University Physics: Waves and Electricity

University Physics: Waves and Electricity

Ch17. Longitudinal WavesLecture 5

Dr.-Ing. Erwin Sitompulhttp://zitompul.wordpress.com

2013

5/2Erwin Sitompul University Physics: Wave and Electricity

Homework 4: Two SpeakersTwo speakers separated by distance d1 = 2 m are in phase. A listener observes at distance d2 = 3.75 m directly in front of one speaker. Consider the full audible range for normal human hearing, 20 Hz to 20 kHz. Sound velocity is 343 m/s.

(a) What is the lowest frequency fmin,1 that gives minimum signal (destructive interference) at the listener’s ear?

(b) What is the second lowest frequency fmin,2 that gives minimum signal?

(c) What is the lowest frequency fmax,1 that gives maximum signal (constructive interference) at the listener’s ear?

(d) What is the highest frequency fmax,n that gives maximum signal?


2 2 23 1 2( ) ( ) ( )d d d

3d 2 23 (2) (3.75)d 4.25 m

3 2L d d 4.25 3.75 0.5 m

sound ,v f sound 343 m sv

Solution of Homework 4: Two Speakers



(b) What is the second lowest frequency fmin,2 that gives minimum signal?

min 686 Hz 0.5,1.5,2.5,f

min,2 686 Hz 1.5f 1029 Hz

(a) What is the lowest frequency fmin,1 that gives minimum signal (destructive interference) at the listener’s ear? Fully destructive

interference

0.5,1.5, 2.5,L

sound min

0.5,1.5,2.5,Lv f

soundmin 0.5,1.5,2.5,vf

L

min,1 686 Hz 0.5f 343 Hz

343 0.5,1.5,2.5,0.5

686 Hz 0.5,1.5, 2.5,



(c) What is the lowest frequency fmax,1 that gives maximum signal (constructive interference) at the listener’s ear? Fully constructive

interference

0,1,2,L

sound max

0,1,2,Lv f

soundmax 0,1,2,vf

L

max,1 686 Hz 1f 686 Hz

343 0,1, 2,0.5

686 Hz 0,1,2,

(d) What is the lowest frequency fmax,1 that gives maximum signal (constructive interference) at the listener’s ear?max 686 Hz 0,1, 2,f

max, 686 Hz 29nf 19894 Hz Highest constructive frequency that still can be listened by human, < 20 kHz


Beats If two sounds whose

frequencies are nearly equal reach our ears simulta-neously, what we hear is a sound whose frequency is the average of the two combining frequencies.

We also hear a striking variation in the intensity of this sound –it increases and decreases in slow, wavering beats that repeat at a frequency equal to the difference between the two combining frequencies.


Beats Let the time-dependent variations of the displacements due to

two sound waves of equal amplitude sm be1 1 1( , ) cos( )ms x t s k x t

2 2 2( , ) cos( )ms x t s k x t

From superposition principle, the resultant displacement is:1 1 2 2( , ) cos( ) cos( )m ms x t s k x t s k x t

1 12 2cos cos 2cos ( )cos ( )

2 cos( ) cos( )2 2mks kx t x t

1 2k k k

11 22 ( )

11 22 ( )k k k k

1 2

2 cos( ) cos( )2 2mks x t kx t

Amplitude modulation, depends on

Δk/2 and Δω/2

Oscillating term, a traveling wave,

depends on k and ω


( , ) 2 cos( ) cos( )2 2mks x t s x t kx t

Beats

cos( )2 2k x t

1 2ampl 2 2

beat ampl 1 22f f f f

1 2ampl 2 2

f fff

In 1 amplitude cycle, we will hear 2 beats (maximum amplitude magnitude)

The "beat" wave oscillates with the frequency average, and its amplitude varies according to the frequency difference


Example: BeatsThe A string of a violin is not properly tuned. Beats at 4 per second are heard when the string is sounded together with a tuning fork that is oscillating accurately at concert A (440 Hz). (a) What are the possible frequencies produced by the string?

(b) If the string is stretched a little bit more, beats at 5 per second are heard. Which of the possible frequencies are the the frequency of the string?

beat 1 2f f f

beat string forkf f f

string4 440f string 436 Hz or 444 Hzf

A string is stretched tighter The frequency will be higher The frequency of beats increases The frequency difference

increases If the string frequency becomes higher and its difference to

440 Hz increases The frequency of the string is 444 Hz.


The Doppler Effect The Doppler Effect deals with the relation between motion and

frequency. The body of air is taken as the reference frame. We measure the speeds of a sound source S and a sound

detector D relatif to that body of air. We shall assume that S and D move either directly toward or

directly away from each other, at speeds less than the speed of sound (vsound = 343 m/s).


The Doppler Effect: D Moving S Stationary If the detector moves toward the source, the number of

wavefronts received by the detector increased. The motion increases the detected frequency.

If the detector moves away from the source, the number of wavefronts received by the detector decreased. The motion decreases the detected frequency.


The Doppler Effect: S Moving D Stationary If the source moves toward the detector, the wavefronts is

compressed. The number of wavefronts received by the detector increased. The motion increases the detected frequency.

If the source moves away from the detector, the distance between wavefronts increases. The number of wavefronts received by the detector decreased The motion decreases the detected frequency.


The Doppler Effect The emitted frequency f and the detected frequency f’ are

related by:D

S

v vf fv v

where v is the speed of sound through the air, vD is the detector’s speed relative to the air, and vS is the source’s speed relative to the air.

D

S

v vf fv v

+ The detector moves toward the source

– The detector moves away from the source

– The source moves toward the detector

+ The source moves away from the detector


The figure indicates the directions of motion of a sound source and a detector for six situations in stationary air. For each situation, is the detected frequency greater than or less than the emitted frequency, or can’t we tell without more information about the actual speeds?

Greater

Checkpoint

(a)zero speed

(b)zero speed

(c)

(d)

(e)

(f)

Source Detector Source Detector

Less

Need more information

Need moreinformation

Greater

Less


Example: The Doppler EffectA toy rocket flies with a velocity of 242 m/s toward a mast while emitting a roaring sound with frequency 1250 Hz. The sound velocity is 343 m/s.(a) What is the frequency heard by an observer who

is standing at the mast?

(b) A fraction of the soundwaves is reflected by the mast and propagates back to the rocket. What is the frequency detected by a detector mounted on the head of the rocket?

242 m s, toward Sv

D

0Dv

D

S

v vf fv v

343 01250343 242

4245 Hz

0Sv

1250 Hzf

4245 Hzf

242 m s, toward Dv

S

D

S

v vf fv v 0

343 2424245343

7240 Hz


Supersonic Speeds

vsource = vsound(Mach 1 - sound barrier)

vsource > vsound(Mach 1.4 - supersonic)

sound

Mach Number vv


Homework 5: Ambulance SirenAn ambulance with a siren emitting a whine at 1600 Hz overtakes and passes a cyclist pedaling a bike at 8 m/s. After being passed, the cyclist hears a frequency of 1590 Hz.(a) How fast is the ambulance moving?(b) What frequency did the cyclist hear before being overtaken

by the ambulance?

Illustration only• Concorde, the supersonic

turbojet-powered supersonic passenger airliner

• Average cruise speed Mach 2.02 or about 2495 kmh


Homework 5A: Bat and Insect

2. (a) A stationary observer hears a frequency of 560 Hz from an approaching car. After the car passes, the observed frequency is 460 Hz. What is the speed of the car? (speed of sound is air is 343 m/s.)(b) A bat, moving at 5 m/s, is chasing a flying insect. If the bat emits a 40 kHz chirp and receives back an echo at 40.4 kHz, at what speed is the insect moving toward or away from the bat?

1. An iron bar produces sound with a frequency of 335 Hz when struck. When the iron bar is struck together with a steel bar, beats with a frequency of 5.5 Hz will be heard. A piece of thread is then tied to the iron bar and its frequency is lowered slightly. When struck at the same time again, both the iron and steel bars now produce a beat with frequency of 8.2 Hz.

(a) What is the frequency of the iron bar after the thread is tied to it?(b) What is the frequency of the steel bar?

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University Physics: Waves and Electricity

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