Chapter 7.6, Problem 59PPS

(a).

To determine

To find: The interval in cents when the frequency changes from $443\text{Hz}$ to $415\text{Hz}$ .

Answer to Problem 59PPS

The interval in cents when the frequency changes from $443\text{Hz}$ to $415\text{Hz}$ is $113.0833$ cents.

Explanation of Solution

Given information:

One octave is equal to $1200cent$

  $n=1200\left({\log }_2\frac{a}{b}\right)$

  $\begin{array}{l}a=443\\ b=415\end{array}$

Calculation:

Consider the difference in cents between the two nodes with frequency a,b is:

  $n=1200\left({\log }_2\frac{a}{b}\right)$

Consider the frequency changes from 443 Hz to 415 Hz, therefore consider

  $\begin{array}{l}a=443\\ b=415\end{array}$

Thus, the corresponding difference is cents is:

  $\begin{array}{c}n=1200\left({\log }_2\frac{443}{415}\right)\\ =1200\left({\log }_{2}1.0675\right)\\ =1200\frac{{\log }_{10}1.0675}{{\log }_{10}2}\end{array}$

Use TI 83 calculator to find the value the screenshot is given as below

  

Hence, $n=113.0833$

Therefore, the interval in cents when the frequency changes from $443\text{Hz}$ to $415\text{Hz}$ is $113.0833$ cents.

(b).

To determine

To find: The final frequency if the interval is 55cents and the beginning frequency is $225\text{Hz}$ .

Answer to Problem 59PPS

The final frequency is $218\text{Hz}$ .

Explanation of Solution

Given information:

The interval is $55\text{cents}$

The frequency is

  $f=225\text{Hz}$

Calculation:

Consider is the frequency changes from $225\text{Hz}$ to $b\text{Hz}$ then the corresponding changes in cent is $55$ . Therefore,

  $\begin{array}{c}55=1200\left({\log }_2\frac{225}{b}\right)\\ \frac{55}{1200}={\log }_2\frac{225}{b}\\ b=217.9643\end{array}$

Therefore, the final frequency is $218\text{Hz}$ .