What is the frequency of note y if it is two sext higher than note x?

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Note x has a frequency of 300 Hz, and the problem involves finding the frequency of note y, which is two sext higher than x. The sext interval ratio is established as 24:40, leading to an initial calculation of y as 500 Hz. However, applying the sext interval relationship twice results in a frequency of 833.33 Hz for note y. The discussion highlights the importance of considering multiple applications of the interval relationship in frequency calculations.
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Homework Statement


Given that note x has frequency 300 Hz. Find the frequency of note y if it is two sext higher than x.


Homework Equations


sext interval --> x : y = 24 : 40


The Attempt at a Solution


I can find the value of y if the interval is sext.

x : y = 24 : 40
300 : y = 24 : 40
y = 500 Hz

two sext higher means that y = 2 x 500 = 1000 Hz?

Thanks
 
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Why not apply the relationship twice?

500/y = 24/40

y = 833.33
 
gneill said:
Why not apply the relationship twice?

500/y = 24/40

y = 833.33

Very good point. Why don't that idea even cross my mind...

Thanks for the help gneill :smile:
 

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