Question

A weather emergency siren is mounted on a tower, 105 m above the ground. On one hand, it would be a good idea to make the siren very loud so that it will warn as many people as possible. On the other hand, safety regulations prohibit the siren from exceeding an intensity level of 101 dB for workers standing on the ground directly below the siren. Assuming that the sound is uniformly emitted, what is the maximum power that the siren can put out

Answer 1

Answer:

Explanation:

101 dB = 10.1 B.

Maximum intensity of sound allowed = 10.1 B

Intensity of sound in terms of W/m² can be found as follows

log (I / I₀) = 10.1

I / I₀ = 10¹⁰°¹

I = I₀ X 10¹⁰°¹

= 10⁻¹² X  10¹⁰°¹

= 10⁻¹°⁹ W/m²

105 m above the ground the this intensity will be 105² times

intensity at source point = 10⁻¹°⁹ x 105²

= 138.79 W/m²

energy of sound from source

= 4π times

= 4 x 3.14 x 138.79

= 1743.28W/m²

To calculate in terms of decibel :

log 1743.28 / 10⁻¹²

= log 1743.28 +12

= 15.24 B

= 152.4 dB .

152.4 dB .

Answer 2

The maximum power that the siren can put out is of 152.4 dB.

Given data:

The height of tower above the ground is, h = 105 m.

Sound intensity from the siren is, I = 101 dB.

We know that the maximum intensity of sound allowed = 10.1 B

Then, the Intensity of sound in terms of W/m² can be found as follows,

log (I / I₀) = 10.1

I / I₀ = 10¹⁰°¹

I = I₀ X 10¹⁰°¹

= 10⁻¹² X  10¹⁰°¹

= 10⁻¹°⁹ W/m²

At 105 m above the ground the this intensity will be 105² times

Then the intensity at source point = 10⁻¹°⁹ x 105²

= 138.79 W/m²

The energy of sound from source is given as,

= 4π times

= 4 x 3.14 x 138.79

= 1743.28 W/m²

To calculate in terms of decibel :

log 1743.28 / 10⁻¹²

= log 1743.28 +12

= 15.24 B

= 152.4 dB .

Thus, we can conclude that the maximum power that the siren can put out is of 152.4 dB.

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