Search
Ask a question
0 0

Music Sequences

Musical tones. The note middle C on a piano is tuned so that the string vibrates at 262 cycles per second or 262 HZ (Hertz). The C note one octave higher is tuned to 524 HZ. The tuning for the 11 notes in between using the method called equal temperament is determined by the sequence an=262*2n/12. Find the tuning for the 11 notes in between. Round to the nearest Hz. Show and explain your work.

PLEASE WALK ME THRU SOLVING THIS ENTIRE PROBLEM.
Tutors, please sign in to answer this question.

1 Answer

In the equal temperament system, every pitch is the same distance apart from both adjacent chromatic pitches. The next note you would need to find above C4 (262 Hz) is C#4. This is calculated using the 12th root of 2 (12√2 ≈ 1.05946). Multiply 262 * 1.05946 to get 278 Hz (rounded). To find D4, you could multiply 278 * 1.05946. You could easily find the rest by using this method, but due to rounding, this wouldn't be a very precise way to do it. Better to use an absolute formula: To find the frequency, Pn, of a note in 12-TET, the following definition may be used: Pn = Pa(12√2)(n − a)
 
Good luck with those 12th roots!