Question

Western music uses a musical scale with equal temperament tuning, which means that any two adjacent notes have the same frequency ratio r. That is, notes n and n + 1 are related by fn+1 = r fn for all n. In this system, the frequency doubles every 12notes—an interval called an octave.

a. What is the value of r?

b. Orchestras tune to the note A, which has a frequency of440 Hz. What is the frequency of the next note of the scale (called A-sharp)?

Short answer

a) The value of the r is 1.0595

b) The frequency of the next scale note is 466 Hz

Step-by-step solution

Discussion about the frequency of the note 

The higher and the lower frequencies waves are responsible for oscillatingthe higher and lower notes.

Understanding the differences between the notes and determining r

The provided information is $f_{n+1}=r{f}_n$

the separation of the two notes are $f_{n+2}=r{f}_{n+1}=r^2{f}_n$

Thus, the separation of the 12th notes are $\frac{f_{N+12}=r^{12}{f}_n}{}$

The double of the twelve-note is$f_{n+12}=2f_n$

based on the simplifaction the remaining value is $2=r^{12}$

Now the r is = $r^{12}=2\Rightarrow r=\sqrt[12]{2}=2\frac{1}{2}\boxed{1.0595}$

Thus, the determination of the r based on the formula islocalid="1649072985238" $fA\#=r{f}_A=1.0595·440=\boxed{446Hz}$