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Additional: Wave Intensity and Energy



Intensity of Waves ($ I \propto A^2\omega^2 v $)

Waves carry and transfer energy. The intensity of a wave is a measure of the power transferred by the wave per unit area. It quantifies how much energy is flowing through a given region of space per unit time.


Definition of Intensity (I)

The intensity ($I$) of a wave is defined as the average power ($\overline{P}$) transmitted by the wave per unit area ($A$) perpendicular to the direction of wave propagation.

$ I = \frac{\overline{P}}{A} = \frac{\text{Energy Transfer}}{\text{Area} \times \text{Time}} $

The SI unit of intensity is Watts per square metre (W/m$^2$).


Relationship between Intensity and Wave Parameters

The intensity of a wave is related to its amplitude, frequency, and the properties of the medium. For a wave of angular frequency $\omega$ and amplitude $A$ travelling with speed $v$ in a medium of density $\rho$, the intensity is generally found to be proportional to:

$ I \propto A^2 \omega^2 \rho v $

More specifically, the average power per unit area transmitted by a harmonic wave is often given by a formula like:

$ I = \frac{1}{2} \rho A^2 \omega^2 v $ (for mechanical waves, where A is displacement amplitude)

or related to pressure amplitude for sound waves:

$ I = \frac{(\Delta P_{max})^2}{2\rho v} $ (for sound waves, where $\Delta P_{max}$ is pressure amplitude)

The key takeaway is the proportionality: the intensity is proportional to the square of the amplitude ($A^2$) and the square of the angular frequency ($\omega^2$). It also depends on the wave speed ($v$) and the density ($\rho$) of the medium. The proportionality mentioned in the subheading, $I \propto A^2\omega^2 v$, is missing the density term $\rho$, which is typically part of the proportionality constant and relates to the medium's impedance. A more complete proportionality for a specific medium would be $I \propto A^2\omega^2$.

The fact that Intensity $\propto A^2$ is very important. Doubling the amplitude of a wave increases its intensity by a factor of four. For sound waves, this means the loudness (related logarithmically to intensity) increases significantly with amplitude.

Example 1. The intensity of a sound wave is $10^{-6}$ W/m$^2$ at a certain point. If the amplitude of the wave is doubled, what is the new intensity at the same point?

Answer:

The intensity of a wave is proportional to the square of its amplitude ($I \propto A^2$).

Let the initial intensity be $I_1$ and the initial amplitude be $A_1$. So, $I_1 = k A_1^2$ for some constant $k$. We are given $I_1 = 10^{-6}$ W/m$^2$.

Let the new amplitude be $A_2 = 2A_1$. The new intensity is $I_2 = k A_2^2$.

Substitute $A_2 = 2A_1$ into the expression for $I_2$:

$ I_2 = k (2A_1)^2 = k (4 A_1^2) = 4 (k A_1^2) $

Since $I_1 = k A_1^2$, we have:

$ I_2 = 4 I_1 $

$ I_2 = 4 \times (10^{-6} \text{ W/m}^2) = 4 \times 10^{-6} $ W/m$^2$.

If the amplitude of the wave is doubled, the new intensity is $4 \times 10^{-6}$ W/m$^2$, which is four times the original intensity.



Energy Density of Waves

Waves carry energy as they propagate. The energy is distributed throughout the medium or region occupied by the wave. The energy density ($u$) of a wave is the average energy per unit volume carried by the wave.


Relationship between Intensity and Energy Density

Intensity is the power (energy per unit time) per unit area. Consider a plane wave travelling in the x-direction with speed $v$. In a time interval $\Delta t$, the wave travels a distance $v \Delta t$. The volume of the medium swept by the wave through a cross-sectional area $A$ is $V = A (v \Delta t)$. If the average energy density in this volume is $\overline{u}$, the energy contained in this volume is $E = \overline{u} V = \overline{u} A v \Delta t$.

The power transmitted through area $A$ is $P = E / \Delta t = (\overline{u} A v \Delta t) / \Delta t = \overline{u} A v$.

Intensity is $I = P/A = (\overline{u} A v) / A = \overline{u} v$.

$ I = \overline{u} v $

So, the intensity of a wave is equal to the average energy density multiplied by the wave speed.

$ \overline{u} = \frac{I}{v} $

The units of energy density are Joules per cubic metre (J/m$^3$).


Energy Density for Specific Waves

For a mechanical wave, the energy is partly kinetic energy of the moving particles and partly potential energy stored in the deformation of the medium. For a harmonic wave, these energies oscillate in time and space, but the average energy density is related to the wave parameters. For a linear harmonic wave with displacement amplitude $A$ and angular frequency $\omega$ in a medium of density $\rho$ and wave speed $v$:

Average kinetic energy density: $ \overline{u}_{KE} = \frac{1}{4} \rho A^2 \omega^2 $

Average potential energy density: $ \overline{u}_{PE} = \frac{1}{4} \rho A^2 \omega^2 $

The total average energy density is the sum: $ \overline{u} = \overline{u}_{KE} + \overline{u}_{PE} = \frac{1}{2} \rho A^2 \omega^2 $.

Using $I = \overline{u} v$, this leads back to the intensity formula $ I = \frac{1}{2} \rho A^2 \omega^2 v $. The consistency between these formulas reinforces the relationship between intensity, energy density, and wave speed.

For electromagnetic waves, the energy is stored in the electric and magnetic fields. The average energy density of an electromagnetic wave in vacuum is $\overline{u}_{EM} = \frac{1}{2}\epsilon_0 E_{rms}^2 + \frac{1}{2\mu_0} B_{rms}^2$. Since $E_{rms} = c B_{rms}$ and $c = 1/\sqrt{\mu_0\epsilon_0}$, this simplifies to $\overline{u}_{EM} = \epsilon_0 E_{rms}^2 = \frac{1}{2}\epsilon_0 E_{max}^2$, where $E_{max}$ is the amplitude of the electric field. The intensity is $I = \overline{u}_{EM} c$, so $I = \frac{1}{2}c\epsilon_0 E_{max}^2 = \frac{E_{max}^2}{2\mu_0 c}$.



Inverse Square Law for Intensity

For waves that spread out uniformly in three dimensions from a point source (like sound waves from a loudspeaker or light waves from a bulb in open space), the intensity of the wave decreases with the distance from the source. This relationship is described by the Inverse Square Law.


Derivation of the Inverse Square Law

Consider a point source S emitting waves uniformly in all directions in an isotropic medium (properties are the same in all directions). Assume the source radiates power $P_{source}$ isotropically.

As the wave propagates outwards, the energy emitted by the source spreads over increasingly larger spherical surfaces centered on the source. If we consider a spherical surface of radius $r$ centered at the source, the area of this surface is $A = 4\pi r^2$.

Diagram illustrating the inverse square law for intensity.

(Image Placeholder: A point source S in the center. Show two concentric spheres around S with radii r1 and r2 (r2 > r1). Show the wave energy spreading over the increasing area of the spheres as it moves outwards. Indicate that the total power passing through the sphere at r1 is the same as the total power passing through the sphere at r2.)

Assuming no energy is absorbed by the medium, the total power passing through any spherical surface centered on the source must be equal to the power radiated by the source, $P_{source}$. The intensity $I$ at a distance $r$ is the power passing through the spherical surface divided by the area of that surface:

$ I(r) = \frac{P_{source}}{\text{Area of Sphere}} = \frac{P_{source}}{4\pi r^2} $

Since $P_{source}$ and $4\pi$ are constants, the intensity $I(r)$ is inversely proportional to the square of the distance $r$ from the source:

$ I(r) \propto \frac{1}{r^2} $

This is the Inverse Square Law for Intensity.

If we compare the intensities at two different distances $r_1$ and $r_2$ from the source:

$ I_1 = \frac{P_{source}}{4\pi r_1^2} $

$ I_2 = \frac{P_{source}}{4\pi r_2^2} $

$ \frac{I_2}{I_1} = \frac{P_{source}/(4\pi r_2^2)}{P_{source}/(4\pi r_1^2)} = \frac{r_1^2}{r_2^2} = \left(\frac{r_1}{r_2}\right)^2 $

So, the ratio of intensities is the inverse ratio of the square of the distances.


Conditions and Limitations

The Inverse Square Law is valid for:

Despite limitations in real media, the Inverse Square Law is a fundamental concept for understanding how intensity diminishes with distance from a source in open space. For sound waves, this means that as you move farther from a sound source, the loudness decreases significantly. For light, the brightness of a light source decreases with distance.

Example 2. The intensity of sound from a point source is measured to be $2 \times 10^{-5}$ W/m$^2$ at a distance of 10 metres from the source. What is the intensity at a distance of 40 metres from the source? Assume the source radiates uniformly and the medium is non-absorbing.

Answer:

Initial distance, $r_1 = 10$ m.

Initial intensity, $I_1 = 2 \times 10^{-5}$ W/m$^2$.

Final distance, $r_2 = 40$ m.

Let the final intensity be $I_2$. Using the Inverse Square Law for intensity from a point source:

$ \frac{I_2}{I_1} = \left(\frac{r_1}{r_2}\right)^2 $

$ I_2 = I_1 \left(\frac{r_1}{r_2}\right)^2 $

Substitute the values:

$ I_2 = (2 \times 10^{-5} \text{ W/m}^2) \times \left(\frac{10 \text{ m}}{40 \text{ m}}\right)^2 $

$ I_2 = (2 \times 10^{-5}) \times \left(\frac{1}{4}\right)^2 $

$ I_2 = (2 \times 10^{-5}) \times \frac{1}{16} $

$ I_2 = \frac{2}{16} \times 10^{-5} = \frac{1}{8} \times 10^{-5} = 0.125 \times 10^{-5} $

$ I_2 = 1.25 \times 10^{-6} $ W/m$^2$.

The intensity at a distance of 40 metres is $1.25 \times 10^{-6}$ W/m$^2$. As the distance increased by a factor of 4, the intensity decreased by a factor of $4^2 = 16$. $(2 \times 10^{-5}) / 16 = 0.125 \times 10^{-5} = 1.25 \times 10^{-6}$.