For every musical tone, f is the frequency, in cycles per second, of the musical tone and w is the...

1. Translate the problem requirements: We need to understand that \(\mathrm{fw} = \mathrm{constant}\) means frequency and wavelength have an inverse relationship. Given middle C (\(\mathrm{f}=260, \mathrm{w}=4\)) and G tone (\(\mathrm{f}=390, \mathrm{w}=?\)), we need to find the unknown wavelength.
2. Establish the constant relationship: Use the given middle C data to find the value of the constant \(\mathrm{fw}\).
3. Apply inverse relationship logic: Since the product must remain constant, use the fact that if frequency increases, wavelength must decrease proportionally.
4. Calculate and verify against answer choices: Solve for the G tone's wavelength and identify the closest answer choice.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what this problem is telling us in plain English. We have musical tones, and for each tone we have two measurements: how fast it vibrates (frequency) and how long each wave is (wavelength).

The key insight is that \(\mathrm{fw} = \mathrm{constant}\). This means frequency times wavelength always gives us the same number, no matter which musical tone we're looking at. Think of it like a seesaw - when one side goes up, the other must go down to keep balance.

We know:

• Middle C: frequency = 260 cycles per second, wavelength = 4 feet
• G tone: frequency = 390 cycles per second, wavelength = ?

We need to find the wavelength of the G tone.

Process Skill: TRANSLATE - Converting the mathematical relationship \(\mathrm{fw} = \mathrm{constant}\) into the intuitive concept of inverse relationship

2. Establish the constant relationship

Since \(\mathrm{fw}\) must be the same for all musical tones, let's find this constant using the middle C information.

For middle C:
\(\mathrm{frequency} \times \mathrm{wavelength} = 260 \times 4 = 1,040\)

So our constant is 1,040. This means for ANY musical tone, when we multiply its frequency by its wavelength, we must get 1,040.

3. Apply inverse relationship logic

Now let's think about what happens with the G tone. We know:

• G tone frequency = 390 cycles per second
• The constant must still be 1,040
• So: \(390 \times \mathrm{wavelength} = 1,040\)

Notice that the G tone has a higher frequency (390) compared to middle C (260). Since the product must stay the same, the wavelength must be smaller than middle C's wavelength of 4 feet. This makes intuitive sense - higher frequency means shorter wavelength.

To find the G tone's wavelength:
\(\mathrm{wavelength} = 1,040 \div 390\)

Let's simplify this step by step:
\(1,040 \div 390 = 104 \div 39\) (dividing both by 10)

Now \(104 \div 39\):
\(39 \times 2 = 78\)
\(104 - 78 = 26\)
So \(104 \div 39 = 2 + \frac{26}{39}\)

Simplifying the fraction \(\frac{26}{39}\):
\(26 = 2 \times 13\) and \(39 = 3 \times 13\)
So \(\frac{26}{39} = \frac{2}{3}\)

Therefore: \(\mathrm{wavelength} = 2 + \frac{2}{3} = 2\frac{2}{3} \text{ feet}\)

4. Calculate and verify against answer choices

Our calculated wavelength for the G tone is \(2\frac{2}{3}\) feet.

Let's verify this makes sense:

• G tone has higher frequency (390 vs 260) than middle C
• G tone has shorter wavelength (\(2\frac{2}{3}\) vs 4) than middle C ✓
• Check: \(390 \times 2\frac{2}{3} = 390 \times \frac{8}{3} = 1,040\) ✓

Looking at the answer choices:

1. \(2\frac{2}{3}\) ✓ (This matches our calculation)
2. \(4\frac{2}{3}\)
3. \(5\frac{1}{2}\)
4. 6
5. \(32\frac{1}{2}\)

Final Answer

The wavelength of the G musical tone is \(2\frac{2}{3}\) feet.

Answer: A

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the constant relationship

Students may incorrectly interpret "fw is a constant" as meaning frequency and wavelength are equal to each other, rather than understanding that their product remains constant across different musical tones. This leads to setting up equations like \(\mathrm{f} = \mathrm{w}\) instead of \(\mathrm{f_1w_1} = \mathrm{f_2w_2}\).

2. Confusing which values belong to which musical tone

With multiple musical tones mentioned (middle C and G tone), students often mix up the given values. They might incorrectly assign the frequency of 390 to middle C or use the wavelength of 4 feet for the G tone when setting up their equations.

3. Not recognizing the inverse relationship pattern

Students may fail to anticipate that since the G tone has a higher frequency than middle C, it should logically have a shorter wavelength. This lack of conceptual understanding means they don't catch obviously wrong answers during their solution process.

Errors while executing the approach

1. Arithmetic errors in division and fraction simplification

When calculating \(1,040 \div 390\), students commonly make errors in long division or struggle with converting the decimal result back to a mixed number. The multi-step simplification from \(\frac{1040}{390}\) to \(\frac{104}{39}\) to \(2\frac{2}{3}\) creates multiple opportunities for computational mistakes.

2. Incorrect fraction reduction

Even when students correctly calculate that the remainder gives them \(\frac{26}{39}\), they often fail to recognize that both 26 and 39 are divisible by 13, missing the simplification to \(\frac{2}{3}\). This leads them to select an answer choice that doesn't match their unreduced fraction.

3. Setting up the wrong equation

Students might incorrectly set up the proportion, such as writing \(\frac{260}{4} = \frac{390}{\mathrm{w}}\) (comparing frequency to wavelength ratios) instead of the correct relationship \(260 \times 4 = 390 \times \mathrm{w}\) (maintaining constant products).

Errors while selecting the answer

1. Converting between mixed numbers and improper fractions incorrectly

Students might correctly calculate \(\frac{8}{3}\) feet but then incorrectly convert this to a mixed number, getting \(2\frac{1}{3}\) instead of \(2\frac{2}{3}\), leading them to select the wrong answer choice or become confused when their result doesn't exactly match any option.

2. Choosing an answer that doesn't make conceptual sense

Without verifying their result against the logical expectation, students might select answer choices like B (\(4\frac{2}{3}\)) or larger values, even though they should recognize that higher frequency must correspond to shorter wavelength compared to middle C's 4 feet.