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A point source emits sound waves with a power output of 135 watts. What is the sound level (in dB) at a distance of 10 m

User Nadean
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1 Answer

4 votes

Answer:

The sound level of the point source at a distance of 10 meters is approximately 110.294 decibels.

Step-by-step explanation:

First, we calculate the intensity (
I), in watts per square meter, by the Inverse-Square Law:


I= (\dot W)/(4\pi\cdot r^(2)) (1)

Where:


\dot W - Power, in watts.


r - Radius, in meters.

And the sound intenity level (
L), in decibels, is expressed by:


L = 10\cdot \log_(10) (I)/(I_(o)) (2)

Where
I_(o) is the reference sound intensity, in watts per square meter.

If we know that
\dot W = 135\,W,
r = 10\,m and
I_(o) = 10^(-12)\,(W)/(m^(2)), then we find that sound level at a distance of 10 meters is:


I= (\dot W)/(4\pi\cdot r^(2))


I = 0.107\,(W)/(m^(2))


L = 10\cdot \log_(10) (I)/(I_(o))


L \approx 110.294\,dB

The sound level of the point source at a distance of 10 meters is approximately 110.294 decibels.

User Akash Karnatak
by
7.9k points
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