Piano key frequencies

This is a list of the fundamental frequencies in hertz (cycles per second) of the keys of a modern 88-key standard or 108-key extended piano in twelve-tone equal temperament, with the 49th key, the fifth A (called A₄), tuned to 440 Hz (referred to as A440).[1][2] Since every octave is made of twelve steps and equals two times the frequency (for example, the fifth A is 440 Hz and the higher octave A is 880 Hz), each successive pitch is derived by multiplying (ascending) or dividing (descending) the previous by the twelfth root of two (approximately 1.059463).[1][2] For example, to get the frequency a semitone up from A₄ (A♯₄), multiply 440 by the twelfth root of two. To go from A₄ to B₄ (up a whole tone, or two semitones), multiply 440 twice by the twelfth root of two (or just by the sixth root of two, approximately 1.122462). To go from A₄ to C₅ (which is a minor third), multiply 440 three times by the twelfth root of two, (or just by the fourth root of two, approximately 1.189207). For other tuning schemes refer to musical tuning.

This list of frequencies is for a theoretically ideal piano. On an actual piano the ratio between semitones is slightly larger, especially at the high and low ends, where string stiffness causes inharmonicity, i.e., the tendency for the harmonic makeup of each note to run sharp. To compensate for this, octaves are tuned slightly wide, stretched according to the inharmonic characteristics of each instrument.[3] This deviation from equal temperament is called the Railsback curve.

The following equation gives the frequency f of the n^th key, as shown in the table:[1]

${\displaystyle f(n)=\left({\sqrt[{12}]{2}}\,\right)^{n-49}\times 440\,{\text{Hz}}\,}$

(a' = A₄ = A440 is the 49th key on the idealized standard piano)

Alternatively, this can be written as:

${\displaystyle f(n)=2^{\frac {n-49}{12}}\times 440\,{\text{Hz}}\,}$

Conversely, starting from a frequency on the idealized standard piano tuned to A440, one obtains the key number by:

${\displaystyle n=12\,\log _{2}\left({\frac {f}{440\,{\text{Hz}}}}\right)+49=}$

${\displaystyle n=39.86\,\log _{10}\left({\frac {f}{440\,{\text{Hz}}}}\right)+49}$