Numerical
A hall of dimension 25×18×12m3 has an average absorption coefficient 0.2. find the reverberation time. If a curtain cloth of area 150m2 is suspended at the centre of Hall with coefficient of absorption 0.75, What will be the reverberation
time?
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Solution
Given Data :- volume of hall = 25 ×18 ×12 = 5400
α = 0.75 αav = 0.2 Scurtain = 100 m2
Formula :- `T_1 = 0.161 × V/(α_(av) × S`)`
`T_2 = 0.161 × V/( α_(av) × S + α_("curtain ") × S`
Calculate :- S = 2[(20 ×15)+(15 ×10)+(10 ×20)] = 1300m2
V = 20×15×10 = 3000 m2
`T_1 = 0.161 × 3000/(0.1×1300) = 3.7 sec`.
∴ Absorption takes place by both the surfaces of the curtain
S’ = 2×100 m2 = 200 m2
`T_2 = 0.161 × 3000/ ((0.1 ×1300)+(0.66 ×200)) = 1.84 sec`
Answer :- Change in reverberation time = 1.85 sec.
Concept: Reflection of Sound(Reverberation and Echo)
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