What is the Power Output of a Bird's Song?

[PHYclueless]

Hello. I was hoping someone could give me a direction to go with this homework problem. I'm having trouble trying to figure it out. Here's the question:

Two people relaxing outdoors listen to a bird sing. One person, only 1.00m from the bird, hears the sound with an intesity of 2.80x10^-6 W/m^2. (a) What is the power output of the bird's song? (b) What intensity is heard by a second person, which is 4.25m from the bird?

The answer I get doesn't correlate with the correct answer given. This is what I did for the first part. I can't solve the second part w/o the correct power from the first.

Intensity=Power/Area
P=IA
P=2.80x10^-6(1m) = 2.80x10^-6

Can anyone give me a clue of another direction to go? The answer for part (a) is 3.52x10^-5.

Thank you!

 

[Doc Al]

Consider the bird as a point source of sound emanating in all directions. At a distance of 1 m, how much area does the sound cover?

 

[PHYclueless]

Thank you so much for your help. So basically I need to use pi(r^2) and r would =.5m.

I have two more questions but will post them seperatly as they relate to thermal processes and heat as it relates to energy.

Thanks again for your help.

 

[lightgrav]

Not at all ... the BIRD is in the CENTER of a sphere of sound (it is emitting),
the listener is at the EDGE of the sphere (at one wave front).
The Power from the bird into sound waves , from 1/340 second ago ,
has spread to cover the entire surface of the sphere.

Did you draw a diagram of this situation?

 

[Curious3141]

Thank you so much for your help. So basically I need to use pi(r^2) and r would =.5m.

No. What is the formula for surface area of a sphere ?

 

[PHYclueless]

Got it! Thank you. I was thinking of it as an area of a circle instead of a sphere and after you said that it made me realize I needed to use a different formula. I=Pav/4(pi)(r^2).

Thanks to you both :). Appreciate it.