Add the signed displacements

Two pulses meeting on a rope do not collide and rebound. They travel through one another and emerge with their original shapes, heights and directions.

While they overlap, the medium shows the sum, and the sum needs signs:

1. A crest of $3$ cm meeting a trough of $2$ cm gives $+3 + (-2) = 1$ cm.
2. A crest meeting a crest gives $+5$ cm.
3. Equal and opposite pulses give zero at the instant of complete overlap.

That last case looks like the end of the story and is not. At that instant the string is flat and every piece of it is moving: the energy is in the transverse motion of the medium, which is why both pulses reappear and carry on. Overlap is a temporary event that the pulses survive.

Adding magnitudes rather than signed values always lands on the larger answer, and deciding that an interaction is "destructive" gives no licence to subtract everywhere else on the string. Constructive and destructive describe the sum at a particular point, and one overlap can show both.

A standing wave is what superposition produces when two waves of the same frequency travel in opposite directions in a confined region. The resulting pattern is stationary:

1. Nodes sit at fixed positions where the displacement is always zero.
2. Antinodes sit at fixed positions where the oscillation is largest.

Nothing about the envelope translates along the string. So timing "how long the pattern takes to cross" is measuring nothing, and the wave speed cannot be found that way.

Get $v$ from the traveling waves that superpose: read $\lambda$ off the standing pattern, take $f$ from the driver, and use $v = f\lambda$. The points on the string oscillate in place, with permanently fixed nodes, while the two waves building the pattern move at $v$ in opposite directions.

The harmonic number comes from how many half or quarter wavelengths fit the region, given what the boundaries require. So mark what each end has to be before counting anything:

1. String fixed at both ends: node at each end.
2. Pipe open at both ends: antinode at each end.
3. Pipe closed at one end: node at the closed end, antinode at the open end.

Those first two cases both give $\lambda = 2L/n$ with every integer $n$ allowed. The third is genuinely different: a node paired with an antinode fits a quarter wavelength into the fundamental, so

$$\lambda = \frac{4L}{n}, \qquad n = 1, 3, 5, \ldots \ \text{(odd only)}.$$

Reusing $\lambda = 2L/n$ for a closed pipe produces harmonics that cannot exist and a fundamental twice as high as the real one. And counting bumps in a picture without checking the ends lands a harmonic off: a closed pipe showing two bumps is the third harmonic, not the second, because the even ones are absent.

Two waves of slightly different frequency drift in and out of phase, so the loudness pulses. The rate of that throbbing is

$$f_{\text{beat}} = |f_1 - f_2|.$$

Two things are audible and they are different quantities:

1. The pitch you hear sits near the average of the two frequencies.
2. The throbbing happens at their difference.

So $440$ Hz and $444$ Hz give a tone near $442$ Hz that pulses 4 times a second. Reporting $442$ answers the pitch question when the beat question was asked, and $884$ is the sum, which corresponds to nothing audible.

Tuning an instrument is this relation used backwards: adjust until the throbbing slows and stops, which drives the difference to zero.