Theory of Intelligence
11TH•Physics•Waves And Vibrations•Intensity (I) of a Wave

7.10 Intensity $(I)$ of a Wave

Energy per unit area per unit time, I ∝ A².

7.10 Intensity (I)(I) of a Wave

Intensity is defined as the amount of energy transmitted per unit area per unit time in the direction of propagation of progressive wave.

It is a measure of the power of a wave and is usually denoted by the symbol "II". It is measured in units of watts per square metre (W/m2)(\mathrm{W/m^{2}}).

A progressive wave or travelling wave is one that travels through a medium in a consistent direction and transferring energy from one point to another. It is a wave that propagates or moves forward, as opposed to a stationary or standing wave. Examples of progressive waves include water waves, sound waves, light waves, etc.

By definition, the intensity of a wave is:

I=EA×tI = \frac{E}{A \times t} I=PA(∵E/t=P)I = \frac{P}{A} \quad (\because E/t = P)

Here:

  • I=I = Intensity of wave in W m−2\mathrm{W\,m^{-2}}
  • E=E = Energy in joules (J)(\mathrm{J})
  • t=t = Time in seconds (s)(\mathrm{s})
  • P=P = Power in watts (W)(\mathrm{W})

We know that in mechanical waves, such as sound waves, water waves, or waves on a vibrating string, energy is stored as kinetic energy and potential energy of the medium's particles. How much energy is stored depends upon the displacement (amplitude) of the particles from the mean position. Therefore, the intensity II of waves is proportional to the square of the amplitude AA, i.e.:

I∝A2I \propto A^{2}

or

I=kA2…(7.14)I = kA^{2} \quad \ldots (7.14)

Here kk is the constant of proportionality and depends upon the physical properties of the wave and the medium.

Example 7.8

(a) A wave has an intensity of 0.5 W m−20.5\,\mathrm{W\,m^{-2}} at a distance of 3.0 m3.0\,\mathrm{m} from the source. What is the power of the wave?

(b) Two progressive waves have intensities of 0.5 W m−20.5\,\mathrm{W\,m^{-2}} and 0.25 W m−20.25\,\mathrm{W\,m^{-2}}. Find total intensities of two waves.

Solution

(a) I=0.5 W m−2I = 0.5\,\mathrm{W\,m^{-2}}, r=3.0 mr = 3.0\,\mathrm{m}, P=?P = ?

∵I=P/A=P/(4πr2)\because I = P/A = P/(4\pi r^{2})

Putting the values: 0.5 W m−2=P/[4×3.14×(3.0 m)2]0.5\,\mathrm{W\,m^{-2}} = P/[4 \times 3.14 \times (3.0\,\mathrm{m})^{2}]

P=0.5×113.04=56.52 WP = 0.5 \times 113.04 = 56.52\,\mathrm{W}

(b) I1=0.5 W m−2I_{1} = 0.5\,\mathrm{W\,m^{-2}}, I2=0.25 W m−2I_{2} = 0.25\,\mathrm{W\,m^{-2}}

Total Intensity=0.5+0.25=0.75 W m−2\text{Total Intensity} = 0.5 + 0.25 = 0.75\,\mathrm{W\,m^{-2}}

Example 7.9

A speaker is emitting sound waves with a power of 5050 watts. If the sound waves are spreading out evenly in all directions and the intensity of the sound waves is measured at a distance of 5 m5\,\mathrm{m} from the speaker, what is the intensity of the sound waves if the area of the sphere (the surface area of a sphere) at that distance is approximately 314 m2314\,\mathrm{m^{2}}?

Solution

Power P=50 WP = 50\,\mathrm{W}

Area A=314 m2A = 314\,\mathrm{m^{2}}

Intensity == Power/Area

I=PA=50 W314 m2≈0.159 W/m2I = \frac{P}{A} = \frac{50\,\mathrm{W}}{314\,\mathrm{m^{2}}} \approx 0.159\,\mathrm{W/m^{2}}

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