Chap 12. Sound Sec. 12.1 - Characteristics of Sound Sound is produced due to source(vibrating object and travels in a medium (londitudinal sound waves and can be heard by a ear (vibrations. Sound waves also called as Pressure Waves. Speed of sound is different in different material. Depends on the elasticity and density of the medium. B, or E T v sound = v string = ρ µ Here B = Bulk modulus of a fluid & E = Elastic modulus of an solid object. µ = is the mass/length of the string. Speed of sound depends on temperature and pressure of the medium. Speed of sound in air at T = 20 c and 1 atm pressure is v(air = 343 m/s The speed of sound does not depend on its frequency or amplitude. Sound waves obey laws of reflection and also interfere to give Standing Waves. Human ear can respond to frequency of 20 Hz to about 20, 000 Hz. Human ear is sensitive to two aspects of sound, loundess (Intensity and the pitch of the sound. Frequencies above 20, 000 Hz are ultrasonic waves. Frequencies below 20 Hz are infrasonic waves. Object travelling above the speed of sound are called supersonic
About Hearing & Sound Measurement 2 Sound Intensity which signifies loudness of the sound is defined as the sound power per unit area. The basic units are watts/m 2 or watts/cm 2. Many sound intensity measurements are made relative to a standard threshold of hearing intensity. The sound level β is given in terms of intensity as β(db = 10 log 10 ( I I o Decibels measure the ratio of a given intensity I to the threshold of hearing intensity, so that this threshold(i o = 10 12 W/m 2 takes the value 0 decibels (0 db. Sound waves obey inverse square law of intensity I (1/r 2. Intensity of sound wave depends of square of Amplitude I (A 2. I 1 I 2 = r2 2 r 2 1 = A2 1 A 2 2 11.10 Relation between Intensity and Amplitude The total energy of the wave is also its maximum potential energy. E = 1 2 k A2 But the angular frequency is given as k m = ω = 2πf k = 4mπ2 f 2 = 4(ρV π 2 f 2 If S is the cross sectional area though which the sound travels with a velocity (v, then V = Sl = Svt
The total energy can thus be written as E = 2(ρS v tπ 2 f 2, A 2 3 The total power is thus P = E t = 2(ρS vπ2 f 2, A 2 Intensity is then I = P S = 2(ρ vπ2 f 2, A 2 Note: Intensity depends on the square of the Amplitude. 12.4 Vibrating Strings The modes of vibration associated with resonance in extended objects like strings and air columns have characteristic patterns called standing waves. Interference of incident and reflected waves is essential to the production of resonant standing waves. Interference has far reaching consequences in sound because of the production of beats between two frequencies which interfere with each other. The behavior of the waves at the points of minimum and maximum vibrations (nodes and antinodes contributes to the constructive interference which forms the resonant standing waves. Standing waves in air columns also form nodes and antinodes, but the phase changes involved must be separately examined for the case of air columns. The medium appears to vibrate in segments or regions and the fact that these vibrations are made up of traveling waves is not apparent - hence the term standing wave.
Standing Waves in a String 4
Harmonics in a String fixed at both ends 5 Fundamental Frequency or First Harmonic: Contains two nodes at the ends and one antinode. L = λ 2 f 1 = v First Overtone or Second Harmonic: Contains 3 nodes at the ends and 2 antinode. L = 2λ 2 f 2 = 2v Second Overtone or third Harmonic: Contains 4 nodes at the ends and 3 antinode. In general L = 3λ 2 L = nλ 2 f 3 = 3v f n = nv
Air Column Open at both ends 6 Fundamental Frequency or First Harmonic: Contains one nodes at the ends and two antinode. L = λ 2 f 1 = v First Overtone or Second Harmonic: Contains 2 nodes at the ends and 3 antinode. L = 2λ 2 f 2 = 2v Second Overtone or third Harmonic: Contains 3 nodes at the ends and 4 antinode. In general L = nλ 2 L = 3λ 2 f 3 = 3v f n = nv...(n = 1, 2, 3...
Air Column closed at one end 7 Fundamental Frequency or First Harmonic: Contains one nodes at the ends and one antinode. L = λ 4 f 1 = v 4L First Overtone or Second Harmonic: Contains 2 nodes at the ends and 2 antinode. L = 3λ 4 f 2 = 3v 4L Second Overtone or third Harmonic: Contains 3 nodes at the ends and 3 antinode. In general L = 5λ 4 f 3 = 5v 4L L = (2n 1λ 4 f n = (2n 1v...(n = 1, 2, 3... 4L
12.6 Interference of Sound - Beats 8 When sound waves interfere with each other they form Beats. Constructive interference occurs when the two waves that interfere are wavelength nλ apart. Where n = 1, 2, 3,... Destructive interference occurs when the two waves that interfere are wavelength mλ apart. Where m = 1 2, 11 2, 21 2, 31 2,... If you are in the path of two sound waves that destructively interfere. Then you will not be able to hear the frequency. When two waves of frequency f 1 and f 2 interfere in time they give rise to Beats. The Beat frequency f is given as 12.7 Doppler Effect f = ±(f 1 f 2 The pitch or the frequency of sound is different if the sound source or observer is 1. approaching 2. stationary or 3. receding with respect to an each other. This phenomenon of change in the pitch or frequency is called as doppler effect. The frequency emitted by the source when the source or observer is moving is not the same f s f o. When a vehicle with a siren passes you, a noticeable drop in the pitch of the sound of the siren will be observed as the vehicle passes. An approaching source moves closer during period of the sound wave so the effective wavelength is shortened, giving a higher pitch since the velocity of the wave is unchanged. Similarly the pitch of a receding sound source will be lowered.
9 There are four cases to study: Source moving towards the stationary observer. Source moving away from the stationary observer. Observer moving towards the stationary source. Observer moving away from the stationary source.
Source moving towards the stationary observer. Let d = λ be the distance two successive crest emitted by the source at rest. Then the period and frequency of the wave at the source is given as T = λ f = λ However the source is moving towards the observer with a velocity v src. It travels a distance d = v src T in time T. So the wavelength is shortened by that amount. The modified wavelength is thus λ = d d λ = d v src T = λ v src = λ(1 v src The velocity of sound remains the same even if the wavelength (λ or frequency f changes ( = λf = λ f. For frequency we get f = f (1 v src v snd f > f Source moving away from the stationary observer. Let d = λ be the distance two successive crest emitted by the source at rest. Then the period and frequency of the wave at the source is given as T = λ f = λ However the source is moving towards the observer with a velocity v src. It travels a distance d = v src T in time T. So the wavelength is shortened by that amount. The modified wavelength is thus λ = d + d λ = d + v src T = λ + v src = λ(1 + v src The velocity of sound remains the same even if the wavelength (λ or frequency f changes ( = λf = λ f. For frequency we get f = f (1 + v src f < f 10
Observer moving towards the stationary source. 11 In this case the wavelength is not shortened but it seems that relative velocity of the wave is increased. Thus f = v λ = + v obs λ But we know that ( = λf. Thus f = v λ = f (v ( snd + v obs = f 1 + v obs f > f Observer moving away from the stationary source. In this case the wavelength is not shortened but it seems that relative velocity of the wave is decreased. Thus f = v λ = v obs λ But we know that ( = λf. Thus f = v λ = f (v ( snd v obs = f 1 v obs f < f Can put all 4 cases together to get ( f vsnd ± v obs = f ± v src Doppler Effect in Light The same effect is seen in light but the frequency shift relation is not identical. The shift to lower frequency is called as Red shift since red colour has lowest frequency for visible light. Galaxies receding from us are red shifted...backs up BIG BANG THEORY.
12.8 Shock Waves or Sonic Booms 12 Object travelling above the speed of sound are called supersonic Mach number is the ratio of the speed of object to the speed of sound. Mach Number = v obj Constructive interference at the nose and the sides of an object moving faster that produces shock wave which a listener hears as a sonic boom. The cone angle of this shock wave is: sinθ = v obj An object needs a tremendous trust to break the sound barrier. 12.9 Applications
Problem 8: What is the intensity of a sound at the pain level of 120 db and compare that of a whisper of 20dB. Soln 8: With I o = 10 12 W/m 2 the sound level is given as β(db = 10 log 10 ( I I o 13 For 120 db ( I 120 = 10 log 10 I o For 20 db ( I 20 = 10 log 10 I o I = I o 10 12 = 1 W/m 2 I = I o 10 2 = 10 10 W/m 2 Problem 18: If the amplitude of the sound level is tripled. a By what factor will the intensity increase? b How many db will the sound level increase. Soln 18: The ratio of the intensity is given as Thus if A 2 = 3 A 1 then I 2 I 1 = A2 2 A 2 1 Thus I 2 I 1 = 32 A 2 1 A 2 1 ( I2 β(db = 10 log 10 I 1 = 9 = 10log 10 9 = 9.54 db
Problem 34: A particular organ pipe can resonate at 264 Hz 440 Hz and 616 Hz. Show what kind of pipe(open/closed is it and obtain its fundamental frequency Soln 34: Let us determine the ratio of the frequency f 2 f 1 = 440 264 = 5 3 f 3 f 2 = 616 264 = 7 3 This is a pipe close at one end and has a fundamental frequency of f o = f 1 /3 = 88 Hz Problem 38: Human ear canal has length of 2.5 cm. Assuming to be a tube closed at one end, estimate the frequencies in the audible range of standing waves. Soln 38: For a pipe closed at one end f n = (2n 1v...(n = 1, 2, 3... 4L 14 f n = (2n 1343...(n = 1, 2, 3... 4 0.025 f n = (2n 1 3430...(n = 1, 2, 3 To be in the audible range 20 Hz to about 20000 Hz we need n max = 3 Frequencies are 3430 Hz, 10290 Hz, 17150 Hz
Problem 48: A source emits sounds of wavelength 2.64 m and 2.76 m in air. How many beats will be heard Soln 48:Find the frequencies f 1 = λ 1 = 343 2.64 = 130Hz f 2 = λ 2 = 343 2.76 = 124Hz Beat frequency= f1 f 2 = 130 124 = 6 Hz Problem 56: A doppler flow meter uses ultrasound waves to measure blood-flow speeds. Suppose the device emits sound at 3.5M Hz and the speed of sound in human tissue is taken to be 1540m/s. What is the expected beat frequency if blood is flowing in the large leg arteries at a velocity of 2.0cm/s directly away from the sound source. Soln 56: The frequency when the observer is receding from the source is f = f ( 1 v obs ( = 3.5 1 0.02 1540 Beat frequency= f f = 3.50 3.49 = 0.01MHz = 3.49MHz Problem 54: Bat problem Soln 54: Velocity of the Bat v src = 5.0 m/s Frequency f = 30kHz We use the fact that the source is moving towards the listener and also the listener is moving towards the source. ( ( f vsnd + v obs 343 + 5 = f = 30 = 30.88kHz v src 343 5 15