## Sound Intensity Formula :

Sound intensity I _{(db) }in decibel is equal to the product of the sound power P_{(w) }in watt is divided by the area A_{(m}^{2}_{)} in meter square. Hence the sound intensity formula has been written as,

**I _{(db) }= P_{(w)} / A_{(m}^{2}_{)}**

Where,

P → sound power in w

A → area in m^{2}

In order to measure the sound intensity level, need to compare the given sound intensity value with standard intensity value.

The formula of Sound Intensity Level is expressed as,

**I _{L}= 10 log_{10 }I /I_{0}**

Where,

I → sound intensity in db

Io → reference intensity in w/m^{2}

The unit of sound intensity is expressed in decibels (dB)

## Sample Problems :

## Problem1 :

Calculate the sound intensity at a distance of 10m? In a person whistles with the power of 2 * 10^{-4} W.

### Solution :

Given,

P (power) = 2*10^{-4}

A (here distance) = 10 m

So by using the sound intensity formula,

I = P / A

I = 2*10^{-4} / 10

I = 2*10^{-5}

Thus the sound intensity in this condition will be 2*10^{-5}W/m^{2}.

## Problem2 :

What is the level of sound sensation in decibels corresponding to an intensity wave 10^{-10} W/m^{-2}?

### Solution :

The first thing to notice is that they are already giving us the intensity of sound, so we can easily calculate the sound sensation.

I = 10^{-10} W/m^{-2}.

S = 10*log(I/I_{0})

= 10*log(10^{-10}W/m^{-2/0-12} W/m^{-2})

= 20 dB

S = 20 dB.