Problem 92 - Precalculus: Mathematics for Calculus

Question

The frequencies of musical notes (measured in cycles per second) form a geometric sequence. Middle C has a frequency of \(256,\) and the C that is an octave higher has a frequency of \(512 .\) Find the frequency of \(\mathrm{C}\) two octaves below middle C.

Step-by-Step Solution

Answer

The frequency of C two octaves below middle C is 64 Hz.

Step 1: Define the Geometric Sequence

In a geometric sequence, each term after the first is found by multiplying the previous term by a constant called the common ratio, \(r\). This can be expressed generally as \(a_n = a_1 \times r^{(n-1)}\), where \(a_n\) is the \(n\)-th term and \(a_1\) is the first term in the sequence.

Step 2: Identify Known Values

We know that the frequency of middle C is \(256\) Hz, which serves as the first term of our sequence (\(a_1=256\)). Additionally, the C an octave higher, which is the second term in the sequence, has a frequency of \(512\) Hz.

Step 3: Calculate the Common Ratio

The frequency one octave higher than middle C is \(512\), making it the second term of the sequence. By using the formula for the second term \(a_2 = a_1 \times r\), we have \(256 \times r = 512\). Solving for \(r\), we find \(r = \frac{512}{256} = 2\).

Step 4: Calculate Frequency Two Octaves Below Middle C

To find the frequency two octaves below middle C, we need to go down by two terms in a sequence where each step is a division by the common ratio \(r\). Being two octaves below middle C corresponds to two terms below it: two divisions by \(r\). Therefore, the frequency is \(a_{-2} = a_1 \times r^{-2} = 256 \times 2^{-2} = 256 \times \frac{1}{4} = 64\).

Step 5: Verify the Solution

Check to ensure the calculations are consistent by analyzing each part of the problem within the geometric sequence's context: moving two octaves below involves dividing by \(2\) twice, following the geometric series properties.

Key Concepts

Musical Frequencies

Musical notes have frequencies which can be described using a geometric sequence. This means each note's frequency is determined by multiplying the previous note's frequency by a fixed number known as the common ratio. In music, moving up an octave means doubling the original frequency. For example, if the middle C note has a frequency of 256 Hz, the C note an octave above has a frequency of 512 Hz. Musical frequencies follow this pattern consistently, making understanding geometric sequences fundamental in music theory.

Common Ratio

The common ratio in a geometric sequence is a vital concept in determining how the sequence progresses. In the context of musical frequencies, the common ratio is the factor you multiply by to get from one frequency to the next in an octave. From the example in the exercise, the common ratio between middle C and the C an octave higher is 2. This is because dividing the frequency of the higher C (512 Hz) by the frequency of middle C (256 Hz) gives us 2. Understanding this common ratio helps calculate other notes in the sequence.

Octave Calculation

Calculating frequencies for octaves involves using the common ratio to move up or down between notes. For example, to find the frequency of a note two octaves below middle C, we can use the formula for a geometric sequence:

• Start with the frequency of middle C at 256 Hz.
• The common ratio for an octave change is 2.
• Going down two octaves means dividing the middle C frequency by 2 twice.
• This results in the frequency: \( 256 \times \left(\frac{1}{2}\right)^2 = 256 \times \frac{1}{4} = 64 \text{ Hz} \).

By understanding octave calculation, we can find any other notes' frequencies relative to a known note.

Precalculus Problem-Solving

Understanding geometric sequences and their application in real-world scenarios like musical frequencies is an important aspect of precalculus problem-solving. Often, exercises in precalculus require understanding the formula for a geometric sequence:

• Identify the first term and the common ratio.
• Use the formula \( a_n = a_1 \times r^{n-1} \) where \( a_1 \) is the first term, \( r \) is the common ratio, and \( n \) is the term number.
• Apply this to find unknowns such as the term number or particular frequency.

Solving precalculus problems enhances logical thinking and helps understand mathematical patterns and relationships better, providing a foundation for future topics in calculus and beyond.