1 Answer

Akash Karnatak · Answer 1 · 2022-08-19T20:50:29+0000

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 (
), 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.

- Radius, in meters.

And the sound intenity level (
), 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.

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