Question

a siren is mounted high in the air and produces 10 W of sound power at 50 Hz. if you are 15 m away from this siren, how many phons would you register at your location?

Answer 1

At a distance of 15 m from the siren emitting 10 W of sound power at 50 Hz, the sound intensity level would be approximately 90 phons.

Step-by-step explanation

Sound intensity level (L) in phons is calculated using the formula:

`\[ L = 10 \cdot \log_(10)\left((I)/(I_0)\right) + 40 \`]

where(
`I \)`\ is the sound intensity and
`\( I_0 \)` is the reference intensity (usually taken as
`\(1 * 10^(-12) \, \text{W/m}^2\)).`

Given that the siren produces 10 W of sound power, the sound intensity
`\( I_0 \)` can be determined using the formula:

`\[ I = (P)/(4\pi r^2) \]`

where
`\( P \)` is the sound power and
`\( r \)`is the distance from the source.

Substituting the values, we get:

`\[ I = (10)/(4\pi \cdot (15)^2) \]`

After calculating
`\( I \),` substitute this value into the first formula to find
`\( L \).` In this case, L is the sound intensity level in phons.

The final result is approximately 90 phons. This means that the perceived loudness at a distance of 15 m is equivalent to the loudness of a sound at 90 phons, as per the psychoacoustic model that relates sound intensity to perceived loudness.