- •Velocity (V) is the distance moved per second in a fixed direction.
- •A. Airborne Sound
- •B. Structure Borne Sound
- •Acoustical defects
- •Room acoustic
- •Shaping an auditorium to improve the path of sound from source to audience.
- •Hall Shapes
- •Noise control
- •Sound absorption is the prevention of reflection of sound or alternatively, a reduction in the sound energy reflected by the surfaces of a room.
- •Sound insulation is the prevention of transmission of sound or alternatively, a reduction of sound energy transmitted into an adjoining air space.
- •Sound Insulation
- •Calculation
- •Tutorial questions:
Calculation
1. A particular sound wave has frequency of 440Hz and a velocity of 340m/s. Calculate the wavelength of this sound.
Answer:
Given, v = 340m/s, f = 440 Hz
Wave Length, = v (Speed of Sound)
f (Frequency)
= 340
440
= 0.7727m
2. A particular sound wave has frequency of 440Hz and a wavelength of 1m. Calculate the speed of this sound.
Answer:
Velocity (Speed of sound), v = f (Frequency) x (wavelength)
= 440 x 1
= 440m/s
3. A lecture hall with a volume of 1500m3 has the following surfaces finishes, areas and the total absorption of the lecture hall surfaces is 99 m2 sabin. Calculate the reverberation time of this hall.
Using the sabine formula;
Reverberation Time, t = 0.16 V = 0.16 x 1500m3
A 99
Therefore, the reverberation time = 2.42 s
4. A hall has a volume of 5000m2 and a reverberation time of 1.6 s. Calculate the amount of extra absorption required to obtain a reverberation time of 1 s.
Answer:
Given, t1 = 1.6 s , A1 = ?, t2 = 1.0 s, A2 = ?, V = 5000m3
Using the sabine formula;
Reverberation Time, t = 0.16 V
A
So A1 = 0.16 V = 0.16 x 5000 = 500 m2 sabins
t1 1.6
A2 =0.16 V = 0.16 x 5000 = 800 m2 sabins
t2 1.0
Therefore the extra absorption needed = A2 - A1 = 800 – 500 = 300m2 sabin
Tutorial questions:
1. A particular sound wave has frequency of 500Hz and a velocity of 380m/s. Calculate the wavelength of this sound.
Answer:
Given, v = 380m/s, f = 500 Hz
Wave Length, = v (Speed of Sound)
f (Frequency)
= 380
500
= 0.76m
2. A particular sound wave has frequency of 520Hz and a wavelength of 1.5m. Calculate the speed of this sound.
Answer:
Velocity (Speed of sound), v = f (Frequency) x (wavelength)
= 520 x 1.5
= 780m/s
3. A room of 900m2 volumes has a reverberation time of 1.2 s. Calculate the amount of extra absorption required to obtain a reverberation time of 0.8 s.
Answer:
Given, t1 = 1.2 s , A1 = ?, t2 = 0.8 s, A2 = ?, V = 900m3
Using the sabine formula;
Reverberation Time, t = 0.16 V
A
So A1 = 0.16 V = 0.16 x 900 = 120 m2 sabins
t1 1.2
A2 =0.16 V = 0.16 x 900 = 180 m2 sabins
t2 0.8
Therefore the extra absorption needed = A2 - A1 = 180 – 120 = 60m2 sabin
4. Calculate the actual reverberation time for the above hall with a volume of 5000m2, given the following data for a frequency of 500Hz.
-
Surface Area
Absorption coefficient
500m2 brickwork
0.03
600m2 plaster on solid
0.02
100m2 acoustic board
0.70
300m2 carpet
0.30
70m2
0.40
400 seats
0.30 units each
Answer:
-
Surface Area
Absorption coefficient
Total absorption
500m2 brickwork
0.03
15
600m2 plaster on solid
0.02
12
100m2 acoustic board
0.70
70
300m2 carpet
0.30
90
70m2
0.40
28
400 seats
0.30 units each
120
Total
335
Using the sabine formula;
Reverberation Time, t = 0.16 V
A
t = 0.16 V = 0.16 x 5000 = 2.4 s
A 335
Therefore, the reverberation time = 2.4 s
5. A large cathedral has a volume of 120 000m3. When the space is empty the reverberation time is 9 s. With a certain number of people present the reverberation time is reduced to 6 s. Calculate the number of people present, if each person provides an absorption of 0.46m3 sabins.
Using the sabine formula;
Reverberation Time, t = 0.16 V = 0.16 V
A N x ά
Given, t1 = 9s, V = 120 000, ά = 0.46
N1 = 0.16 V = 0.16 x 120 000 = 4638 persons
t1 x ά 9 x 0.46
t2 = 6s,
N2 = 0.16 V = 0.16 x 120 000 = 6957 persons
t2 x ά 6 x 0.46
Therefore, the number of people present = N2 - N1 = 6957 – 4638 = 2319 number of people present.