Chapter 13, Problem 104QAP

To determine

(a)

Show that the ratio of the frequency of any semitone to the frequency of the note just below it is 2¹/12

Answer to Problem 104QAP

The ratio semitone to the frequency of the note just below is 2¹/12

Explanation of Solution

Given:

The 12 semitones of octave are as follow.

  $\text{C},{\text{C}}^{*}/{\text{D}}^b,\text{D},{\text{D}}^{\#}/{\text{E}}^b,\text{E},\text{F},{\text{F}}^{\#}/{\text{G}}^b,\text{G},{\text{G}}^{\#}/{\text{A}}^b,\text{A},{\text{A}}^{\#}/{\text{B}}^b,\text{B},{\text{C}}^{\prime }$

Introduction:An octave or perfect octave is the interval between one musical pitch and another with double its frequency.

  $\begin{array}{l}\frac{{\text{f}}_{\text{C'}}}{{\text{f}}_{\text{C}}}\text{=}\text{2}\\ where,\\ {\text{f}}_{\text{C}}\text{=first}\text{frequency}\text{of}\text{actave}\\ {\text{f}}_{\text{C'}}\text{=second}\text{frequency}\text{of}\text{octave}\end{array}$

Further octaves of a note occur at times the frequency of that note (where n is an integer), such as 2, 4, 8, 16, 32, 64 etc. and the reciprocal of that series.

The octave is divide into 12 semitone of equal ratio r.

  $\begin{array}{l}\left(\frac{f_{C^{\prime }}}{f_{\text{B}}}\right)=\left(\frac{f_{\text{B}}}{f_{{\text{A}}^{*}}}\right)=\left(\frac{f_{{\text{A}}^{*}}}{f_{\text{A}}}\right)=\left(\frac{f_{\text{A}}}{f_{{\text{G}}^{*}}}\right)=\left(\frac{f_{{\text{G}}^{*}}}{f_{\text{G}}}\right)=\left(\frac{f_{\text{G}}}{f_{{\text{F}}^{*}}}\right)=\left(\frac{f_{{\text{F}}^{*}}}{f_{\text{F}}}\right)=\left(\frac{f_{\text{F}}}{f_{\text{E}}}\right)\\ =\left(\frac{f_{\text{E}}}{f_{{\text{D}}^{*}}}\right)=\left(\frac{f_{\text{D}}}{f_{\text{D}}}\right)=\left(\frac{f_{\text{D}}}{f_{{\text{C}}^{*}}}\right)=\left(\frac{f_{{\text{C}}^{*}}}{f_{\text{C}}}\right)=r\end{array}$

Calculation:

  $\begin{array}{l}\frac{f_C}{f_{\text{C}}}=\left(\frac{f_{C^{\prime }}}{f_{\text{B}}}\right)\left(\frac{f_{\text{B}}}{f_{{\text{A}}^{*}}}\right)\left(\frac{f_{{\text{A}}^{*}}}{f_{\text{A}}}\right)\left(\frac{f_{\text{A}}}{f_{{\text{G}}^{*}}}\right)\left(\frac{f_{{\text{G}}^{*}}}{f_{\text{G}}}\right)\left(\frac{f_{\text{G}}}{f_{{\text{F}}^{*}}}\right)\left(\frac{f_{{\text{F}}^{*}}}{f_{\text{F}}}\right)\left(\frac{f_{\text{F}}}{f_{\text{E}}}\right)\left(\frac{f_{\text{E}}}{f_{{\text{D}}^{*}}}\right)\left(\frac{f_{\text{D}}}{f_{\text{D}}}\right)\left(\frac{f_{\text{D}}}{f_{{\text{C}}^{*}}}\right)\left(\frac{f_{{\text{C}}^{*}}}{f_{\text{C}}}\right)\\ \Rightarrow \frac{f_C}{f_{\text{C}}}=r^{12}\\ \Rightarrow 2=r^{12}\\ \Rightarrow r=2^{1/12}\end{array}$

Conclusion:

The ratio semitone to the frequency of the note just below is $2^{1/12}$

To determine

(b)

The frequency of F# in the tempered scale, If $A$ is 440 Hz.

Answer to Problem 104QAP

The frequency of F# is 370 Hz

Explanation of Solution

Formula used:

The ratio of the frequency of any semitone to the frequency of the note just below it is $2^{1/12}$

  $\left(\frac{f_{\text{A}}}{f_{{\text{G}}^{*}}}\right)=\left(\frac{f_{{\text{G}}^{*}}}{f_{\text{G}}}\right)=\left(\frac{f_{\text{G}}}{f_{{\text{F}}^{*}}}\right)=2^{\frac{1}{12}}$

Calculation:

  $\begin{array}{l}\frac{f_{\text{A}}}{f_{{\text{F}}^{*}}}=\left(\frac{f_{\text{A}}}{f_{{\text{G}}^{*}}}\right)\left(\frac{f_{{\text{G}}^{*}}}{f_{\text{G}}}\right)\left(\frac{f_{\text{G}}}{f_{{\text{F}}^{*}}}\right)={\left(2^{\frac{1}{12}}\right)}^3=2^{\frac{1}{4}}\\ f_{{\text{F}}^{*}}=\left(2^{-\frac{1}{4}}\right)f_{\text{A}}=\left(2^{-\frac{1}{4}}\right)(440\text{Hz})=370\text{Hz}\end{array}$

Conclusion:

The frequency of F# is 370 Hz.

To determine

The value by which tension should change so that the string tuned from B_b to B.

Answer to Problem 104QAP

The tension should be increased by a factor of 1.12

Explanation of Solution

Given:

  $\frac{f_{\text{B}}}{f_{{\text{B}}^b}}=2^{\frac{1}{12}}$

Formula used:

  $\begin{array}{l}f=\frac{v}{\lambda }=\sqrt{\frac{F}{\mu }}\\ where,\\ f=requency\\ v=velocity\\ F=tension\\ \mu =linearmassdensity\end{array}$

Calculation:

  $\begin{array}{l}\frac{f_{\text{B}}}{f_{{\text{B}}^b}}=\frac{\left(\frac{v_{\text{B}}}{\lambda }\right)}{\left(\frac{v_{{\text{B}}^b}}{\lambda }\right)}=\frac{\left(\sqrt{\frac{F_{\text{B}}}{\mu }}\right)}{\left(\sqrt{\frac{F_{{\text{B}}^6}}{\mu }}\right)}=\sqrt{\frac{F_{\text{B}}}{F_{{\text{B}}^t}}}=2^{\frac{1}{12}}\\ \frac{F_{\text{B}}}{F_{{\text{B}}^b}}={\left(2^{\frac{1}{12}}\right)}^2=2^{\frac{1}{6}}\\ F_{\text{B}}=\left(2^{\frac{1}{6}}\right)F_{{\text{B}}^b}=1.12F_{{\text{B}}^b}\end{array}$

Conclusion:

The tension should be increased by a factor of 1.12 so a string on a musical instrument is tuned from B^b to B