Psource is 22.56 X 10⁻³ W
Explanation:
Given-
Sound Intensity level, β = 85dB
Position at which intensity is measured, x = 1.5m
y = 2.5m
We know,
Intensity, I = Power (P) / Area (a)
For a spherical wave, the intensity at distance r is:
I = P (source) / 4πr²
The sound intensity level in decibels is expressed as:
β = (10dB) log₁₀ (I/I₀)
where, I₀ = 1 X 10⁻¹² W/m²
[tex]85dB = 10 X log_1_0 (\frac{I}{1 X 10^-^1^2} )\\\\I = 10^8^.^5 X 10^-^1^2\\I = 3.16 X 10^-^4 W/m^2[/tex]
To calculate the distance r from the source to the position at which the Intensity is measured.
[tex]r = \sqrt{x^2 + y^2} \\\\r = \sqrt{(1.5)^2 + (2.5)^2} \\\\r = 2.92m[/tex]
It is given that 75% of the energy is emitted in forward hemisphere. So,
P' = 75% Po/50%
P' = 1.5P₀
By plugging in the values in equation I = P (source) / 4πr², we get
[tex]3.16 X 10^-^4 = \frac{1.5P_o}{4\pi (2.92)^2 } \\\\P_o = \frac{(3.16 X 10^-^4) X 4\pi (2.92)^2}{1.5} \\\\P_o = 22.56 X 10^-^3 W[/tex]
Thus, Psource is 22.56 X 10⁻³ W
