Acoustics PDF

ACOUSTICS  Contents : Acoustics and its importance Reverberation time Absorption coefficient Method to measure absorption coefficient Sabine’s and Eyring’s formula Applications of acoustics- Designing of hall for speech, concert, and opera  ACOUSTICS Deals with the production, propaga...

ACOUSTICS  Contents : Acoustics and its importance Reverberation time Absorption coefficient Method to measure absorption coefficient Sabine’s and Eyring’s formula Applications of acoustics- Designing of hall for speech, concert, and opera  ACOUSTICS Deals with the production, propagation, transmission, detection of sound waves. Classification of sound: (i) Infrasonic 20 Hz (Inaudible) (ii) Audible 20 to 20,000Hz (Music and Noise) (iii) Ultrasonic 20,000Hz (Inaudible)  Acoustics Properties : Sound is a mechanical wave and therefore requires a medium to travel. So, It is reflected, transmitted, or absorbed by the materials it encounters. Architectural Acoustics: Deals with the planning of a building or a hall so that there is most advantageous flow of properly diffused sound to entire audience without affecting the speech intelligibility and tonal qualities of music.  Reverberation When sound waves are produced inside a building, they spread out, scatter and get partly reflected (and partly absorbed) from various surfaces such as walls, ceiling and floor of the hall. Audience will receive a direct sound from the source followed by series of sounds reflected and traveling towards him. These successive sounds will be of diminishing intensity. Therefore listener will continue to receive the sound even after the source of sound has stopped emitting. This is called as reverberation. Definition: The persistence or prolongation of sound in a hall even though the sound source is stopped called Reverberation. The time during which sound persists in the hall is called reverberation time. It is measured from time that the original sound is produced, or in case of continuous note from time that source stopped sounding to time at which sound becomes inaudible.  Reverberation Time The time taken by the sound wave to fall below the minimum audibility level after the source is stopped. Or i.e. to fall to one millionth of its initial intensity, after the source is stopped. Here, I = (1/106)I0 I and I is the initial & final sound intensity level 0 If Reverberation Time is too low: Sound disappear quickly and hence less loud. If Reverberation Time is too high: Sound exist for a long period of time - an overlapping of successive sounds results in unclear information. Therefore, for the good audibility: Reverberation time should be optimum. Dry speech : 0.8 s Reverberation time : 2 s Reverberation time : 5 s Reverberation time : Growth and decay of sound density in room : Sound density Time Optimum reverberation time 10 Acoustical demands of an auditorium: ❑Sound should be sufficient loud and intelligible at every point in room. ❑Sound of each syllable should soon decay so that the succeeding syllable is heard distinctly (no excessive reverberations.) ❑No more echoes than necessary for maintaining continuity. ❑No undesirable focusing or zones of silence or regions of poor audibility due to walls, ceiling etc. ❑No unpleasant reinforcement so that tonal quality is unaffected. ❑ Extraneous noise and echelon effects (regular successive echoes) within building should be avoided. Absorption coefficient (α) : ❑ It is defined with respect to a standard of absorption which is open window unit (OWU). ❑ Sabine chose one sq ft of an open window as standard of absorption (OWU) as all sound falling on it will pass through it and can be said to be absorbed. This is also called as sabin. ❑ One person is equal to 4.7 sq ft of open window. ❑ Absorption coefficient (α) is defined as the ratio of the sound energy absorption by that surface to that of an equal area of perfect absorber such as open window. In case of n number of surfaces S1, S2, S3…………Sn with absorption coefficient α1, α2, α3…….αn respectively, average absorption coefficient ( ) is defined as 1S1 +  2 S 2 +  3 S3 +.............. n S n = S1 + S 2 + S3 +..............S n   Si =   i Si i i  Absorption of sound The property of a surface by which sound energy is converted into other form of energy (heat) is known as absorption. It is mainly due to two causes: 1. Porosity: In the process of absorption sound energy is converted into heat due to frictional resistance inside the pores of the material. The fibrous and porous materials absorb more sound energy as compared to other solid materials. 2. Flexural vibration: When sound waves fall on flexible materials not rigidly mounted, the material of course is set into vibration and the damping forces called into paly dissipate the incident sound energy into heat. 14  Sabine’s Formula for Reverberation Time Prof. W. C. Sabine (1868-1919) determined the reverberation times of empty halls and furnished halls of different sizes and arrived at the following conclusions. The reverberation time depends on the reflecting properties of the walls, floor and ceiling of the hall. The reverberation time depends on the absorbing power of the various surfaces (carpets, cushions, curtains etc). The reverberation time is independent of the positions of the source and the listener and the shape of the room. The reverberation time is directly proportional to volume of the hall. The reverberation time is inversely proportional to effective absorbing surface area. Sabine’s Formula for Reverberation Time Prof. Sabine summarized his results that Reverberation time (T) is T Volume of the hall Effective absorbing surface area T =K V A eff where K is a proportionality constant. K= 0.05 (if V is in ft3 and Aeff in ft2) K = 0.165 (if V is in m3 and Aeff in m2) This Equation is known as Sabine’s formula for reverberation time. 16  T =K V =K V A S eff Where α is the absorption coefficient and S is geometrical surface area. If there are n number of surfaces then V T =K 1S1 +  2 S 2 +  3 S3 +....... +  n S n where Sn are different surfaces of absorption coefficients an as V, S and  can be calculated from plans and specifications, so it is possible for an architect to design an auditorium with any desired time of reverberation. 17 Derivation of Reverberation time : Statistically, distance travelled by sound between two successive 4V reflections = (V is volume and S is surface area) S 4V if v is velocity of sound, then time between two successive reflections = Sv Svt Average no. of reflections in time t = 4V If  is average absorption coefficient, then after first reflection (1 -  ) is reflected. After 2 reflections (1 -  ) 2 is reflected and so on...... If I 0 is the initial intensity and I t after time t , then Svt I t = I 0 (1 −  ) 4V Svt It  = (1 −  ) 4V (1) I0 According to definition of reverberation time (T), when t=T It −6 = 10 (2) I0 Putting eq (1) = (2) SvT 10 −6 = (1 −  ) 4V SvT  ln(10 −6 ) = ln(1 −  ) 4V −6 4V  T = ln(10 ) Sv ln(1 −  ) 4V  T = 2.3026 [log10 (10 )] -6 Sv ln(1 −  ) Putting v = 1120 ft/sec, 2.3026  (-6)  4V T= S 1120  ln(1 −  ) V Eyring’s Formula (V in ft 3, = −0.05 S  ln(1 −  ) S in ft2)  0.05V T =− S ln(1 −  ) 2 3   if is much less than 1, then ln(1 −  ) = − − − −........ 2 3 0.05V 0.05V 0.05V T =−  =  2  3 S  Si i S (− − − −........) 2 3  Limitation of Sabine’s Formula 1.It is good for only small values of absorption coefficient ( < 0.2) 2.It is not valid for higher values of  This is because for  = 1, T should be zero, whereas Sabine’s formula fines T = kV/Aeff, a non zero value. 3. For higher values of absorption coefficient, the Sabine’s formula gives higher value of reverberation time than its actual value. Therefore, Eyring’s Formula came in existence  Sabine’s vs Eyring’s Formula 0.05V 0.05V T ( Sabine) = T(Eyring) = - S S ln(1 -  ) 1. Both formulae gave identical value when  is small. 2. However, for large value of  (for ex.  = 1) two gave different values 0.05V T ( Sabine) = T(Eyring) = 0 S 3. When  = 0, then both Sabine and Eyring’s formula give correct result T (Sabine) =  T(Eyring) =   Methods to measure Absorption Coefficient Two important methods : 1. Reverberation chamber method 2. Stationary wave method (Measures only for normal incidence. Material can be used in small samples and impossible to mount sample in similar conditions to those when material is mounted in bulk.)  Reverberation Chamber method : 1. First the reverberation time of the room with object is measured. Then object is removed and the extent of open window is adjusted until the reverberation time is same as before. Absorption of sound by the object is equal to absorption of sound by window. Area of the window Absorption coefficient of the object (α) = Total area of the object α will always be a fraction. Reverberation Chamber method : 2. In second method, reverberation time of room is measured with the object of interest and without it. 0.05V where   i Si   i Si Reverberation time without object (T1) = i i is absorption in hall without object. 0.05V   i Si + cS Reverberation time with object (T2) = where c is absorption coefficient of object and S is isurface area in ft2. 1 1 cS − = T2 T1 0.05V 0.05V  1 1 c=  −  S  T2 T1  Reverberation Chamber method : 3. Third method can be used for calculation of average absorption coefficient (𝛼ത) of a room. 𝐼𝑡 = 𝐼0 1 − 𝛼ത 𝑆𝑣𝑡/4𝑉 Taking log on both side 𝐼𝑡 𝑆𝑣𝑡 𝑙𝑛 = 𝑙𝑛 1 − 𝛼LJ 𝐼0 4𝑉 𝐼𝑡 𝑆𝑣𝑡 𝛼LJ2 ⇒ = −𝛼LJ −.... 𝐼0 4𝑉 2 𝐼𝑡 −𝑆𝑣𝑡𝛼LJ ⇒ 𝑙𝑛 ≈ 𝐼0 4𝑉 𝐼𝑡 −𝑆𝑣𝑡𝛼LJ ⇒ = 𝑒 4𝑉 𝐼0 −𝑆𝑣𝑡𝛼LJ ⇒ 𝐼𝑡 = 𝐼0 𝑒 4𝑉 Suppose there are 2 sources with power P0 and P1 giving maximum intensities I0 and I1 respectively. Let Im be maximum audible intensity and T0 and T1 respective times of decay to threshold of audibility −𝑆𝑣T0 𝛼LJ For first source : (3) 𝐼m = 𝐼0 𝑒 4𝑉 −𝑆𝑣T1 𝛼LJ For second source : (4) 𝐼m = 𝐼1 𝑒 4𝑉 ഥ 𝑆𝑣𝑇 𝛼 ഥ 𝑆𝑣𝑇 𝛼 Putting (3) = (4) − 4𝑉 0 − 4𝑉 1 𝐼0 𝑒 = 𝐼1 𝑒 ഥ 𝑆𝑣 𝛼 𝐼0 (𝑇 −𝑇1 ) ⇒ = 𝑒 4𝑉 0 𝐼1  ഥ 𝑆𝑣 𝛼 𝐼0 (𝑇0 −𝑇1 ) =𝑒 4𝑉 𝐼1 Taking log on both sides 𝐼0 𝛼𝑆𝑣 ത ln = (𝑇0 − 𝑇1 ) 𝐼1 4𝑉 4𝑉 ln(𝐼0 /𝐼1 ) 𝛼ത = 𝑆𝑣(𝑇0 − 𝑇1 ) 4𝑉 ln(𝑃0 /𝑃1) 𝐼0 𝑃0 = 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 = 𝑆𝑣(𝑇0 − 𝑇1 ) 𝐼1 𝑃1 4𝑉[ln 𝑃0 − ln 𝑃1] = 𝑆𝑣(𝑇0 − 𝑇1 ) Hence value of can be calculated if P0, P1, T0, T1, S, v and V are known. Stationary wave method : A loudspeaker produces an acoustic wave which travels down the pipe and reflects from the test sample. The phase interference between the waves in the pipe which are incident upon and reflected from the test sample will result in the formation of a standing wave pattern in the pipe. The pressure amplitudes at nodes and antinodes are measured with a microphone probe attached to a car which slides along a graduated ruler. Incident Wave : 𝑦1 = 𝑎 sin(𝜔𝑡 − 𝑘𝑥) Reflected Wave : 𝑦2 = 𝑚𝑎 sin(𝜔𝑡 + 𝑘𝑥) [m : Reflection coefficeint] Resultant displacement : 𝑦 = 𝑦1 + 𝑦2 = 𝑎 sin 𝜔𝑡 cos 𝑘𝑥 − 𝑎 cos 𝜔𝑡 sin(𝑘𝑥) +𝑚𝑎 sin 𝜔𝑡 cos 𝑘𝑥 + 𝑚𝑎 cos 𝜔𝑡 sin(𝑘𝑥) 𝑦 = 𝑎 1 + 𝑚 sin 𝜔𝑡 cos 𝑘𝑥 − 𝑎 1 − 𝑚 cos 𝜔𝑡 sin 𝑘𝑥 2𝜋 2𝜋 = 𝑎 1 + 𝑚 sin 𝜔𝑡 cos( 𝑥) − 𝑎 1 − 𝑚 cos 𝜔𝑡 sin( 𝑥) 𝜆 𝜆 For the first term in above expression, amplitude is given by 2𝜋 𝑎 1 + 𝑚 cos( 𝑥) and for second term the amplitude is given 𝜆 2𝜋 by 𝑎 1 + 𝑚 sin( 𝑥). 𝜆  2𝜋 2𝜋 𝑦 = 𝑎 1 + 𝑚 sin 𝜔𝑡 cos( 𝑥) − 𝑎 1 − 𝑚 cos 𝜔𝑡 sin( 𝑥) 𝜆 𝜆 For first term, maximum amplitude is given by a(1+m) at points x = 0, -λ/2, -λ......... For second term, maximum amplitude is given by a(1-m) at points x = -λ/4, -3λ/4, -5λ/4......... Positions of maximum displacement in first term are positions of minimum displacements in second term and vice-versa. So, there will be equally spaced planes of maximum and minimum vibrations. If displacement at antinode is given by a1 and at node by a2 then 1 + 𝑚 𝑎1 = [m : Reflection coefficeint] 1−𝑚 𝑎2 𝑎1 − 𝑎2 ⇒ 𝑚= 𝑎1 + 𝑎2 Coefficient of reflection for intensity will be 2 2 𝑎1 − 𝑎2 𝑅=𝑚 = [𝑖𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 ∝ 𝐴𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒2 ] 𝑎1 + 𝑎2 Coefficient of absorption for intensity will be 4𝑎1 𝑎2 𝛼 = 1−𝑅 = 2 𝑎1 + 𝑎2 Conditions for good acoustical design of rooms : 1. Control of reverberation : a) Few open windows. b) Walls covered with absorbent materials. c) Upholstered seats. d) Good audience. 2. Shape of walls and ceiling (a) No curved walls or corners to avoid focusing and dead spaces. (b) splayed ceilings Conditions for good acoustical design of rooms : 3. Avoid concave surfaces such as domes, curved arches etc. to eliminate focusing effects, non distribution of sound and echoes. 4. Floor plan with diverging walls : helps in higher sound level at the rear of the hall. 5. Seats : Gradually elevated to promote free flow of direct sound from source to listener. An anechoic chamber is a space in which there are no echoes or reverberations. The surfaces absorb all sound, and reflect none.