For an open-open tube with length L = 1.0 m and wave speed v = 340 m/s, what is the fundamental frequency?

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Multiple Choice

For an open-open tube with length L = 1.0 m and wave speed v = 340 m/s, what is the fundamental frequency?

Explanation:
Open-open tubes form standing waves with a node of displacement (antinode of displacement) at each open end, so the fundamental pattern fits half a wavelength into the tube. That means L = lambda/2, or lambda = 2L. The frequency is f = v / lambda, so f = v / (2L). With v = 340 m/s and L = 1.0 m, you get f = 340 / 2 = 170 Hz. This is the lowest frequency that can form a standing wave in this tube; higher harmonics come from shorter effective wavelengths that still fit into the tube with additional half-wavelength segments. The other options would require a wavelength or pattern that doesn’t fit the open-open boundary conditions for this tube length.

Open-open tubes form standing waves with a node of displacement (antinode of displacement) at each open end, so the fundamental pattern fits half a wavelength into the tube. That means L = lambda/2, or lambda = 2L. The frequency is f = v / lambda, so f = v / (2L). With v = 340 m/s and L = 1.0 m, you get f = 340 / 2 = 170 Hz. This is the lowest frequency that can form a standing wave in this tube; higher harmonics come from shorter effective wavelengths that still fit into the tube with additional half-wavelength segments. The other options would require a wavelength or pattern that doesn’t fit the open-open boundary conditions for this tube length.

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