Standing Waves
Practice
practice problem 1
solution
Answer it.
practice problem 2
solution
Answer it.
practice problem 3
The ionosphere is a layer in the Earth's upper atmosphere where a large portion of the atoms and molecules have been ionized by exposure to the ultraviolet radiation of the sun. With so many charged particles free to roam around, the ionosphere is a reasonably good conductor of electricity. The surface of the Earth is also a reasonably good conductor. This should be somewhat obvious since 70% of the Earth's surface is covered in saltwater, which will short out electrical equipment as everyone knows, and the remaining 30% is exposed rock or soil, the stuff that electrical circuits are grounded to. The layer of atmosphere in between these two conductors is ordinary, non ionized air, which is transparent to radio waves. For extremely low frequency (ELF) radiation, the gap between the Earth and its ionosphere acts as a spherical wave guide — a kind of racetrack for radio waves. Lightning and other natural phenomena generate ELF waves at all sorts of different frequencies. Those frequencies that are just right will travel around the Earth, meet themselves in phase, and form standing waves. The set of frequencies that will do this are known as the Schumann resonances in honor of Winfried Schumann (1888–1974), the scientist who predicted their existence in 1952.
- Complete the following table…
Schumann resonances harmonic λ (km) fpredicted (Hz) fobserved (Hz) Δf/fobserved first 7.8 second 14 third 20 fourth 26 fifth 33 sixth 39 seventh 45 - Do the predicted Schumann resonances agree with the observed values to a reasonable degree? Account for any significant discrepancies.
solution
Start with a picture. Here's what the 5th harmonic looks like as an example…
For the wavelength, divide the circumference of the Earth by the number of the harmonic.
λ = C = 2πr n n Use the wave speed equation to get the frequency.
f = c ⇐ c = fλ λ Compute the relative uncertainty with the given equation.
relative uncertainty = Δf fobserved relative uncertainty = |fpredicted − fobserved| fobserved Repeat the procedure over and over again. I suggest letting a computer do the work for you. Here's a suggested method using the 5th harmonic as an example …
(2*pi*radius of earth)/5 = 8014.95684 km
speed of light/(8014.95684 km) = 37.4041263 Hz
(37.4041263Hz-33 Hz)/(33 Hz) = 0.133458373Compile your results. (Please report only a reasonable number of significant digits.) You should get something like this …
Schumann resonances harmonic λ (km) fpredicted (Hz) fobserved (Hz) Δf/fobserved first 40,100 7.48 7.8 0.041 second 20,000 15.0 14 0.069 third 13,400 22.4 20 0.12 fourth 10,000 29.9 26 0.15 fifth 8,010 37.4 33 0.13 sixth 6,680 44.9 39 0.15 seventh 5,720 52.4 45 0.16 The disagreement between theory and observation is significant (more than 10% in most cases). The flaw lies in our use of a one-dimensional wave model. The earth is a sphere, an object whose surface is a two-dimensional object. A proper solution to this problem is beyond the scope of this book (and the skills of the author).
practice problem 4
solution
Answer it.