\(T=\frac{0.16 V }{\sum a s}\)
Where \(V\) is volume of hall in \(m^3\)
\(\sum a s=a_1 s_1+a_2 s_2+\ldots \ldots=\) Total absorption of the hall (room)
Here \(s_1, s_2, s_3 \ldots \ldots \ldots\) are surface area of the absorbers and \(a_1, a_2, a_3 \ldots \ldots \ldots\) are their respective absorbption coefficients
\(\frac{T^{\prime}}{T}=\frac{V^{\prime}}{s^{\prime}} \times \frac{s}{V}=\frac{(2)^3}{(2)^2}=\frac{8}{4}=2\)
Hence, \(T^{\prime}=2 T=2 \times 1=2\; s\).
${y}=1.0\, {mm} \cos \left(1.57 \,{cm}^{-1}\right) {x} \sin \left(78.5\, {s}^{-1}\right) {t}$
${x}>0$ ના ક્ષેત્રમાં ઉગમબિંદુથી નજીકનું નિસ્પંદ બિંદુ ${x}=\ldots \ldots \ldots\, {cm}$ અંતરે હશે.