Why is wave intensity defined as displacement^2 rather than just its magnitude?

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SUMMARY

Wave intensity is defined as the square of displacement due to its direct relationship with energy transfer. Intensity, defined as power per unit area, correlates with the energy carried by a wave, which is proportional to the square of its amplitude. This relationship is particularly evident in mechanical waves, where each point undergoes simple harmonic motion (SHM). Thus, using displacement squared ensures that intensity remains a positive value, essential for accurate calculations in classical physics.

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  • Understanding of wave mechanics
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  • Knowledge of simple harmonic motion (SHM)
  • Basic principles of energy transfer in waves
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jeebs
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pretty straightforward question. for a long time I've been blindly calculating that intensity is (displacement from zero)^2, but never questioned why.

So why is this? I get that by squaring you get a positive value, which helps in, say, quantum probability calculations, but what's wrong with just taking a magnitude of a wave at a given point and calling that the intensity?
is it just one of our arbitrary conventions in physics or is there some real reason?
 
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Intensity is a classical concept, so let's leave the quantum realm alone for now.
To answer your question, we have to look back at what intensity is defined as: crudely, it is power delivered per unit area (albeit perpendicular to the propagation), which works out to energy per unit time per unit area.

Now, the energy carried by a wave is proportional to the square of its amplitude. This can best be illustrated with mechanical waves, such as a wave on a rope, in which every point is essentially in SHM. Consequently, intensity of a wave is also proportional to (amplitude^2) of the wave.
 

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