To expand on Xcheckr's answer:
The full equation for a single-frequency traveling wave is
$$f(x,t) = A \sin(2\pi ft - \frac{2\pi}{\lambda}x).$$
where $f$ is the frequency, $t$ is time, $\lambda$ is the wavelength, $A$ is the amplitude, and $x$ is position. This is often written as
$$f(x,t) = A \sin(\omega t - kx)$$
with $\omega = 2\pi f$ and $k = \frac{2\pi}{\lambda}$. If you look at a single point in space (hold $x$ constant), you see that the signal oscillates up and down in time. If you freeze time, (hold $t$ constant), you see the signal oscillates up and down as you move along it in space. If you pick a point on the wave and follow it as time goes forward (hold $f$ constant and let $t$ increase), you have to move in the positive $x$ direction to keep up with the point on the wave.
This only describes a wave of a single frequency. In general, anything of the form
$$f(x,t) = w(\omega t - kx),$$
where $w$ is any function, describes a traveling wave.
Sinusoids turn up very often because the vibrating sources of the disturbances that give rise to sound waves are often well-described by
$$\frac{\partial^2 s}{\partial t^2} = -a^2 s.$$
In this case, $s$ is the distance from some equilibrium position and $a$ is some constant. This describes the motion of a mass on a spring, which is a good model for guitar strings, speaker cones, drum membranes, saxophone reeds, vocal cords, and on and on. The general solution to that equation is
$$s(t) = A\cos(a t) + B\sin(a t).$$
In this equation, one can see that $a$ is the frequency $\omega$ in the traveling wave equations by setting $x$ to a constant value (since the source isn't moving (unless you want to consider Doppler effects)).
For objects more complicated than a mass on a spring, there are multiple $a$ values, so that object can vibrate at multiple frequencies at the same time (think harmonics on a guitar). Figuring out the contributions of each of these frequencies is the purpose of a Fourier transform.
The quantum wavefunction is not a wave in the classical sense. In particular, it is not a wave obeying $E = \hbar \omega$, as electromagnetic waves do.
The term wavefunction is, in this sense, just a bad naming decision. It is no wave, there is nothing oscillating, and it has no connection whatsoever except the mathematical form to physical waves.
Best Answer
The confusion lies in the fact that you are mixing up the sinusoidal behaviour of the pendulum in time and the physical distance covered.
It’s simpler explain in terms of a simple harmonic oscillator. Here the displacement $x$ as a function if time is given by $$x(t)=A\sin\left(\omega t\right)$$ As you can see, the displacement is sinusoidal in time. For this wave, the “wavelength” will be the time interval after which the wave returns to the original position. Which is nothing but the time period $T$. So when we talk about the time period of the oscillations we are essentially talking about the “wavelength” of the wave.