Resonance at frequencies different from natural frequency
A sine wave has three properties, which describe it:

\$f(t) = A * \sin(\omega*t + \phi)\$


Amplitude \$A\$
Frequency \$\omega\$
Phase \$\phi\$


As such, there is no resonant frequency as an inherent property of a sine wave. If you add two sine waves there are certain rules which have to be respected. You may only add the amplitude if frequency and phase are the same. If they differ, you will get another waveform:


This picture shows what happens if you add \$\sin(x)\$ (red) to \$\sin(2*x)\$ (green).

A resonant frequency is a property of a system. That is:







simulate this circuit &ndash; Schematic created using CircuitLab

You have an input \$x(t)\$ and your system will respond with \$y(t)\$. The transfer function describes how the system will react. In this transfer function a resonant frequency can occur.

Let's see what happens with a system we know quite well, the humble RLC-circuit:







simulate this circuit

In this case the voltage source V1 acts as the input of the LTI system \$x(t)\$. And with the current meter AM1, we observe the output \$y(t)\$. If we simulate this circuit in the frequency domain (that is we vary the frequency of our voltage V1, the amplitude stays the same) we will get this as a result:



I have varied the value of the resistor: 1k\$\Omega\$(teal), 10\$\Omega\$(red), 1\$\Omega\$(blue) and 0.01\$\Omega\$(green). 

As can be seen, for different values of R, we get different responses. With higher values of R, there is nothing really happening, the oscillation is damped very strong. For a very low R (very low damping) like in the green curve, you actually have a quite pronounced resonance peak in your response.

For a RLC-Circuit we know the solution to calculate the resonant frequency:

$$\omega_0 = \frac{1}{\sqrt{LC}} \rightarrow f_0 = \frac{1}{2\pi\sqrt{LC}}$$

Which turns out to be 159kHz for these parameters. The resistance plays no role for calculating the resonant frequency, but it has an influence if you can actually measure it. For the case with the 1k\$\Omega\$ resistor, you wouldn't really notice that there is a resonance in that system. But with only 0.01\$\Omega\$ the resonance is so pronounced, that it could damage your components if you haven't considered it (there are some famous examples around for resonance catastrophes, Tacoma bridge to name one).



So to have a more mathematical view on this problem, it's easier to understand if we take a look at the frequency space. In the laplace domain we get for the current:

$$ I(s) = \frac{s}{ L \left ( s^2 + {R \over L}s + \frac{1}{LC} \right ) } V(s) $$

What we are interested in for resonance is the sinusoidal steady state, or more exactly the amplitude of the sinusoidal steady state. The steady state is represented for \$s=j\omega\$ and so we end up with:

$$| I(s=j \omega) | = \frac{1}{\sqrt{ R^2 + \left ( \omega L - \frac{1}{\omega C} \right )^2 }} | V(j \omega) |$$

What you get is a multiplication, not an addition of the input signal. Inside the system happens a frequency dependent complex multiplication of your input signal. Which will turn out to a modified amplitude and a modified phase of the input signal. If you approach the resonant frequency your amplitude will reach it's maximum.

This happens when \$\left ( \omega L - \frac{1}{\omega C} \right )^2\$ is equal to zero. If you solve that equation you end up with the known \$\omega_0 = \frac{1}{\sqrt{LC}}\$.

So to state this clearly again: resonance is not adding sine waves together, it's a complex multiplication of the input signal.