What happens if a string with both ends fixed is excited with sinusoidal wave with $\lambda \neq 2 L/n $? We know that if $\lambda = \frac {2L}{n}$ then stationary waves are formed.
So, if a wave generator is connected to the string and we set $\lambda = 2L $ then we will see stationary waves. Maybe  we will even hear those waves, if $\lambda $ is within our hearing range.
Then, we may decrease the wavelength until we reach $\lambda = L$ and again we will see stationary waves.
But during the transition from $\lambda=2L $ to $\lambda=L$. What happens with the generated waves, what do we see? Will we hear those waves also? (again assuming that $\lambda$ is within our hearing range.)
 A: The "key idea" here is that when you excite a system with an external source you are in the "area" of the forced oscillations. If we restrain ourselves to the linear regimes, the frequency of the oscillations "in" the system will always coincide with the frequency of the source. This means that the system will oscillate with the given frequency.
Now, when the source frequency matches one of the natural frequencies of the system (two of them are those you mention) then we have what we call a resonance. At this frequency the amplitude of oscillation becomes maximum. Of course, neighboring frequencies will also exhibit large amplitude but the extend of the frequencies (bandwidth) of high amplitude (as well as the maximum amplitude at resonance) depends on the characteristics of the system (damping).
From limited personal experience I can say that you may or may not hear those frequencies (or see them with an oscilloscope or waveplot of some kind). Whether you hear them depends on the amplitude of the source and the damping of the system (as well as other factors such as leakage) and whether you will be able to see the signal depends on the noise of the measuring system.
