Respuesta :
Answer:
Sound Intensity at microphone's position is [tex]9.417\times 10^{- 4} W/m^{2}[/tex]
The amount of energy impinging on the microphone is [tex]9.417\times 10^{- 8} W/m^{2}[/tex]
Solution:
As per the question:
Emitted Sound Power, [tex]P_{E} = 32.0 W[/tex]
Area of the microphone, [tex]A_{m} = 1.00 cm^{2} = 1.00\times 10^{- 4} m^{2}[/tex]
Distance of microphone from the speaker, d = 52.0 m
Now, the intensity of sound, [tex]I_{s}[/tex] at a distance away from the souce of sound follows law of inverse square and is given as:
[tex]I_{s} = \frac{P_{E}}{Area} = \frac{P_{E}}{4\pi d^{2}}[/tex]
[tex]I_{s} = \frac{32.0}{4\pi (52.0)^{2}} = 9.417\times 10^{- 4} W/m^{2}[/tex]
Now, the amount of sound energy impinging on the microphone is calculated as:
If [tex]I_{s}[/tex] be the Incident Energy/[tex]m^{2}/s[/tex]
Then
The amount of energy incident per 1.00 [tex]cm^{2} = 1.00\times 10^{- 4} m^{2}[/tex] is:
[tex]I_{s}(1.00\times 10^{- 4}) = 9.417\times 10^{- 4}\times 1.00\times 10^{- 4} = 9.417\times 10^{- 8} J[/tex]
