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# Intensity Formula

In a quiet forest, we can sometimes hear a single leaf fall to the ground. We may hear our blood pulsing through our ears when we are in the bed at night. On the other hand, when a passing motorist has his stereo turned up, we cannot even hear what the person beside you in your car is saying. These examples are giving ideas about the loudness of sounds and hence the intensity of sound. Also, these are related to how energetically the source is vibrating. The relevant physical quantity is sound intensity, and this concept is valid for all sounds whether or not they are in the audible range. In this article, we will discuss the sound intensity and intensity formula with examples. Let us learn the concept!

Source: en.wikipedia.org

## Intensity Formula

### Concept of the intensity of the wave:

Intensity is the quantity of energy which the wave conveys per unit time across the surface of the unit area. Also, it is equivalent to the product of energy density and wave speed.

We normally measure it with units of watts per square meter. The magnitude of intensity will depend on the strength and amplitude of a wave under propagation. We represent Intensity by the symbol I.

Sound intensity is sometimes also referred to as acoustic intensity. It is computed as the power carried by the sound waves per unit area in the direction perpendicular to that area. The SI unit of intensity i.e. sound intensity is the watt per square meter.

The intensity of one sound can be compared to the intensity of another sound source with the same frequency. It can be done by taking the ratio of power values of these. When this ratio is ten, the difference in intensity of the sounds is measured as one bel.

### Intensity Formula

The formula for intensity is:

$$I = \frac {P}{A}$$

Where

 I Intensity P Power A Area of cross-section

## Solved Examples

Q.1: Calculate the intensity of a wave propagating with the power of 20 KW. The area of the cross-section of the surface is $$35 \times 10^6$$ square meters.

As given in the problem known measures are as follows:

P = 20 KW

P = $$20 \times 10^3 W$$,

A = $$35 \times 10^6$$ square meter.

Intensity formula is:

I =$$\frac {P}{A}$$

= $$\frac {20 \times 10^3 }{ 35 \times 10^6}$$

= $$0.57 \times 10 ^ {-3} Watt per square meter.$$

Therefore the intensity will be $$0.57 \times 10 ^{-3}$$ Watt per square meter.

Q.2: Find out the power of a wave whose intensity is given as $$30 \times 10 ^{-5}$$ Watt per square meter. And the area of cross-section is 100 square meters.

Known quantities in the problem are,

I = $$30 \times 10^{-5}$$

and Area of cross section, A = 100 square meter

Then we may  use the Intensity formula as:

I = $$\frac {P}{A}$$

Rearranging the formula we get,

P = $$I \times A$$

Substituting the known values,

P = $$30 \times 10 ^{-5} \times 100$$

P = $$30 \times 10^{-3} Watt$$

Therefore power will be $$30 \times 10^{-3}$$ Watt.

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