Question 187713: f(x)=k2^x/12 can be used to determine the frequency, in cycles per second, of a musical note that is x half-steps above a note with frequency k.
The frequency of concert A for a trumpet is 440 cycles per second. Find the frequency of the A that is two octaves (24 half-steps) above concert A.
I tried to plug in the numbers but it does not seem right:
x=440(2)^24/12
x=440(2)^2
x=440(4)
x=1760
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! f(x)=k2^x/12 can be used to determine the frequency, in cycles per second, of a musical note that is x half-steps above a note with frequency k.
The frequency of concert A for a trumpet is 440 cycles per second. Find the frequency of the A that is two octaves (24 half-steps) above concert A.
I tried to plug in the numbers but it does not seem right:
x=440(2)^24/12
x=440(2)^2
x=440(4)
x=1760
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It's correct. An octave up doubles the frequency.
A is 440 Hz, 880 Hz, 1760 Hz, etc.
It's also 220 Hz, 110 Hz, etc.
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