If L = 1.0 m and v = 340 m/s, what is the fundamental frequency f1 = v/(2L)?

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Multiple Choice

If L = 1.0 m and v = 340 m/s, what is the fundamental frequency f1 = v/(2L)?

Explanation:
The fundamental frequency for a string fixed at both ends comes from the idea that the fundamental mode fits half a wavelength along the length: lambda1 = 2L, so f1 = v / (2L). With L = 1.0 m and v = 340 m/s, f1 = 340 / (2 × 1.0) = 170 Hz. So the fundamental frequency is 170 Hz. The other numbers would come from different boundary conditions or higher harmonics (for example, v/L would give 340 Hz, which isn’t the fundamental for a string fixed at both ends; v/(4L) would give 85 Hz for a pipe closed at one end; 510 Hz would be a higher harmonic).

The fundamental frequency for a string fixed at both ends comes from the idea that the fundamental mode fits half a wavelength along the length: lambda1 = 2L, so f1 = v / (2L). With L = 1.0 m and v = 340 m/s, f1 = 340 / (2 × 1.0) = 170 Hz. So the fundamental frequency is 170 Hz. The other numbers would come from different boundary conditions or higher harmonics (for example, v/L would give 340 Hz, which isn’t the fundamental for a string fixed at both ends; v/(4L) would give 85 Hz for a pipe closed at one end; 510 Hz would be a higher harmonic).

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