Chapter 15, Problem 4SP

For standard tuning, concert A is defined to have a frequency of 440 Hz. On a piano. A is five white keys above C, but 9 half steps above C counting both the white and black keys. (See fig. 15.22.) A full octave consists of 12 half steps (semitones). In equally tempered tuning, each half step has the ratio of 1.0595 above the preceding step. (This ratio is the 12th root of 2.0.)

a.    What is the frequency of A-flat, one half step below A for equal temperament?

b.    Working down, find the frequency of each succeeding half step until you get down to C. (Carry your computations to four figures, to avoid rounding errors. For each half step, divide by 1.0595.)

c.    In just tuning, middle C has a frequency of 264 Hz. How does your result in part b compare to this value?

d.    Working up, find the frequency of C above concert A in equal temperament. Is this frequency twice that obtained in part b for middle C?

To determine

What is the frequency of A-flat.

Answer to Problem 4SP

The frequency of A flat is $415.3\text{Hz}$.

Explanation of Solution

Given info: The frequency of A is $440\text{Hz}$

Write the formula to calculate the frequency of A flat.

$f_A=\left(\frac{f}{1.0595}\right)$

Here,

$f_A$ is the frequency of A flat

f is the frequency of A

Substitute $440\text{Hz}$ for f in the above equation to calculate $f_A$.

$\begin{array}{c}{f}_A=\left(\frac{440\text{Hz}}{1.0595}\right)\\ =415.3\text{Hz}\end{array}$

Conclusion:

Therefore, the frequency of A flat is $415.3\text{Hz}$.

To determine

Find the each succeeding half up to reach C.

Answer to Problem 4SP

The succeeding frequencies are $415.29\text{Hz}$, $391.97\text{Hz}$, $369.96\text{Hz}$, $349.18\text{Hz}$, $329.57\text{Hz}$, $311.06\text{Hz}$, $293.59\text{Hz}$, $277.10\text{Hz}$ and $261.54\text{Hz}$.

Explanation of Solution

Given info: The frequency of A is $440\text{Hz}$ and C is the 9 half steps below C.

Since C is the 9 half steps below A, to find the succeeding frequencies divide the frequency of A by 1.0595 to get the one step below of A and go on up to nine times. The 9^th frequency is corresponding to the frequency of C.

The succeeding frequencies are $415.29\text{Hz}$, $391.97\text{Hz}$, $369.96\text{Hz}$, $349.18\text{Hz}$, $329.57\text{Hz}$, $311.06\text{Hz}$, $293.59\text{Hz}$, $277.10\text{Hz}$ and $261.54\text{Hz}$.

Conclusion:

Therefore, the succeeding frequencies are $415.29\text{Hz}$, $391.97\text{Hz}$, $369.96\text{Hz}$, $349.18\text{Hz}$, $329.57\text{Hz}$, $311.06\text{Hz}$, $293.59\text{Hz}$, $277.10\text{Hz}$ and $261.54\text{Hz}$.

To determine

Compare the given tuned frequency of C with part (b).

Answer to Problem 4SP

The difference in frequency is $\text{2}\text{.5}\text{Hz}$.

Explanation of Solution

Given Info: The tuned frequency of C is $264\text{Hz}$ and frequency of C is $261.54\text{Hz}$.

Write the expression to calculate the difference between the tuned frequency of C with calculated.

$\Delta f=f_C-f_C{}^{\prime }$

Here,

$\Delta f$ is the difference in frequency

$f_C$ is the tuned frequency of C

$f_C{}^{\prime }$ is the calculate frequency of C

Substitute $264\text{Hz}$ for $f_C$ and $261.54\text{Hz}$ for $f_C{}^{\prime }$ in the above equation to calculate $\Delta f$.

$\begin{array}{c}\Delta f=264\text{Hz}-261.54\text{Hz}\\ =2.46\text{Hz}\\ \approx \text{2}\text{.5}\text{Hz}\end{array}$

Conclusion:

Therefore, the difference in frequency is $\text{2}\text{.5}\text{Hz}$.

To determine

The frequency of C above A and will this frequency is twice as compared with part (b).

Answer to Problem 4SP

The frequency of C is $523.3\text{Hz}$ and which is almost twice as found before with a slight difference of $0.2$.

Explanation of Solution

Given info: The frequency of A is $440\text{Hz}$

In this case the C is five keys above A including white and black keys and three keys above without including black keys or halves. Therefore to find the frequency of C, the frequency of A is multiplied by the value 1.0595 by three times.

Therefore the frequency of C is $523.3\text{Hz}$ and this frequency is almost twice as that of the frequency of C which is found to be $261.54\text{Hz}$ with a slight difference of $0.2$.

Conclusion:

Therefore, the frequency of C is $523.3\text{Hz}$ and which is almost twice as found before with a slight difference of $0.2$.