In a tube closed at one end, the wavelength of the fundamental harmonic (n=1) is which of the following in terms of length L?

In a tube closed at one end, the wavelength of the fundamental harmonic (n=1) is which of the following in terms of length L?

• A. 2L
• B. L
• C. 4L/3
• D. 4L

Answer: (D)

A tube closed at one end supports standing waves with a node at the closed end and an antinode at the open end. For the fundamental mode, the tube length is one-quarter of a wavelength. So L = λ/4, which gives λ = 4L. This is the wavelength that fits exactly in the tube for the lowest standing wave. For higher odd harmonics, the length relates to wavelength by L = (2n−1)λ/4, so the next allowed wavelength would be 4L/3, and so on.

A tube closed at one end supports standing waves with a node at the closed end and an antinode at the open end. For the fundamental mode, the tube length is one-quarter of a wavelength. So L = λ/4, which gives λ = 4L. This is the wavelength that fits exactly in the tube for the lowest standing wave. For higher odd harmonics, the length relates to wavelength by L = (2n−1)λ/4, so the next allowed wavelength would be 4L/3, and so on.