A theatre of volume $100 \times 40 \times 10\text{ m}^3$ can accommodate 1000 visitors. The reverberation…

A theatre of volume $100 \times 40 \times 10\text{ m}^3$ can accommodate 1000 visitors. The reverberation time of the theatre when empty is $8.5\text{ s}$. If the theatre is now filled with 500 visitors, occupying the front-half seats, the reverberation time changes to $6.2\text{ s}$. The average absorption coefficient of each visitor is nearly [EAMCET 2011]
  1. 0.6
  2. 0.5
  3. 0.45
  4. 0.7

Solution

Number of visitors = 1000 Volume of theatre = $100 \times 40 \times 10\text{ m}^3$ Case I Volume acquired by one visitor $= \frac{100 \times 40 \times 10}{1000} = 40\text{ m}^3$ Reverberation time = 8.5 s Case II Volume acquired by one visitor $= \frac{100 \times 40 \times 10}{500} = 80\text{ m}^3$ Reverberation time = 6.2 s The average absorption coefficient, $\eta = \frac{40 \times 8.5}{80 \times 6.2} = \frac{85}{124} \approx 0.7$

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