What is the frequency of the note nine half-steps below middle A (which has a frequency of $437 \mathrm{cps}$ )? eleven half-steps below middle $A$ ? The frequency nine half-steps below middle A is $260 \mathrm{cps}$. (Round to the nearest integer as needed.) The frequency eleven half-steps below middle $A$ is cps. (Round to the nearest integer as needed.)
Solution
Step 1 ：We are given that the frequency of middle A (\(f_0\)) is 437 cps, and we are asked to find the frequency of the note nine half-steps below middle A (\(n = -9\)) and eleven half-steps below middle A (\(n = -11\)).
Step 2 ：The frequency of a note n half-steps from a given note can be calculated using the formula: \[f = f_0 * (2^{1/12})^n\] where f is the frequency of the note n half-steps away, f_0 is the frequency of the given note, and n is the number of half-steps. If n is negative, the note is below the given note; if n is positive, the note is above the given note.
Step 3 ：Substitute \(f_0 = 437\), \(n = -9\) into the formula, we get \(f_1 = 437 * (2^{1/12})^{-9} = 259.84175462809446\).
Step 4 ：Round \(f_1\) to the nearest integer, we get \(f_1 = 260\) cps.
Step 5 ：Substitute \(f_0 = 437\), \(n = -11\) into the formula, we get \(f_2 = 437 * (2^{1/12})^{-11} = 231.49268611750594\).
Step 6 ：Round \(f_2\) to the nearest integer, we get \(f_2 = 231\) cps.
Step 7 ：Final Answer: The frequency of the note nine half-steps below middle A is \(\boxed{260}\) cps. The frequency of the note eleven half-steps below middle A is \(\boxed{231}\) cps.
