How Do Nodes Function in Standing Waves with Fixed and Open Ends?

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In standing waves, when a wave is fixed at both ends, the fundamental frequency (1st harmonic) has 2 nodes, including the fixed ends. For a wave fixed at one end and open at the other, the frequency equation f(n) = n*v / 4L uses odd integers for n, representing the harmonic level rather than the number of nodes. The first harmonic has 1 node at the fixed end, while the absence of a second harmonic occurs because n can only be odd. The concept of n relates to the number of half-wavelengths fitting into the length L, with odd harmonics resulting in a node at one end and an antinode at the other. Understanding this distinction clarifies why harmonic numbers and node counts differ in various wave configurations.
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Homework Statement


I don't have an actual problem to solve, it's more of a conceptual problem.

I'm trying to understand harmonics and standing waves, and how the nodes work out, but it's very confusing.

(1) If a wave is FIXED at both ends, for the fundamental or 1st harmonic, does it have 0 nodes, 1 node, or 2 nodes? Do you count both "end nodes" where the wave is fixed?

(2) If a wave is FIXED at one end, and OPEN at the other end, I know that the equation for frequency is f(n) = n*v / 4L, where n=1,3,5,7...
Does "n" represent the # nodes, or the harmonic level?
This is super confusing for me, because for f(1), there is 1 node at the fixed end, which makes sense, because this is the first harmonic.
However, for f(2), there are 2 nodes, but "n" can only equal odd numbers. So there is no second harmonic... Why is that? So to find values of n, we have to go by harmonic number, and not the number of nodes, for all situations (ie. open, closed, open-closed waves)?


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The Attempt at a Solution

 
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kathyt.25 said:
(1) If a wave is FIXED at both ends, for the fundamental or 1st harmonic, does it have 0 nodes, 1 node, or 2 nodes? Do you count both "end nodes" where the wave is fixed?
Yes you do.

kathyt.25 said:
(2) If a wave is FIXED at one end, and OPEN at the other end, I know that the equation for frequency is f(n) = n*v / 4L, where n=1,3,5,7...
Does "n" represent the # nodes, or the harmonic level?
This is super confusing for me, because for f(1), there is 1 node at the fixed end, which makes sense, because this is the first harmonic.
However, for f(2), there are 2 nodes, but "n" can only equal odd numbers. So there is no second harmonic... Why is that? So to find values of n, we have to go by harmonic number, and not the number of nodes, for all situations (ie. open, closed, open-closed waves)?
Think of it this way: n represents the number of half-wavelengths that fit into the length L. A wave that consists of an odd number of half-wavelengths has a node at one end and an antinode at the other end, which fits perfectly into a pipe that's open at one end and closed at the other. Alternatively, for a wave that has an even number of half-wavelengths, there are two possibilities: either it has a node at both ends, which fits a pipe that's closed at both ends, or it has an antinode at both ends, which fits a pipe that's open at both ends. That's why for pipes that are open at both ends or closed at both ends, n only takes even values.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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