# What is the wavelength of middle "C" on a piano as it travels through air at standard temperature and pressure?

Mar 3, 2016

Wavelength $= 1.3 m$ rounded to one place of decimal

#### Explanation:

Let's find out frequency of note Middle C to begin with.

A modern piano has 88 keys and is tuned to twelve-tone equal temperament.

It has its 49th key, the fifth A, also called A4, tuned to a frequency of 440 Hz (referred to as A440).

Due to twelve-tone equal interval, frequency of each successive key is derived by multiplying frequency (also called pitch) of lower key (or dividing pitch of higher key) by a factor of the twelfth root of two.

General expression which gives the frequency $f$ of the ${n}^{t h}$ key is
$f \left(n\right) = {\left(\sqrt[12]{2}\right)}^{n - 49} \times 440 , \textrm{H z}$

Given note is Middle C, also called C4, is the 40th key. Inserting this value in general expression we obtain its pitch as

$f \left(40\right) = {\left(\sqrt[12]{2}\right)}^{40 - 49} \times 440$ ,
Remembering that $\left(\sqrt[12]{2}\right)$ can also be written as ${2}^{\frac{1}{12}}$
$f \left(40\right) = {\left({2}^{\frac{1}{12}}\right)}^{40 - 49} \times 440$
$= {\left(2\right)}^{\frac{40 - 49}{12}} \times 440$
$= 261.626$ $\textrm{H z}$, rounded to three places of decimal.

Taking speed of sound at STP (Temp 0"^@text{C} and pressure 1 bar) as 331.5 metres per second, we find the required wavelength from

$v = n \times \lambda$
$331.5 = 261.626 \times \lambda$
$= 1.267$
$= 1.3 m$ rounded to one place of decimal