FFT usually requires power-of-2-sized windows, so let's say just DFT (alternatively use $1024$ samples).
A sine wave with a frequency of $6\:\mathrm{Hz}$ is not orthogonal to any of the $0\:\mathrm{Hz}$, $10\:\mathrm{Hz}$, ... waves with respect to the $L^2$ scalar product over an interval of $100\:\mathrm{ms}$, so it would in fact appear in all of the bins. That's the problem with a rectangular window function: unless you happen to start with a perfect superposition of only the quantized frequency values, you will end up with a horrible smear across the whole frequency range. To avoid having to introduce a nontrivial window function here, let's talk about compactly supported signals, like wavelets, centered in our DFT window. As you certainly know, such functions always have an intrinsic frequency indeterminacy, which essentially means that the (infinite) Fourier transform consists not of sharply defined (dirac) peaks but of Gaussian-like bell peaks. If the time confinedness was $100\: \mathrm{ms}$, the frequency indeterminacy will be more than $\frac1{100\:\mathrm{ms}}=10\:\mathrm{Hz}$. So as you see, the width of the DFT bins is not just a technical issue with the specific Fourier transform algorithm, it represents the general inability to define the frequency of a "correctly processable" signal more precisely than the bin width.
You probably know this already. Anyway, let's have a look now at a wavelet with a frequency centered about $60\:\mathrm{Hz}$, like you would get when window-functioning mains hum. Assuming the freq indeterminacy gets no bigger than necessary, this will give you a pretty sharp peak in the $55$ to $65\:\mathrm{Hz}$ bin, with only small values in the neighbouring bins - so we can approximate the total energy, that is, the $L^2$ norm of our signal, by just the square of the value in this bin (that's due to Bessel's equality). Likewise, if you were interested in the energy between $45$ and $105\:\mathrm{Hz}$, you would just sum up the squared values of those bins and get, correctly, the total energy of the wavelet. Where it gets interesting is when you want to know about the energy in the range $61$ to $63\:\mathrm{Hz}$. According to your proposal, this should be calculated as $\tfrac15$ of the squared value in the $55$ to $65\:\mathrm{Hz}$ bin, that is, $\tfrac15$th of the total energy of our wavelet. And that's pretty good actually, because as we said the energy of this wavelet is actually smeared over an interval of $10\:\mathrm{Hz}$ about $60\:\mathrm{Hz}$, so it's quite a reasonable approximation to say $\tfrac15$th of it is in the range $61$ to $63\:\mathrm{Hz}$!
What about a wavelet centered about $65\:\mathrm{Hz}$? If we DFT this, it will appear in both the $55$ to $65\:\mathrm{Hz}$ and $65$ to $75\:\mathrm{Hz}$ bins, with each values of $\sqrt{\tfrac12}$ of the total amplitude. If you now calculate the energy between $45$ and $105\:\mathrm{Hz}$, you will get
$$
0 + \sqrt{\tfrac12}^2E + \sqrt{\tfrac12}^2E + 0 + 0 = E
$$
so that's again the total energy, correctly. If you want the energy between $55$ and $65\:\mathrm{Hz}$, you get
$$
\sqrt{\tfrac12}^2E = \frac{E}2
$$
which is pretty reasonable, because in fact only about half of the signal energy lies in this band.
But where you start getting weird results is when you calculate the energy between $55$ to $56\:\mathrm{Hz}$, which results in
$$
\frac1{10}\sqrt{\tfrac12}^2E = \frac{E}{20}
$$
and compare it with the energy between $65$ to $66\:\mathrm{Hz}$, for which you would obviously get the same result. But then, the actual wavelet does not really have any notable frequency component at $55$ to $56\:\mathrm{Hz}$ at all, while $65$ to $66\:\mathrm{Hz}$ is where it is strongest!
In conclusion: it does make sense to do this interpolation, but it should be handled with care.
As I just notice, what you do is in fact not a linear interpolation, but just a $0$th order domain extension. A linear or higher-order interpolation would suffer less from the problem I just explained.
The insight and engineering rule of thumb is that
we need to observe at least half of a sine wave to determine its amplitude and frequency.
Of course, as a rule of thumb it is a drastic cutoff, but it is quite intuitive and close to "real".
Now take the sum of two sinusoids with near frequencies
$$
\sin \left( {2\pi \left( {f - \Delta f/2} \right)t} \right) + \sin \left( {2\pi \left( {f + \Delta f/2} \right)t} \right) = 2\cos \left( {2\pi \,\Delta f/2\,t} \right)\sin \left( {2\pi \,f\,t} \right)
$$
we obtain a sine of frequency $f$ amplitude-modulated by a (co)sine of frequency $\Delta f/2$.
(a well known phenomenon of which I do not remember the name in english).
According to the given rule, we need to "see" at least half of the modulating wave to be able
and reconstruct the signal, and thus the difference in frequency. If the duration , i.e. "persistance" of the
signals, is much less we will only see one sinusoid with the average frequency and double amplitude.
Calling $T$ the duration of the observation window we shall have
$$
{1 \over 2}{1 \over {\Delta f/2}} < T\quad \Rightarrow \quad 1 < T\Delta f
$$
Passing to a more rigorous mathematical analysis, consider that
the Box function (a window of duration $T$)
$$
R(t,T) = \left\{ {\matrix{
1 & { - T/2 \le t \le T/2} \cr
0 & {otherwise} \cr
} } \right. = U(t + T/2) - U(t - T/2)
$$
where $U$ is step function, has a frequency spectrum (bilateral Fourier Transform) given by
$$
G(f,T) = T{\rm sinc}(f\,T) = {{{\rm sin}(\pi f\,T)} \over {\pi f}}
$$
A sinusoid of duration $T$ is the product of the Box function for
a sine function.
The spectrum will be the convolution of a Dirac at frequency $f$ (apart that at $-f$)
with the $G(f,T) =T{\rm sinc}(f\,T) $ function , i.e. the same centered at $f$.
Then, approximating the $G(f,T)$ to a box function with a cutoff at $f \pm 1/(2T)$
leads to that we can distinguish between frequencies with a separation
of at least $1/T$, i.e. to the formula given above.
Best Answer
compare it with the FFT of $\theta (0.3-t)$ and of $$H(t) = \theta (0.3-t)[\exp(2i \pi \ 30 \ t)+\exp(2 i \pi \ 100 \ t) + 3.0]$$
use a window (hanning)
if you understand what's happening, you will understand nearly everything of the FFT.