Intensity and its relation with amplitude I was going through the texts of my book and read that intensity of a wave is proportional to its amplitude squared. I don't know how this relation came?
I thought for it but didn't come with any solution. My intuition was this :
Amplitude of fields is proportional to the energy supplied to an oscillating charge because the more energy is given the more change should come. Also the more energy is given, the more energy will be released. From the above statements we can conclude that energy released per unit time and per unit area i.e. intensity must be proportional to the amplitude not its square.
What's wrong in the this? Can the well known relation be proved?
 A: Amplitude of an electromagnetic wave field amounts to the average electric field strength: the average of the absolute value of the electric field.  From electrostatics we know that the energy density of an electric field is proportional to the square of the electric field's absolute value.  So the energy density of an EM wave field is proportional to the square of its amplitude.  "Intensity" is energy density (or power density to be more precise), so is proportional to the square of amplitude.
A: It is easy to see the relation between amplitude and energy in an elastic wave travelling in the $x$ direction in a string.
Each small portion of the string oscillates transversely (say in the $y$ direction) as the wave passes by. If the density is $\rho$:
$$E_k = \frac{1}{2}\rho v_y^2$$ is the energy per unit of volume.
But the equation for the wave displacement is:
$$y = A\cos(k(x - vt))$$
$$v_y = Akv\sin(k(x-vt)) = A\omega \sin(k(x-vt))$$
$$v_y^2 = A^2\omega^2 \sin^2(k(x-vt))$$
$$E_k = \frac{1}{2}\rho A^2\omega^2 \sin^2(k(x-vt))$$
So, the energy is proportional to the square of the amplitude.
