 9.1 msfbeat = 20 Hz Tbeat = 0.05 s (loud pulsation every 0.05 s)

What is the physics of this image?

CP 495

DOPPLER EFFECT - motion related frequency changes

Doppler 1842, Buys Ballot 1845 - trumpeters on railway carriage

Source (s) Observer (o)

oo s

s

v vf f

v v

Applications: police microwave speed units, speed of a tennis ball, speed of blood flowing through an artery, heart beat of a developing fetous, burglar alarms, sonar – ships & submarines to detect submerged objects, detecting distance planets, observing the motion of oscillating stars.

CP 495

note: formula is very different to textbook

formula different to textbook

Doppler Effect

• Consider source of sound at frequency fs, moving speed vs, observer at rest (vo = 0)

• Speed of sound v • What is frequency fo heard by observer?

On right - source approaching• source catching up on waves• wavelength reduced• frequency increased

On left - source receding• source moving away from waves• wavelength increased• frequency reduced

CP 495

v = f

oo s

s

v vf f

v v

v (source) = 0 v (source) = v (wave)

v (source) = v (wave) / 2 v (source) = 1.25 v (wave)

CP 495

oo s

s

v vf f

v v

source vs observer vo observed frequency

fo

stationary stationary = fs

stationary receding < fs

stationary approaching > fs

receding stationary < fs

approaching stationary > fs

receding receding < fs

approaching approaching > fs

approaching receding ?

receding approaching ?

CP 595

oo s

s

v vf f

v v

Shock Waves – supersonic waves

b o a t m o v i n g t h r o u g h w a t e r : s p e e d o f b o a t > s p e e d o f w a t e r w a v e c r e a t e d

f a s t m o v i n g p o w e r b o a t

s a i l i n g b o a t r o c k e t “ v i o le n t l y ” -w a v e c r e s t s a d d t o g i v e “ la r g e w a v e ”

M a c h c o n e

s o n i c b o o o o m

s h o c k w a v e - b u n c h i n g o f w a v e f r o n t s - - - > a b r u p t i v e r i s e a n d f a l l o f a i r p r e s s u r e

p la n e t r a v e l l i n g a t s u p e r s o n ic s p e e d s

M a c h N u m b e r = v / v s v s s p e e d o f s o u n d i n a i r

e g M a c h N u m b e r = 2 . 3 s p e e d i s 2 . 3 t im e s t h e s p e e d o f s o u n d

CP 506

CP 506

Shock Waves – supersonic waves

Problem 9.1

A train whistle is blown by the driver who hears the sound at

650 Hz. If the train is heading towards a station at 20.0 m.s-1,

what will the whistle sound like to a waiting commuter? Take

the speed of sound to be 340 m.s-1.

[Ans: 691 Hz]

Problem 9.2

The speed of blood in the aorta is normally about 0.3000 m.s-1.

What beat frequency would you expect if 4.000 MHz ultrasound

waves were directed along the blood flow and reflected from the

end of red blood cells?

Assume that the sound waves travel through the blood with a

velocity of 1540 m.s-1.

Solution 9.2 I S E E

fs1 = 4.0 MHz = 4.0x106 Hz

v = 1.54x103 m.s-1 fo1 = ? Hz

fs2 = fo1fo2 = ? Hz

vs2 = 0.30 m.s-1

v01= 0.30 m.s-1

reflected wave

Doppler Effect Beats oo s

s

v vf f

v v

beat 2 1f f f

3

6 6o1o1 s1 3

s1

1.54 10 0.304 10 3.999221 10 Hz

1.54 10

v vf f

v v

3

6 6o2o2 s2 3

s2

1.54 103.999221 10 3.998442 10 Hz

1.54 10 0.30

v vf f

v v

Blood is moving away from source observer moving away from source fo < fs

Wave reflected off red blood cells source moving away from observer fo < fs

Beat frequency = | 4.00 – 3.998442| 106 Hz = 1558 Hz

In this type of calculation you must keep extra significant figures.

An ambulance travels down a highway at a speed of 33.5 m.s-1, itssiren emitting sound at a frequency of 4.00x102 Hz. What frequencyis heard by a passenger in a car traveling at 24.6 m.s-1 in the oppositedirection as the car and ambulance: (a) approach each other and (b) pass and move away from each others?

Speed of sound in air is 345 m.s-1.

(a) 2 345 24.6

(4.00 10 ) Hz 475 Hz345 33.5

OO S

S

v vf f

v v

(b) 2 345 24.6)(4.00 10 ) Hz 339 Hz

345 33.5O

O SS

v vf f

v v

Problem 8.3

Solution