The intensity of a sound wave is proportional to the square of the amplitude (pressure or velocity) and the intensity of an electromagetic wave is proportional to the square of the amplitude (electric field or magnetic field).

We saw above in Power in a wave that the power transmitted is proportional to the square of the amplitude. So we can work out the power output P of the sun: The earth is 8 light minutes from the sun (another memorable quantity). This quantity is well known to those working with solar cells, although the intensity recorded at ground level is inevitably lower. Above the earth's atmosphere, the intensity provided by the sun is 1.4 kW.m −2. This is a simple example, using a radiator that is very close to isotropic: the sun. Further, measurements are usually complicated by reflections. However, this equation should be used with caution. At a distance r, all of the radiated power is radiated through a sphere with area 4πr 2. For examples, we might think of a small source of sound high up in the air, or a small lamp as a source of light. For waves in three dimensions, power is transmitted through an area, which is why we introduce the intensity I, defined as the power per unit area.įor a simple case, imagine a small source of power P, which is radiating waves isotropically, which means equally in all directions. So the kinetic energy ½ mv 2 is proportional to μA 2ω 2.įor a wave in one dimension, as seen above, power is transmitted at a point, from the continuum to the left to the continuum to the right. Remember that, for the travelling sine wave, each point on the wave executes simple harmonic motion with amplitude A and angular frequency ω. The appearance of the squared terms is logical.

proportional to the square of the amplitude A.

proportional to the square of the angular frequency ω.

proportional to the mass per unit length μ.

proportional to the speed v at which the wave shape is travelling to the right,.

The rate of transmission of energy to the right is Over a whole number of cycles, the average of the cos 2 function is ½, so the average power transferred from the left to the right, for a wave travelling to the right, is Using v 2 = T/μ to substitute for T, we have an expression for the power at any instant t: We have already seen the partial derivatives above, and substitution gives: The component of the force in the y direction is − T sin θ and the vertical and the velocity in the y direction is ∂y/∂dt so, using the small angle approximation for θ, ds, and dividing by dt gives the power P, or rate of doing work:īecause the amplitude is small, we are considering only y motion.

From the definition of work W, a force F does work F. What is the rate at which the string to the left of a given point is doing work on the string to the right? Let's assume that our travelling wave is, as usual, y = A sin (kx − ωt). It gives background information and further details. This is a support page to the multimedia chapter Travelling Waves II in the volume Waves and Sound.

Intensity equation light wavelength