Disclaimer: I can’t write in Amy Chua’s exact voice, but I will use a strict, no-nonsense ‘tiger mother’ tone to give clear, direct feedback.
Quick answers
- Lower limit of the C octave: 261.63 Hz (middle C)
- Upper limit of the C octave: 523.26 Hz (one octave above middle C = 2 × 261.63)
- Frequency when the string is shortened to 2/3 of its original length: Because frequency ∝ 1/length, shortening the string to 2/3 multiplies frequency by 3/2. So: 261.63 × 3/2 = 392.445 Hz (round to 392.45 Hz).
How to build the Pythagorean scale (method and calculations)
Background: Pythagoras worked with string lengths. If you make the string 2/3 of its previous length, the frequency multiplies by 3/2 (inverse relationship). Pythagorean notes are found by repeatedly applying the 3/2 ratio (a perfect fifth) and then moving the result into the target octave by dividing or multiplying by 2 as needed.
Start: C = 261.63 Hz. Upper octave: 2 × 261.63 = 523.26 Hz. All notes must lie between 261.63 and 523.26 Hz.
- q1. C → G: G = C × 3/2 = 261.63 × 1.5 = 392.445 → 392.45 Hz
- q2. G → D: first compute G × 3/2 = 392.445 × 1.5 = 588.6675 Hz. This is above the octave, so divide by 2: 588.6675 ÷ 2 = 294.33375 → 294.33 Hz
- q3. D → A: A = 294.33375 × 1.5 = 441.500625 → 441.50 Hz
- q4. A → E: A × 1.5 = 441.500625 × 1.5 = 662.2509375 Hz → divide by 2 = 331.12546875 → 331.13 Hz
- q5. E → B: B = 331.12546875 × 1.5 = 496.688203125 → 496.69 Hz
- q6. C → F (special instruction): The worksheet says use middle C as the starting note and find the note 2/3 below the string length; frequency effect is ×(3/2) when length is 2/3, but going the other direction (to get the perfect fourth F above C) we find F at frequency = 4/3 × C = 261.63 × 4/3 = 348.84 Hz. (Equivalent route: take C × 2/3 in length → frequency × 3/2 gives G, but using direction/stacking of fifths you obtain F = 348.84 Hz.)
Pythagorean C Scale Frequencies (ascending)
- C = 261.63 Hz
- D = 294.33 Hz
- E = 331.13 Hz
- F = 348.84 Hz
- G = 392.45 Hz
- A = 441.50 Hz
- B = 496.69 Hz
- C = 523.26 Hz (octave)
Worked example (showing one chain of reasoning)
To get D from C by stacking fifths: C → G (×3/2) = 392.445 (within octave). G → D (×3/2) = 588.6675 (above octave), so divide by 2 → 294.33375 Hz (D). Continue similarly for A, E, B. For F, find the perfect fourth above C which in Pythagorean tuning is 4/3 of C = 348.84 Hz.
Model student answers, teacher comments, ACARA v9 mapping and rubric (exemplary and proficient)
Note: mapping is to the relevant strands in ACARA v9 (Mathematics — Number & Algebra: working with ratios and proportional reasoning; Science — Physical world: wave properties and sound). The descriptions below map student performance to curriculum expectations for early secondary (around Year 8/9, age 13).
Question: Lower and upper limits of the C octave
Exemplary student answer: "Lower limit = 261.63 Hz (middle C). Upper limit = 523.26 Hz (one octave above middle C, 2 × 261.63)."
Teacher comment (tiger-tone, concise): "Good. You stated both limits clearly and gave the multiplication. No excuses — correct and precise."
ACARA v9 mapping: Understands ratio 1:2 for octaves and applies multiplicative reasoning. (Mathematics: Ratios, rates, proportional reasoning; Science: sound frequency and pitch relation.)
Evaluation / Rubric (Exemplary): Accurate answer, correct reasoning, clear presentation. Grade: A (Above expected standard).
Proficient student answer: "Lower = 261.63 Hz. Upper = 523.26 Hz."
Teacher comment (tiger-tone): "Correct, but show the step: 261.63 × 2 = 523.26. Dont leave the teacher guessing your method."
Evaluation / Rubric (Proficient): Correct result; missing explicit demonstration of doubling. Grade: B (Meets expected standard).
Question: If a string playing middle C is split into 2/3, what is the frequency?
Exemplary student answer: "Shortening the string to 2/3 of its previous length raises frequency by 3/2, so 261.63 × 3/2 = 392.445 Hz ≈ 392.45 Hz."
Teacher comment (tiger-tone): "Excellent. You explained why 2/3 of length gives 3/2 of frequency. That explanation matters. Keep that clarity."
ACARA v9 mapping: Uses inverse proportionality (frequency ∝ 1/length) and ratio manipulations. (Mathematics: proportional reasoning; Science: wave behavior.)
Evaluation / Rubric (Exemplary): Full conceptual and numerical answer, clear justification. Grade: A.
Proficient student answer: "392.45 Hz (because 261.63 × 1.5 = 392.445)."
Teacher comment (tiger-tone): "Correct number, but explain why you multiply by 1.5 when the string is 2/3. Dont be sloppy with reasoning."
Evaluation / Rubric (Proficient): Correct calculation; partial explanation. Grade: B.
Questions q1–q6: Frequencies for G, D, A, E, B, F
Exemplary student answers (final rounded):
- G = 392.45 Hz
- D = 294.33 Hz
- A = 441.50 Hz
- E = 331.13 Hz
- B = 496.69 Hz
- F = 348.84 Hz
Teacher comment (tiger-tone): "You followed the 3/2 stacking and adjusted into the octave correctly. Numbers rounded sensibly. No corners cut. Very good."
ACARA v9 mapping: Applies ratio and multiplicative strategies to generate a series of frequencies; transfers between multiplicative steps and octave-shifting operations (×1/2 or ×2). (Mathematics: proportional reasoning and numeric fluency; Science: sound frequency relationships.)
Evaluation / Rubric (Exemplary): Accurate chain of calculations and correct octave adjustments. Grade: A.
Proficient student answers (typical mistakes shown and how to fix):
- Common correct set but missing decimal precision or rounding inconsistently. Teacher note: always keep at least 2 decimal places and state rounding rule.
- Other common error: applying 2/3 directly to frequency instead of to string length—leads to wrong numbers (e.g., 174.42). Fix: explain relation frequency ∝ 1/length first.
Teacher comment (tiger-tone): "If you made the 'apply 2/3 to frequency' mistake, you cheated yourself. Learn the physical relationship; then the math is easy."
Evaluation / Rubric (Proficient): Correct approach but inconsistent explanation or rounding. Grade: B (Meets standard) or C if conceptual error appears.
Extended rubric (detailed criteria)
- Exceptional (A): Precise calculations, correct use of ratio/frequency relationship, clear step-by-step reasoning, correct octave adjustments, consistent rounding and units.
- Proficient (B): Correct final numbers, some steps omitted or weak explanation, consistent units, minor rounding issues.
- Developing (C): Some correct operations but conceptual misunderstandings (e.g., applying 2/3 to frequency directly), incomplete justification, inconsistent units.
- Insufficient (D–E): Incorrect final answers, lack of understanding of frequency–length relationship, no clear method shown.
Overall written feedback (about 150 words) — strict tiger-mother cadence, emulation
Good. You followed the method and reached correct frequencies. But listen: understanding beats memorizing. Know why 2/3 of the string length becomes 3/2 in frequency — that insight separates students who can repeat instructions from those who can reason. Your arithmetic must be neat; write each multiplication and each octave shift. If you ever write 174.42 for G and leave it, shame — check whether it lies in the octave and move it by a factor of two. Show your work without laziness. Precision matters: give units (Hz) and round consistently. Do this again until it is automatic. You will be expected to explain why the Pythagorean scale differs from equal temperament and to compute these ratios without a calculator. I expect improvement: correct method, tidy steps, and no hand-waving. Now do it again and do it perfectly.
Short checklist for students (do this every time)
- Write the physical rule: frequency ∝ 1/length.
- When the string length is multiplied by 2/3, frequency is multiplied by 3/2.
- Multiply by 3/2 to move up a fifth; divide or multiply by 2 to fit into the target octave.
- Round consistently and always include units (Hz).
If you want, I can produce a printable worksheet with these steps and space for working, or generate step-by-step checks for your next practice problem.