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Sound Intensity Formula

In a quiet forest, we may sometimes hear a single leaf fall to the ground. Similarly, after settling into the bed, we may hear our blood pulsing through our ears. On the other hand, when a passing motorist has his stereo turned up, then we cannot even hear what the person next to us in our car is saying. We are all familiar with the loudness of sounds. Here, the relevant physical quantity is the sound intensity. This is the concept that is valid for all sounds whether or not they are in the audible range. The student will learn the sound intensity formula. Let us learn the interesting concept!

Sound Intensity Formula 

Sound intensity Formula

What is the sound intensity?

The intensity of sound is defined as the power of sound per unit area. The usual context is the measurement of the intensity of sound in the air. It also depends on the surface of the sound source. The increasing amplitude of the source and that of the vibrating surface causes the kinetic energy of the mass of air.

This kinetic energy increases with the mass of air into vibration and with its average speed. The intensity of perception of a sound by the ear also depends on the distance from the sound source. Intensity also depends on the nature of the elastic medium between the source of a sound and the ear. Non-elastic media, like wool, felt, etc., weaken the sounds.

The formula for sound Intensity:

It is defined mathematically as:

Sound intensity = \(\frac {acoustic power}{ normal area to the direction of propagation}\)




I Sound intensity.
P Acoustic power.
A Normal area to the direction of propagation


The intensity of sound is measured in decibel unit (dB). The scale of sound sensation is logarithmic, it means an increase of 10 dB will correspond to an intensity 10 times greater. For example, the noise of the waves on the coast is 1,000 times more intense than a whisper, which equals an increase of 30 dB.

The conversion between intensity and decibels follows the equation:

The intensity in decibels = \(10 \times log_{10} \frac {intensity}{ intensity of zero decibels}\)


S = \(10\log_{10}left\frac{I}{I_0}right\)

S intensity in decibels.
I sound intensity.
\(I_0\) sound intensity of zero decibels= \(10^{-12} W per m^2\)

Thus sound intensity is applicable as sound per unit area perpendicular to the direction of sound waves. The standard text is the measurement of the sound intensity of the noise in the air at the location of the listener as a sound energy quantity.

Solved Example

Q.1: What will be the level of sound sensation in decibels corresponding to an intensity wave \(10^{-10} W per m^2\)?

Solution: Here,

I = \(10^{-10} W per m²\)

Thus using formula,

S = \(10\log_{10}\frac{I}{I_0}S = 10\log_{10}\frac{I}{I_0}\)

= 20 dB

= 20 dB.

The level of sound sensation will be 20 dB.

Q.2:  A person whistles with the power of 0.9 \(\times 10^{-4}\) W. Determine the sound intensity at a distance of 7 meters.


Given parameters are:

Sound power, P = 0.9 \(\times 10^{-4}\) ,

Area, A = 7

Sound intensity formula is given as,

I=\(\frac{P}{A} \)

I=\(\frac{0.9 \times 10^{-4} }{7} \)

I= 1.28 \(\times 10^{-5} W per m^2\).

Thus sound intensity will be 1.28 \(\times 10^{-5} W per m^2\).

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