Speakers and Changes in Temperature Let's say that there is a speaker that oscillates the same way. Now, let's say there is a sudden drop in temperature.
I know the speed of sound would drop. But, what will drop, the wavelength or the frequency?
In my head, the frequency should stay the same as the oscillations still stay the same. But, then, the wavelength too as there is the same type of oscillation, please help.
 A: If the speaker oscillates at the same frequency then the sound will have the same frequency - but it will arrive more slowly so the wavelength will become shorter.
Interesting question: if the change in temperature is really sudden what happens a during the drop?
To answer that we need to know about the air movement during the temperature drop - if pressure and volume stay the same and temperature drops, more air must be coming from somewhere.
If you want an answer to that second question please clarify with appropriate constraints.
A: Floris, great answer. The confusion in the question is a really common one that isn't emphasized enough in teaching.
Intuitive View
Here's an easy way to remember this:


*

*When the speaker pushes, the air touching the speaker moves one way. When the speaker pulls, the air touching the speaker moves the other way. Put more formally, for each movement of the speaker, there is a corresponding movement of the air touching the speaker.

*Furthermore, this air movement sets up a chain reaction, so that the disturbance propagates through the air away from the speaker. This is wave propagation and it occurs at a constant speed (the speed of sound), and this speed is a fixed property of the air at its given temperature and pressure.

*The frequency is determined by the speaker and cannot be influenced by the air. (As an aside, we can also realize that the wavelength is thus a property of both the air and the speaker, and does not owe itself to either exclusively.)


Let's say, hypothetically, that the air and speaker might be able to vibrate at different frequencies. If this were true, there would need to be some instant in time when the air moves opposite to the speaker, contrary to the first point above.
Once we understand the speaker and the air acting together like this, it's easy to see how they can't operate at different frequencies.
Linearity
Another way to understand this is by realizing (to a very good approximation) that  air is a linear medium. A function $f$ is linear if:
$$\begin{align}f(a\cdot x) &= a\cdot f(x)\\
f(x+y) &= f(x)+f(y)\end{align}$$
In this case, when I say the air is a linear medium, take $f$ to be the air's response to some signal $x(t)$ generated at the speaker.
Another Proof By Contradiction
If $f$ were able to change a frequency $\omega$ into $\phi$, where $\omega\ne\phi$, say $f(a\sin(\omega t))=a\sin(\phi t)$, then the question is, how can one produce this result with linear functions alone? The short answer is there's no way to do it.
