In a tube closed at one end, the fundamental frequency f1 is related to v and L by f1 = v/(4L). If v = 340 m/s and L = 0.85 m, what is f1 when L doubles to 1.70 m?

In a tube closed at one end, the fundamental frequency f1 is related to v and L by f1 = v/(4L). If v = 340 m/s and L = 0.85 m, what is f1 when L doubles to 1.70 m?

• A. 50 Hz
• B. 100 Hz
• C. 200 Hz
• D. 25 Hz

Answer: (A)

Fundamental frequency for a tube closed at one end scales as f1 = v/(4L). Doubling the length halves the frequency because f1 ∝ 1/L. With v = 340 m/s and L = 1.70 m, f1 = 340/(4×1.70) = 340/6.8 ≈ 50 Hz. So the fundamental frequency is 50 Hz.

Fundamental frequency for a tube closed at one end scales as f1 = v/(4L). Doubling the length halves the frequency because f1 ∝ 1/L. With v = 340 m/s and L = 1.70 m, f1 = 340/(4×1.70) = 340/6.8 ≈ 50 Hz. So the fundamental frequency is 50 Hz.