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Gnom [1K]
3 years ago
6

As the distance of a sound wave from its source quadruples, how is the intensity changed?

Physics
2 answers:
timurjin [86]3 years ago
8 0
Through the 1/r law, it was concluded that for the sound intensity or pressure is directly proportional to distance or radius. That is,
                                 p = k/r 
where k is the proportionality constant. If the distance of a sound wave is quadrupled then, the intensity of the sound is decreased to 1/4 of its original value. 
Ainat [17]3 years ago
6 0

Answer: Hello mate!

Here we have a sound wave, which expands itself like a sphere that starts in the source.

the model that describes the intensity of the wave with respect of the distance to the source is:

A(r) = A₀/(4πr^2)

where A₀ is the initial intensity, π is 3.14159265 and r is the distance to the source.

then if you quadriply the distance, you now have:

A(4r) = A₀/(4π(4r)^2) = A₀/(4π16r^2) = A(r)/16

this means that if you quadruply the distance to the source, the intensity of the sound wave is 16 times smaller, this is because the intensity decreases with the distance squared.

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