contestada

A concert loudspeaker suspended high off the ground emits 39 w of sound power. a small microphone with a 1.0 cm2 area is 60 m from the speaker. part a what is the sound intensity at the position of the microphone?

Respuesta :

As we know that intensity of sound is defined as power received per unit area

so here the power of source is given as

[tex]P = 39 W[/tex]

distance of microphone is given as

[tex]d = 60 m[/tex]

now if loudspeaker is considered as spherical source then we will have

[tex]Intensity = \frac{power}{area}[/tex]

[tex]Intensity = \frac{39}{4\pi r^2[/tex]

here

r = d = 60 m

[tex]Intensity = \frac{39}{4\pi (60^2)}[/tex]

[tex]Intensity = 0.86 \times 10^{-3} W/m^2[/tex]