I do not really understand the question but all I can say is:
geometric sequence
fn = a r^(n-1)
fn = 27.5 r^(n-1)
log ( fn /27.5)= (n-1) log r
The first key on a piano keyboard corresponds to a pitch with a frequency of 27.5 cycles per second. With every successive key, going up the black and white keys, the pitch multiplies by a constant. The formula for the frequency of the pitch sounded when the nth note up the keyboard is played is given by log _(2)((f)/(27.5))
A. a note has a frequency of 830 cycles per second. How many notes up the piano keyboard is this?
2 answers
Not familiar with your formula, but here is another way to look
at the frequencies of the piano keys
when f(n) is the frequency of a certain key n
it can be found by
f(n) = 440* 2^( (n-49)/12 )
or
if you take log (f(n)) - log (f(n-1)), where n is the key number on the piano,
you get a constant 0.025085832
440* 2^( (n-49)/12 ) = 830
2^(n-49)/12 = 1.8863636..
((n-49)/12) log2 = log 1.8863636..
(n-49)/12 = .915607812
n-49 = 10.98729375
n = 59.987 , looks like the key # 60 on the piano
According to a chart of frequencies, there is no key with a frequency of exactly 830.
www.askinglot.com/open-detail/609252
at the frequencies of the piano keys
when f(n) is the frequency of a certain key n
it can be found by
f(n) = 440* 2^( (n-49)/12 )
or
if you take log (f(n)) - log (f(n-1)), where n is the key number on the piano,
you get a constant 0.025085832
440* 2^( (n-49)/12 ) = 830
2^(n-49)/12 = 1.8863636..
((n-49)/12) log2 = log 1.8863636..
(n-49)/12 = .915607812
n-49 = 10.98729375
n = 59.987 , looks like the key # 60 on the piano
According to a chart of frequencies, there is no key with a frequency of exactly 830.
www.askinglot.com/open-detail/609252
10 answers