SOUND WAVES

Slide 1

SOUND WAVESwaves_041

fluteclarinetclick for sounds

waves_04: MINDMAP SUMMARY - SOUND WAVES2Sound waves, ultrasound, compressional (longitudinal) waves, pressure, particle displacement, medical imaging, superposition principle, stadning waves in air columns, constructive interference, destructive interference, boundary conditions, pipe open and closed ends, nodes, antinodes, speed of sound in air, period, frequency, wavelength, propagation constant (wave number), angular frequency, normal modes of vibrations, natural frequencies of vibration, fundamental, harmonics, overtones, harmonic series, frequency spectrum, radian, phase, sinusoidal functions, wind musical instruments, beats, beat frequency, Doppler Effect, Doppler radar, shock waves

3SOUND WAVES IN AIRLongitudinal wave through any medium which can be compressed: gas, liquid, solidFrequency range 20 Hz - 20 kHz (human hearing range)Ultrasound: f > 20 kHz

Infrasound: f < 20 HzAtoms/molecules are displaced in the direction of propagation about there equilibrium positions

For medical imaging f = 1-10 MHzWhy so large?34 0 max 0 max 0 displacement min 0 max 0 min pressure

Displacement of molecules Pressure variation SOUND WAVES IN AIR45Ultrasonic sound waves have frequencies greater than 20 kHz and,as the speed of sound is constant for given temperature and medium, they have shorter wavelength. Shorter wavelengths allow them to image smaller objects and ultrasonic waves are, therefore, used as a diagnostic tool and in certain treatments.

Internal organs can be examined via the images produced by the reflection and absorption of ultrasonic waves. Use of ultrasonic waves is safer than x-rays but images show less details. Certain organs such as the liver and the spleen are invisible to x-rays but visible to ultrasonic waves.

Physicians commonly use ultrasonic waves toobserve fetuses. This technique presents far less risk than do x-rays, which deposit more energy incells and can produce birth defects.

ULTRASOUND

What is the physics of this image?CP 49567

Flow of blood through the placentaULTRASOUND The speed of a sound wave in a fluid depends on the fluids compressibility and inertia. Speed of sound wave in a fluid

B : bulk modulus of the fluidr : equilibrium density of the fluid

Speed of sound wave in a solid rod

Y : Youngs modulus of the rodr : density of the fluid Speed of sound wave in air

v = 343 m.s-1 at T = 20oCAbove formulae not examinable8 The loudest tolerable sounds have intensities about 1.0x1012 times greater than the faintest detectable sounds. Intensity level in decibelEnergy and Intensity of Sound waves The sensation of loudness is approximately logarithmic in the human ear. Because of that the relative intensity of a sound is called the intensity level or decibel level, defined by:

I0 = 1.0x10-12 W.m-2 : the reference intensitythe sound intensity at the threshold of hearing

Threshold of painThreshold of hearing9Not examinable

Problem 1A noisy grinding machine in a factoryproduces a sound intensity of 1.00x10-5 W.m-2. (a) Calculate the intensity level of the single grinder.

(b) If a second machine is added, then:

(c) Find the intensity corresponding to an intensity level of 77.0 dB.

10Not examinable11Problem 2A point source of sound waves emits a disturbance with a power of 50 W into a surrounding homogeneous medium. Determine the intensity of the radiation at a distance of 10 m from the source. How much energy arrives on a little detector with an area of 1.0 cm2 held perpendicular to the flow each second? Assume no losses.Solution

P = 50 W r = 10 m I = ? W.m-2 A = 1.0 cm2 = 1.010-4 m2 W = ? J

I = P / (4 r2) = 4.010-2 W.m-2 W = I A = 4.010-6 J

11 A small source emits sound waves with a power output of 80.0 W.

(a) Find the intensity 3.00 m from the source.

(b) At what distance would the intensity be one-fourth as much as it is at r = 3.00 m?

(c) Find the distance at which the sound level is 40.0 dB?

Problem 312Loud-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 pitchBEATSCP 521313BEATS 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 difference of the two frequencies.

1414

Superimpose oscillations of equal amplitude, but different frequenciesModulation of amplitudeOscillation at the average frequencyfrequency of pulses is | f1-f2 |BEATSCP 5271515

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 52716

CP 527f1 = 100 Hz f2 = 104 Hz frapid = 102 Hz Trapid = 9.8 msfbeat = 4 Hz Tbeat = 0.25 s (loud pulsation every 0.25 s)17

click for sound

CP 527f1 = 100 Hz f2 = 110 Hz frapid = 105 Hz Trapid = 9.5 msfbeat = 10 Hz Tbeat = 0.1 s (loud pulsation every 0.1 s)18click for sound

CP 527f1 = 100 Hz f2 = 120 Hz frapid = 110 Hz Trapid = 9.1 msfbeat = 20 Hz Tbeat = 0.05 s (loud pulsation every 0.05 s)1924 25 click for sound

19DOPPLER EFFECT - motion related frequency changes

Doppler 1842, Buys Ballot 1845 - trumpeters on railway carriage

Source (s) Observer (o)

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, weather (Doppler radar), plaet detectionCP 495note: formula is very different to textbookformula different to textbook2021

DOPPLER RADAR

DOPPLER EFFECTConsider 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 approachingsource catching up on waveswavelength reducedfrequency increasedOn left - source recedingsource moving away from waveswavelength increasedfrequency reducedCP 495v = f

2222

CP 495

23Source frequency

fs = 1000 Hz

click for soundsfo > 1000 Hzfo < 1000 Hz

source vsobserver voobserved frequency fostationarystationary= fsstationaryreceding< fsstationaryapproaching> fsrecedingstationary< fsapproachingstationary> fsrecedingreceding< fsapproachingapproaching> fsapproachingreceding?recedingapproaching?CP 595

24SHOCK Waves supersonic waves

CP 50625

bullet travelling at Mach 2.45

CP 506Shock Waves supersonic waves2627DOPPLER EFFECT

Stationary Sound Source

Source moving with vsource < vsound ( Mach 0.7 )

Source moving with vsource = vsound ( Mach 1 - breaking the sound barrier )

Source moving with vsource > vsound (Mach 1.4 - supersonic)

26 27 28 29 Problem 4

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-1.28Solution

fs = 650 Hz vs = 20 m.s-1 vo = 0 m.s-1 v = 340 m.s-1 fo = ? Hz (must be higher since train approaching observer).

28Problem 5

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.

29Solution 5

Doppler Effect Beats

30

Blood is moving away from source observer moving away from source fo < fsWave reflected off red blood cells source moving away from observer fo < fsBeat frequency = | 4.00 3.998442| 106 Hz = 1558 Hz

In this type of calculation you must keep extra significant figures.31An 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)

(b)

Problem 6Solution 632Problem 7

An ultrasonic wave at 8.000104 Hz is emitted into a vein where the speed of sound is about 1.5 km.s-1. The wave reflects off the red blood cells moving towards the stationary receiver. If the frequency of the returning signal is 8.002104 Hz, what is the speed of the blood flow?

What would be the beat frequency detected and the beat period? Draw a diagram showing the beat pattern and indicate the beat period.

333034

Solutionfs = 8.000104 Hz fo = 8.002104 Hz v = 1.5103 m.s-1 vb = ? m.s-1 Need to consider two Doppler shifts in frequency blood cells act as observer and than as source.Red blood cells (observer) moving toward sourceRed blood cells (source) moving toward observerfbeat = |f2-f1| = (8.002-8.000)104 Hz = 20 Hz

Tbeat = 1/fbeat = 0.05 s

tbeat35If we try to produce a traveling harmonic wave in a pipe, repeated reflections from an end produces a wave traveling in the opposite direction - with subsequent reflections we have waves travelling in both directionsThe result is the superposition (sum) of two waves traveling in opposite directionsThe superposition of two waves of the same amplitude travelling in opposite directions is called a standing waveSTANDING WAVES IN AIR COLUMNS (PIPES)3536LClosed end: displacement zero (node), pressure max (antinode)Open end: displacement max (antinode), pressure zero (node)

CP 516

Closed at both ends

Closed at one endopen at the other

Open at both endsFlute & clarinet same length, why can a much lower note be played on a clarinet?STANDING WAVES IN AIR COLUMNS (PIPES)3637

Organ pipes are open at both ends

Sound wave in a pipe with one closed and one open end (stopped pipe)

38STANDING WAVES IN AIR COLUMNS (PIPES)39

CP 516STANDING WAVES IN AIR COLUMNS (PIPES)40

Search google or YouTube for

Rubens or Rubins tube

41L

LCP 516STANDING WAVES IN AIR COLUMNS (PIPES)41Normal modes in a pipe with an open and a closed end (stopped pipe)

42STANDING WAVES IN AIR COLUMNS (PIPES)Fundamental3rd harmonic or 2nd overtone

43

CP 52344Musical instruments wind

An air stream produced by mouth by blowing the instruments interacts with the air in the pipe to maintain a steady oscillation.

All brass instruments are closed at one end by the mouth of the player.

Flute and piccolo open at atmosphere and mouth piece (embouchure) covering holes L f

Trumpet open at atmosphere and closed at mouth covering holes adds loops of tubing into air stream L f

Woodwinds vibrating reed used to produce oscillation of the air molecules in the pipe.CP 51645Woodwind instruments are not necessarily made of wood eg saxophone, but they do require wind to make a sound. They basically consist of a tube with a series of holes. Air is blow into the top of the tube, either across a hole or past a flexible reed. This makes the air inside the tube vibrate and give out a note. The pitch of the note depends upon the length of the tube. A shorter tube produces a higher note, and so holes are covered. Blowing harder makes a louder sound. To produce deep notes woodwind instruments have to be quite long and therefore the tube is curved.

Brass instruments (usually made of brass) consist of a long pipe that is usually coiled and has no holes. The player blows into a mouthpiece at one end of the pipe, the vibration of the lips setting the air column vibrating throughout the pipe. The trombone has a section of pipe called a slide that can be moved in and out. To produce a lower note the slide is moved out. The trumpet has three pistons that are pushed down to open extra sections of tubing. Up to six different notes are obtained by using combinations of the three pistons. CP 51646

Boundary conditions

Reflection of sound wave at ends of air column: Open end a compression is reflected as a rarefaction and a rarefaction as a compression ( phase shift). Zero phase change at closed end.Natural frequencies of vibration (open closed air column)

Speed of sound in air (at room temperature v ~ 344 m.s-1) v = f

odd harmonics exit: f1, f3, f5, f7 , CP 51647Problem 8

A narrow glass tube 0.50 m long and sealed at its bottom end is held vertically just below a loudspeaker that is connected to an audio oscillator and amplifier. A tone with a gradually increasing frequency is fed into the tube, and a loud resonance is first observed at 170 Hz. What is the speed of sound in the room?48Solution 8Pressure distribution in tube for fundamental modepressure nodepressure antinodef = 170 HzL = 0.50 mv = ? m.s-1

speed of wave

NA49Problem 9

What are the natural frequencies of vibration for a human ear? Why do sounds ~ (3000 4000) Hz appear loudest?

Solution

Assume the ear acts as pipe open at the atmosphere and closed at the eardrum. The length of the auditory canal is about 25 mm. Take the speed of sound in air as 340 m.s-1.

L = 25 mm = 0.025 m v = 340 m.s-1For air column closed at one end and open at the otherL = 1 / 4 1 = 4 L f1 = v / 1 = (340)/{(4)(0.025)} = 3400 Hz

When the ear is excited at a natural frequency of vibration large amplitude oscillations (resonance) sounds will appear loudest ~ (3000 4000) Hz.

RESONANCE When we apply a periodically varying force to a system that can oscillate, the system is forced to oscillate with a frequency equal to the frequency of the applied force (driving frequency): forced oscillation. When the applied frequency is close to a characteristic frequency of the system, a phenomenon called resonance occurs. Resonance also occurs when a periodically varying force is applied to a system with normal modes. When the frequency of the applied force is close to one of normal modes of the system, resonance occurs.

5051Why does a clarinet play a lower note than a flute when both instruments are about the same length ?A flute is an open-open tube.

A clarinet is open at one end and closed at the other end by the players lips and reed.

openopenopenclosedProblem 1052Solution 10FLUTE is open at both ends particle displacement antinodes at the open ends

AA

Fundamental (1st harmonic)

2nd harmonic (1st overone)

A

All harmonics can be excited

Particle displacement variations53Solution 10CLARINET is open at one and closed at the other pressure node at open end & antinode at closed end

AN

Fundamental (1st harmonic)

3nd harmonic (1st overone)

All odd harmonics can be excited

pressure variationsANN

The sound waves generated by thefork are reinforced when the lengthof the air column corresponds to oneof the resonant frequencies of thetube. Suppose the smallest value ofL for which a peak occurs in thesound intensity is 90.0 mm. Problem 11ResonanceLsmalles t= 9.00 cmFind the frequency of the tuning fork.

(b) Find the wavelength and the next two water levels giving resonance.

5455Solution 11 (alternative)1st resonanceL1= 90.0 mm = 90.010-3 m v = 343 m.s-1 f1 = ? Hz = 4 L = 0.360 m v = f f1 = v / = (343)/(0.36) Hz = 958 Hz2nd and 3rd resonancesL3 = 3/4 = (3)(0.36)/(4) m = 0.270 mL5 = 5/4 = (5)(0.36)/(4) m = 0.450 mTube closed at one end and open at the other odd harmonics can be excited pressure variation

The frequency of the tuning fork is fixed and hence the wavelength also has a fixed value. The length of the tube is variable.

ANL1= /4L3= 3/4L5= 5/4NANN56Why does a chimney moan ?Chimney acts like an organ pipe open at both endsPressure nodePressure nodeNNAFundamental mode of vibrationSpeed of sound in air v = 340 m.s-1

Length of chimney L = 3.00 m

L = / 2 = 2 L v = f

f = v / = 340 / {(2)(3)} Hz

f = 56 Hz low moanProblem 12

Natural frequencies of a trumpet closed at mouth and open at the flared end

Fundamental f1 = 60 Hz

1st overtone 163 Hz2nd overtone 247 HzNOT a harmonic sequence for the natural frequencies of vibration58Simulation of the human voice tract natural frequencies

Pipe closed at one end and open at the otherclosed end

particle displacement zero

( node

open end

max particle displacement

( antinode