The $k^\text{th}$ element of the result will be the coefficient of the wave that has $k-1$ full periods in the complete sample. Thus if the length of your sample is $t$ time units, the $k^\text{th}$ element of the result will correspond to frequency $\frac{k-1}{t}$, regardless of the sample rate.

It should be noted that due to aliasing, element $k$ of a result of length $n$ will correspond to the same frequency as $n-k+2$ for $k=2..n$. Usually one is interested in amplitudes only, not phase, i.e. you're working with (Abs@Fourier[sample])^2. In this case you only need to use elements $k = 2..\lceil \frac{n}{2}\rceil$.

The $k^\text{th}$ element of the result will be the coefficient of the wave that has $k-1$ full periods in the complete sample. Thus if the length of your sample is $t$ time units, the $k^\text{th}$ element of the result will correspond to frequency $\frac{k-1}{t}$, regardless of the sample rate.

It should be noted that due to aliasing, element $k$ of a result of length $n$ will correspond to the same frequency as $n-k+2$ for $k=2..n$. Usually one is interested in amplitudes only, not phase, i.e. you're working with (Abs@Fourier[sample])^2. Because the input is real you only need to use elements $k = 2..\lceil \frac{n}{2}\rceil$.

The $k^\text{th}$ element of the result will be the coefficient of the wave that has $k-1$ full periods in the complete sample. Thus if the length of your sample is $t$ time units, the $k^\text{th}$ element of the result will correspond to frequency $\frac{k-1}{t}$, regardless of the sample rate.

The $k^\text{th}$ element of the result will be the coefficient of the wave that has $k-1$ full periods in the complete sample. Thus if the length of your sample is $t$ time units, the $k^\text{th}$ element of the result will correspond to frequency $\frac{k-1}{t}$, regardless of the sample rate.

It should be noted that due to aliasing, element $k$ of a result of length $n$ will correspond to the same frequency as $n-k+2$ for $k=2..n$. Usually one is interested in amplitudes only, not phase, i.e. you're working with (Abs@Fourier[sample])^2. In this case you only need to use elements $k = 2..\lceil \frac{n}{2}\rceil$.