An intuitive question on the meaning of bins in a power spectrum of FFT
So for the first bin I will have 1Hz 2Hz 3Hz and 4Hz


No.

The FFT computes the Discrete Fourier Transform (DFT). The DFT is the integral
$$
X_k = \sum_{n=0}^{N-1} x_n \left(\cos(2\pi kn/N) - j\sin(2\pi kn/N)\right)
$$
Just to make things simpler, I'm going to ignore the sine terms, so we can talk about the sum
$$
X'_k = \sum_{n=0}^{N-1} x_n \cos(2\pi kn/N)
$$

This sum has a couple of nice properties. If our original function was just the cosine
$$
x_n = \cos(2\pi \mathbf{1} n/N)
$$
then the sum is
$$
X'_k = \begin{cases} 
N / 2, &amp; n = \mathbf{1} \\
0,     &amp; n \neq \mathbf{1}  \end{cases}
$$
In real life, this means that your 4 Hz signal will not appear in any of the other bins (8 Hz, 12 Hz, etc). Nice - we can detect a 4 Hz signal by looking in the bin $k = 1$. This pattern repeats if you replace the bolded \$\textbf{1}\$ with any other integer - the signal will only appear in one bin.

Things fall apart when you stop using integers. If I replace the \$\textbf{1}\$ with a fraction like \$\textbf{1/4}\$ or \$\textbf{6/4}\$, then the sum is much harder to calculate. If you do this in Matlab or some other programming language, you'll find that these coefficients are not all zero! Some of your 6 Hz signal finds its way into all of these coefficients.

To get a rough idea of how the coefficients are spread out, they'll look like a scaled sinc function:



This happens to all of your frequencies that are not a multiple of 4 Hz. Yuck! To avoid this, you can try filtering the signal before you take the FFT by using a window function, which will cause this spectral leakage to shrink. However, you can't totally remove it unless you can manage to take the FFT with 1 Hz bins. (Note that you'll still be stuck with all of the 0.5 Hz noise leaking across the intervals - there'll always be something left over.)

This doesn't really answer your question as stated, but I hope it pushes you in the right direction.