Why does pitch increase when you blow harder into a whistle?

I don't believe the other answers are correct. FGSUZ describes pushing air out of a tube, which sort of plays a little part, but not the whole story.

The way woodwind instruments produce sound, is they cause a column of air within the instrument to vibrate. This is done by splitting the air stream. Instruments such as the sax or clarinet use a reed to do this. A concert flute or a wine bottle blows air across a sharp edge, and a recorder or a whistle uses something called a fipple.

In any case, that splitting of the air causes a pressure differential in the stream. One side of the split goes into free air, the other side goes into the body of the instrument. Additionally, virtually all of the air you blow goes out into free air, very little goes into the body*. We know from Bernoulli's principle that the moving air is at a slightly lower pressure. In an attempt to equalize, the column of air in the body will begin to move to fill the low pressure zone. Because the air has some mass and momentum, it will overshoot, and a newly-created high pressure zone will push the column of air back the other way, and the process will repeat.

Pressing keys or (un)covering different holes will change the effective length of that air column, which you can think of as changing its mass**, which results in different pitches sounding.

So when you blow with a greater airspeed, you will create a slightly more intense pressure differential, and so will create a little bit more relative energy to oscillate the air column. Blow a little slower, and the pitch will go down a little bit. Smoothly alternate between and you may have a nice vibrato.

What's really important here, is it's not the volume of air that is important, but the air's speed.

This phenomenon is also why many wind instruments tend to sound sharp at high notes, and flat on low notes, and the player needs to correct by varying their airspeed, as the keys or holes on the instrument alone are not enough to get the right pitch.

In the case of a concert flute, which uses a sharp edge, rather than a fipple or reed, the player can aim their air, and directly control that pressure relationship, by varying the proportion of how much goes into the embouchure hole and how much goes over it. As a result, a skilled flutist can bend notes often more than a whole step up or down, based on air stream control alone, without changing anything about the flute itself, or without changing airstream velocity.

Lastly, if you produce enough power in your airstream, you can overblow and play 1 or more octaves above the note as fingered. When playing in the upper registers, the tendencies for the instruments to sound increasingly sharp as it goes higher becomes more dramatic.

Edit: I want to mention, but couldn't figure out where to work it into the answer above, but air speed is really important. Especially on the concert flute, it is important to the extent of massive frustration to newcomers. A fishing-line sized stream of air over the mouthpiece at the right speed will speak louder and clearer than 100 times more air if it's uncontrolled and slower. New flute players are often taught to think about "hot" vs "cold" air when learning to control their air stream. And, ultimately, when a player has attained sufficient skill, they can play quiet notes, by carefully blowing very small amounts of air, at very high speeds, and sound out even the highest notes quietly. If the physics of the instrument was about pushing air out of the body of the instrument, this would be impossible. It's not, because that tiny bit of air at the right speed is still enough to create that pressure differential, no matter how small.

*Not true for reed instruments; the air-splitting behavior is caused by the reed itself, but the rest of the concepts are still true.

**Massive oversimplification that borders on being completely wrong, but frankly it doesn't really matter.

This is a very interesting phenomenon.

Roughly speaking, the thing is that pressure affects the "effective length" of the tube.

Let me explain, tubes are not as easy as strings. A string has a fixed length, and then the speed of sound determines its frequency uniquely.

On the other hand, open tubes behave differently. Since we're talking about longitudinal waves, we're talking about "pressure waves", it's a series of compression and expansion of the air molecules inside the tube.

But there's one issue: those stationary waves are not excited in the same wave as you move a rope. You excite the sound waves by blowing air. That means, blowing mass with a velocity, so you're carrying momentum, air makes a force, and that force pushes molecules away.

When you blow air into the whistle, mass conservation is forcing that air to come out from somewhere else. The air you blow inside comes out from the other end. But that air coming out pushes the surrounding air molecules back. In other words, when you blow air, you are displacing the surrounding air.

In other words, the air you blow encounters no high resistance against the previous air that was already there. So you are pushing surrounding air molecules back.

Those molecules go back only a certain distance. At some point, molecules bounce back to the whistle again. You can blow molecules away until the pressure of the air is the same as the one you're forcing through the whistle .

Obviously, that distance depends on how strong you blow, but it is of the order of 1cm.

And what does this have to do with all that? Well, what happens here is that the wave doesn't "need" to bounce back until it reaches that "pressure barrier". So, instead of bouncing back right in the tube end, it bounces back a little later.

So, to sum up, the fact that it is an open end allows air to bounce back a little after the tube ends. So you have the same effect as if you had a "longer ideal tube", and a longer tube implies a different $\lambda$, and different harmonicks.

typically when you resonate an object (or mass of air) it goes through a full vibration in one length of your object. This is because it likes to (for the example of a closed end tube) be at the end of its wave when it gets to the end of the tube (boundary conditions). But if you put enough energy into it, it will go through 2 vibrations (and still fulfill the boundary conditions). Your higher pitch will typically be twice the frequency of your lower pitch. Try it out!