Music Sequences

Question

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.

Answer 1

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!