Chapter 11.3, Problem 14P

To determine

To find: the frequency of the middle C and show that the common ratio is $2^{\frac{1}{12}}$ .

Answer to Problem 14P

The frequency of middle of $C$ is $261\text{Hz}$ .

Explanation of Solution

Given:

There are $12$ steps in the chromatic scale.

  

Concept used:

Generic geometric sequence:

  $a_1,a_2,a_3\mathrm{.....}{a}_{n-1}.$

There is common ratio:

  $c.r=\frac{a_n}{a_{n-1}}$ .

The geometric progression:

  $a,ar,a{r}^2\mathrm{.....}a{r}^{n-1}.$

Last term or nth term of the geometric progression will be:

  $t_n=a{r}^{n-1}$ .

  $s_n=\frac{a\left(1-r^n\right)}{1-r};s_{\infty }=\frac{a_1}{1-r}$ .

Calculation:

Set the start amount to $220\text{ Hz}$ .

Multiply by given common ratio which is $12$ times to get from the A key to the left to the A key on other right.

The count includes the white and black keys.

The result should be $440\text{ Hz}$ .

  $220{\left(2^{\frac{1}{12}}\right)}^{12}=220\left(2\right)=440\text{Hz}$ .

Middle C is $3$ keys away from the left A so, the exponent of the common ratio to $3$ .

  $220{\left(2^{\frac{1}{12}}\right)}^3=220\left(2^{\frac{3}{12}}\right)=220\left(2^{\frac{1}{4}}\right)=261.6\text{Hz}$ .

Hence, the middle of the $C$ is $261\text{Hz}$ .