Mumbai University > First Year Engineering > Sem 1 > Applied Physics 1

Marks: 3M

Year: Dec 2015

Given:

$V = 20 \times 15 \times 5 m^3$

Reverberation time=3.5 sec

Find: Total absorption of its surface and average absorption co-efficient

Solution:-

$T = \frac{0.161V}{\sum \alpha S}$

$\therefore \hspace{1cm} \sum\alpha S = \frac{0.161(20 \times 15 \times 5)m^3}{3.5 s} = 69$

The surface areas of the walls, eciling and floor of the room = $2(20 \times 15 + 15 \times 5 + 5 \times 20) m^2$

$\alpha_{average} = \frac{69}{2(20 \times 15 + 15 \times 5 + 5 \times 20)}=\frac{69}{950} = 0.07$