acoustics

Physics – Acoustics Dr. A. Claude

1

ACOUSTICS OF BUILDINGS The branch of science that deals with the planning of a building or hall to provide best audible sound to the audience within the hall is called acoustics of buildings. The basic requirements of a hall to be acoustically good are the following.

1. The sound heard must be sufficiently loud in every part of the hall 2. No echoes should be present 3. The quality of sound (i.e., combinations of frequency) should be

uniform throughout the hall.It should remain more or less the same irrespective whether the hall is filled with audience or not.

4. The successive syllables spoken must be clear without overlapping 5. Reverberation should be optimum 6. There should be no concentration of sound in any part of the hall 7. The boundary of the hall should be sufficiently sound proof so as to

exclude noise from outside the hall 8. There should be no echelon effect 9. There should be no resonance within the building

Factors affecting Acoustics of Building and their remedies

1. Reverberation

The prolonged presence or persistence of decaying sound in a hall due to reflection of sound from walls, ceilings, floors etc is known as reverberation. It may be defined as the persistence of audible sound even after the source has stopped to emit sound. The time taken for the sound to fall from its average value to the level of inaudibility is called reverberation time. The standard reverberation time is the time taken by sound to fall below one millionth of its original intensity just after the source is cut off. If the reverberation time of a hall is very large, then overlapping of successive syllables results. If it is too small, then the loudness of sound becomes inadequate. The optimum value of reverberation time is 0.5s for speech and between 1 and 2s for music programs. Experimentally it is found that reverberation time depends on size of hall, loudness of sound and kind of music for which the hall is used. Reverberation tome can be controlled by the following factors.

1. Providing windows and ventilators that can be closed and opened. 2. Decorating the walls with pictures and maps 3. Using heavy curtains with folds 4. Lining the walls with sound absorbing materials and Carpeting the

floors


2

2. Adequate loudness Sufficient loudness throughout the hall is an important factor for good acoustics of the building. Sometimes the use of sound absorbent materials to reduce reverberation time results in the decrease of loudness. To get adequate loudness, good quality loudspeakers should be installed to minimize the loss of sound energy from speakers, large sounding boards may be used behind the speaker and facing the audience. Large polished concave surfaces above the speaker will also improve loudness. The radius of curvature of the concave surface should be about twice the height of the building. This will avoid the bad focusing of sound. Low ceiling will also help in the reflection of sound towards audience.

4. Echoes Echoes are caused by reflected sounds if they reach a listener with

a time interval ≥ 1/7 s between the direct (original) and reflected signal. An echo causes confusion in listening. These should be avoided by lining the distant walls and high ceilings with appropriate sound absorbing materials. Concave surfaces reflect and focus sound to a particular place. To minimize such selective focusing (echo), radius of curvature of the concave surface in ceiling should be about twice the height of the building. A fan shaped floor of an auditorium gives rise to favorable reflections from the walls. . 4. Interference Effects (Pockets of maxima and minima) Any concave surface with a small radius of curvature reflects and focusses sound to a particular place.Hence sound is concentrated in that region when compared to other regions. In addition, the interference taking place between the direct and reflected waves produces similar


3

effects by forming interference maxima and minima of sound in different regions. Hence this must be avoided. 5. Echelon effect

A staircase or a similar structure withequal spacing will produce a musical note due to a regular succession of echoes of the original sound to a listener. This effect is known as echelon effect which makes the original sound to be unintelligible. Such structures should be avoided inside the hall. If unavoidable, they must be covered by suitable sound absorbing materials such as thick carpets. 6. Resonance effects:

Hollow spaces, cave formations and crevices in a hall produce resonance effect. They reinforce the sound at their natural frequency thereby producing an unpleasant humming sound. Resonance and echoes are reduced by convex cylindrical segments on walls and ceilings.

SABINE’S FORMULA FOR REVERBERATION TIME Reverberation

Sound produced by a source travels in all directions. The sound waves undergo a number of repeated reflections by the objects present in the hall during which time its intensity reduces gradually. The persistence or the prolonged presence of decaying sound in a hall due to reflection of sound from walls, ceilings, floors etc is known as reverberation. It may be defined as the persistence of audible sound even after the source has stopped to emit sound. The time taken for the sound to fall from its average value to the level of inaudibility is called reverberation time. The standard reverberation time is the time taken by sound to fall below one millionth of its original intensity just after the source is cut off. If the reverberation time of a hall is very large, then overlapping of successive syllables results. If it is too small, then the loudness of sound becomes inadequate. The optimum value of reverberation time is 0.5s for speech and between 1 and 2s for music programs. Experimentally it is found that reverberation time depends on size of hall, loudness of sound and kind of music for which the hall is used. Sabine derived a formula for reverberation time based on an assumption that there is an uniform distribution of sound energy in the hall The derivation of the formula involves three steps:


4

1. Finding the rate of incidence of sound energy (in terms of the energy density of sound, E) on walls and other surfaces and hence the rate at which it is absorbed.

2. Finding the steady state value of E in terms of the power (rate of emission of sound energy) P at which the sound energy is emitted from the source

3. Finding the reverberation time from the study of decay of sound energy after the source is cut off. Consider a hall of volume V. Let P is the power at which a source in the hall generates sound energy and E be the uniform sound energy per unit volume present in the hall at a given time. AB is the plane wall on which a small element of area ds is considered. From the center of O of area ds and with radii r and r+dr draw two semicircular arcs such that they lie in the plane containing the normal ON to the surface ds. Consider the

shaded portion of area CDEF between the two circular arcs. θ and θ + dθ are angles made by the two radial lines bounding the shaded area with the normal ON.

Area of the shaded portion, ds ≈ (CD . DE) = ( r.dθ).dr

The whole figure is rotated by an angle dϕ about the normal ON. As the

radius of rotation is GC = rsinθ, the distance through which the shaded portion has been moved

∆l = rsinθ . dϕ

∴Volume dV traced by the area CDEF on moving through a distance

∆l = Area of CDEF x ∆l

i.e., dV = ( r.dθ).dr x rsinθ . dϕ

dV = r2dr.sinθdθ.dϕ ……………(1)

∴Energy present in the volume dV = E.dV

= E r2dr.sinθdθ.dϕ…….…….(2)


5

This sound energy is traveling through the volume element in all

directions. If dΩ is the solid angle subtended by the elemental area dS on the wall at the volume element dV, then The sound energy from volume element traveling towards the area dS on the wall

= E.dV. (dΩ / 4π). ……….(3) But the solid angle subtended by the area dS at the volume element is

dΩ = (dS.cosθ / r2).

∴ substituting this in eqn. (3), we get, Energy traveling from volume element to the area

dS = E.dV. (dΩ / 4π)

= (E.dS./ 4π).dr.sinθcosθdθ.dϕ ….(4) Let C be the velocity of sound. To find the amount sound energy incident on the area dS on the wall, eqn.(3) integrated within the limits of r from 0

to C, θ from 0 to π/2 and ϕ from 0 to 2π. i.e., sound energy received at dS per unit time

/ 2 2

0 0 0

.sin cos

4

cE dSdr d d

π π

θ θ θ ϕπ

= ∫ ∫ ∫

[ ]

/ 2

0 0

.sin cos . 2

4

cE dSdr d

π

θ θ θ ππ

= ∫ ∫

/ 2

0 0

.2sin cos

4

cE dSdr d

π

π θ θ θπ

= ∫ ∫

= ( )

∫∫

2/

00

2sin..4/.π

θθddrdSEc

= (E.dS/4).[C].[1] = EC.dS /4 …………..(5) If ‘a’ is the absorption coefficient which is defined as the ratio of energy absorbed by a surface to that incident on it, then, the energy absorbed by the area dS per unit time

= EC.a.dS / 4.

∴Total rate of absorption of sound energy by the entire surface of the

wall = EC.Σ(a.dS) / 4 = ECA / 4 ……(6)

where A = .Σa.dS = . (a1S1 + a2S2 + a3S3 + ……) is the total absorption on all surfaces of the wall. The total energy present in the Hall = energy density x volume = E.V


6

∴The rate of growth of sound energy in the hall = ( )

dt

dEV

dt

VEd.

. =

But Also, Rate of growth of sound energy = rate of power supply – rate of absorption

i.e., 4

.ECA

Pdt

dEV −=

Dividing through out by V

α.4

EV

P

V

ECA

V

P

dt

dE −=−= ,

Where α = (CA / 4V)

Rearranging this, and using 1 / V = 4α / CA, we get

4

. .dE P

Edt CA

α α+ = ………..(7)

When the sound energy density reaches a steady state (maximum) value Em, (dE / dt) =0

∴Using (dE / dt) =0 when E = Em in eqn (7), we get Em = ( 4P / CA ) …….(8)

Multiplying eqn. (7) throughout by eαt,

4

. . . . .t t tdE Pe E e e

dt CAα α αα α+ = . ….(7a) (or)

( ). 4

. .t

td E e P

edt CA

ααα= as LHS of eqn.(7a) = d(E.eαt) / dt

i.e., ( ) 4. . . .t tP

d E e e dtCA

α αα= . …(6). Integrating eqn. (6),

. 4

. .t tPE e e K

CAα α= + ……….(9)

In eqn. (9) K is the constant of integration which can be found by setting the initial conditions.

Growth of sound energy

Suppose the source of sound starts producing sound energy at a power rate of P from time t =0. Assuming that energy density E = 0 at t = 0, eqn. (7) becomes, 0 = ( 4P / CA ) .1 + K i.e., K = - ( 4P / CA ) = -Em ……(10) [Using eqn. (8)] Substituting eqn. (10) back in eqn. (9), and also using ( 4P / CA ) = Em from eqn.(8), we get,


7

E.eαt = Em.( eαt - 1)

Dividing throughout by eαt, we get,

E = Em.( 1 – e-αt)………….(11)

Eqn. (11) indicates that The sound intensity E starts increasing Exponentially from zero at time t= 0 when the source starts emitting sound; It reaches the steady, maximum Em at time t very large. Decay of sound energy

Suppose the source stops emitting energy after the energy density in the hall reaches a steady value, Em. If it stops producing sound at a time t = 0, then the manner in which E decreases with time from Em can be found by substituting E = Em and P = 0 at t = 0 in eqn. (9).

∴Eqn. (9) becomes Em.1 = 0 + K

⇒ K = Em Substituting K = Em in eqn (9) and

Dividing throughout by eαt we get,

E = Em .e-αt …….(12) Eqn. (12)

indicates that sound energy density E decreases exponentially from Em to zero after the source stops producing sound Reverberation Time

Reverberation time is defined as the time taken for sound intensity to decrease from the steady value to one millionth of the steady value from the time when the source stops to emit sound. Let T be the reverberation time. Then eqn. (12) becomes

10-6Em = Em. e- α T (or) e- α T = 10 –6 Taking log on both sides,

-αT = - 6. loge10 ⇒ αT = 6 x 2.3026 (or) T = 6 x 2.3026 / α

Substituting α = ( CA / 4V ) we get T = (6 x 2.3026 x 4 x V) / CA = (6 x 2.3026 x 4 x V) / (340 x

ΣaS )

i.e., T = 0.161V / Σ aS .

where C = 340 ms-1 is the velocity of sound and A = Σ aS


8

Measurement of absorption coefficient of a sound absorbing material :

Consider a sound absorbing material of area S1. In order to find the absorption coefficient of this surface, the reverberation T of a Hall is determined without placing the above material inside the hall. Then the above material is placed or suspended (if it is like a screen) inside the hall and its reverberation time is found again. Let it be T1.

T = 0.161V / Σ aS (or)

1 / T = ΣaS / 0.161V. Similarly,

1 / T1 = (ΣaS + a1S1) / 0.161V = ΣaS / 0.161V + a1S1 / 0.161V i.e., 1 / T1 = 1 / T + a1S1 / 0.161V (or) ( 1 / T1 - 1 / T ) = a1S1 / 0.161V Rearranging this, the unknown absorption coefficient (a1 ) of the surface S1 is found by using a1 = 0.161V ) x ( 1 / T1 - 1 / T ) / S1 .

Note:

The solid angle suntended by the area dS at the volume element

dV is = dS.Cosθ / r2

Where dS.Cosθ is the component of dS which is normal to the radius r drawn from Center O of dS to dV. Remember that this radius is at an

angle θ to the normal ON to the elemental surface of area dS on the wall AB.

Total Solid angle subtended by a Sphere of radius r is given by

(surface area of sphere )

Ω = ----------------------------- (square of radius)

= (4πr2 / r2 )

= 4π

dΩ = dS.cosθ / r2


9

acoustics

Documents