Why Is Wave Intensity Proportional to the Square of Amplitude?

[Pyroadept]

Homework Statement

Q. Explain why the intensity of a sound or light wave is proportional to A^2, where A is the amplitude of the wave.

Homework Equations

S = (1/μ_0)ExB ... Poynting vector
S = (1/μ_0)EB
B = E/c
E = Asin(kx - ωt)

The Attempt at a Solution

Hi everyone,

So, for light, I used the pointing vector as above, substituting in B = E/c to get
S = (1/cμ_0)E^2

and then E = Asin(kx - ωt) to get
S = (1/cμ_0)A^2sin^2(kx - ωt)

And I used I = S_average

So I = (1/cμ_0)[A^2sin^2(kx - ωt)]_avg

From which we can conclude that I α A^2.

Is this correct?

As regards sound, I really don't know what to do, as it's not an electromagnetic wave - is it still possible to use the Poynting vector?
Please help!

 

[anathema]

Hi there,

I can confirm that your approach for light is correct. The intensity of a light wave is indeed proportional to the square of its amplitude, as shown by the Poynting vector equation.

For sound waves, the intensity is not directly related to the amplitude. Instead, it is related to the square of the pressure amplitude. This is because sound waves are longitudinal waves, meaning that the particles in the medium vibrate back and forth in the direction of the wave's propagation. The pressure amplitude is directly related to the amplitude of these vibrations, which in turn affects the intensity of the sound wave.

So, while the Poynting vector may not be applicable to sound waves, the concept of intensity being proportional to A^2 still holds true. I hope this helps clarify things for you. Keep up the good work in your studies!

 

[nlightner]

Hello,

Your approach for light is correct. The Poynting vector is defined as the average power per unit area, and it is proportional to the square of the electric field. This is because the energy carried by a wave is proportional to the square of its amplitude, and the power is the rate at which energy is transferred per unit time.

For sound waves, the intensity is defined as the average power per unit area, just like for light waves. However, the Poynting vector is specific to electromagnetic waves and cannot be used for sound waves. Instead, we can use the equation I = P/A, where P is the power of the wave and A is the cross-sectional area through which the wave is passing. For sound waves, the power is proportional to the square of the amplitude of the wave, just like for light waves. Therefore, the intensity is also proportional to the square of the amplitude for sound waves.

I hope this helps clarify the relationship between wave intensity and amplitude for both light and sound waves. Let me know if you have any further questions.