Xylophone Mystery

As a $C$ scale, the original xylophone contained no sharps or flats. We can also use this chart to find the frequency of each note, and the fact that the frequency is inversely proportional to the square of the bar length.

If the longest bar at $30.17 \text{ cm}$ is a $C$ at $261.63 \text{ Hz}$ , then the medium bar at $26.11 \text{ cm}$ would produce a frequency of $\frac{30.17^2 \cdot 261.63}{26.11^2} \approx 349 \text{ Hz}$ or an $F$ , and the shortest bar at $23.95 \text{ cm}$ would produce a frequency of $\frac{30.17^2 \cdot 261.63}{23.95^2} \approx 415 \text{ Hz}$ or a $G \#$ . However, this is not possible since there are no sharps in the $C$ scale, so the longest bar is not a $C$ .

If the longest bar at $30.17 \text{ cm}$ is a $D$ at $293.66 \text{ Hz}$ , then the medium bar at $26.11 \text{ cm}$ would produce a frequency of $\frac{30.17^2 \cdot 293.66}{26.11^2} \approx 392 \text{ Hz}$ or an $G$ , and the shortest bar at $23.95 \text{ cm}$ would produce a frequency of $\frac{30.17^2 \cdot 293.66}{23.95^2} \approx 466 \text{ Hz}$ or a $A \#$ . However, this is not possible since there are no sharps in the $C$ scale, so the longest bar is not a $D$ .

If the longest bar at $30.17 \text{ cm}$ is an $E$ at $329.63 \text{ Hz}$ , then the medium bar at $26.11 \text{ cm}$ would produce a frequency of $\frac{30.17^2 \cdot 329.63}{26.11^2} \approx 440 \text{ Hz}$ or an $A$ , and the shortest bar at $23.95 \text{ cm}$ would produce a frequency of $\frac{30.17^2 \cdot 329.63}{23.95^2} \approx 523 \text{ Hz}$ or a $C$ . This is possible, so the longest bar must be an $\boxed{E}$ .

For completeness, if the longest bar at $30.17 \text{ cm}$ is an $F$ or higher at $349.23 \text{ Hz}$ or higher, then the shortest bar at $23.95 \text{ cm}$ would produce a frequency of $\frac{30.17^2 \cdot 349.23}{23.95^2} \approx 554 \text{ Hz}$ or higher, which is out of range on the $C$ scale from $C_4$ to $C_5$ .