Solving Sound Intensity with Inverse Square Law & Door Dimensions

[dura]

I had a question regarding a sound with a given Intensity and a given radius (say 0.1m from point source) for that intensity that travels through a door with dimensions given as well. It was gives a second radius (say 30m from point source) asking what the acoustic power was at a second radius after traveling through the door.

I know I am using the inverse square law to relate the second intensity, but how do I figure the door into the question?

Im sorry I can't give the exact question, it was in a quiz and I don't have any similar examples with me. I am trying to remember the question off the top of my head.

Any help would be excellent.

Regards,
Julie

 

[Valerie Park]

The answer to this question would depend on the type of door. Walls, doors and other obstacles can absorb, reflect and/or diffuse sound, which affects how much power is present at the second radius. If the door is made of a material that absorbs or reflects sound, then the acoustic power at the second radius will be lower than the original intensity at the first radius. On the other hand, if the door is made of a material that diffuses sound, then the acoustic power at the second radius may be higher than the original intensity at the first radius.