Explaining Wavelength Change in Shallow Water

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The discussion centers on the relationship between wave speed, wavelength, and frequency as water waves transition from deep to shallow water. It highlights that the reduction in wavelength occurs because the wave speed decreases in shallow water, leading to a shorter wavelength while maintaining energy conservation. Participants argue that as waves approach the shore, their amplitude appears to increase due to the shorter wavelength, despite the total energy remaining constant. The analogy of cars slowing down on a motorway effectively illustrates how wavelengths decrease as wave speed diminishes. The conversation also touches on resonance in air columns, questioning how motion antinodes can form at open ends despite reflections.
Craig113
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Hello!
This quote is from my physics textbook

“ From the formula speed = lambda * Frequency we can conclude that the reduction in wavelength is because water travels with a slower speed in when moving from deep to shallow water”

This does not make any sense at all. You can’t explain the change of any factor in a formula
By the fact that the product of the factors have changed. I just mean that speed in this formula is directly dependent on the lambda and the frequency, while the wavelength and its frequency are not dependent of the speed, they rather determent it together.

I think this explanation sucks. Do you agree? And if so, can you provide me with a better one on why the wavelength change when a water wave cross from depth to shallow water?

I am counting on you guys.
 
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Although it's certainly not one the areas that I'm knowledgeble in, I think that it's a sort of wedge effect. The total moving mass of water is forced into a smaller space, thereby increasing the energy density. The result is a shorter wavelength and higher amplitude. Since the bottom of the wave is leveraged upward, and the water is incompressible, the wave itself gets taller. The overall energy content can't increase, so the wavelength shortens to compensate.
I know that I didn't state that very well.
 
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Try this for an explanation on wave velocity - http://hyperphysics.phy-astr.gsu.edu/hbase/watwav.html

I believe the reason that waves slow down in moving from deep to shallow water is that in shallow water, the amplitude is greater, i.e. the height of the wave increases, so the water is pushed up and there is more inertial resistance. Out at sea, the amplitude of the wave is very small - tsunamis maybe only s few feet. In shore the wave increases in height. For a given momentum, if the mass increases, then the velocity must decrease.

http://electron4.phys.utk.edu/141/dec8/December%208.htm

http://www.es.flinders.edu.au/~mattom/IntroOc/notes/lecture09.html
 
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The Amplitude does not change otherwise energy conservation is violated, since enrgy delivered by wave is proportional to (amplitude)^2. This may appear to be the case when waves on the sea approach the beach... but actually what is happeneing is the waves are becoming steeper as the wavelength is less so they appear to become taller.

The distance between two wave crests (wavelength) decreases when wave speed decreases. An analogy is to think about cars on a motorway driving from a 70 mph zone to a 40mph zone... all of the cars bunch up (intercar length decreases) due to the decrease in speed.
 
It is not the water that is travelling: the water particles are only moving upwards and downwards. The formula you quoted is for phase speed:
v = \nu \lambda = \frac{\lambda}{T}
because a wave propagates exactly one wavelength \lambda far in the period T=1/\nu

In general, the velocity of propagation for a water wave is
v = \sqrt{\frac{g \lambda}{2 \pi} (tanh\frac{2 \pi h}{\lambda})}

so the velocity depends on the gravitational acceleration g = const, the depth h of the water and the wavelength. If h changes, lambda has to change.
 
I have another question; it’s about resonance in air columns when resonance appears in a air column with a open end. Its seems that the sound wave reflect at the ending between the open air and the air stored in the column. But if it’s so, then how can one explain motion antinodes that appear at the open end. I would seem that the only interference that should appear on that spot should be a motion nod.

It should not matter if that spot of the column is "open" or not. The only thing that opening should mean is that its "possible" for a motion antinod to appear there, if the two waves add up together in the right way. And this is impossible if the incident wave reflect exactly at the limit surface between the free air and the air in column, or at the "opening".

How does this work anyway?
 
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