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MCAT Physics Question — Fundamental Wavelength
- by
- Jan 24, 2017
- MCAT Question of the Day
The fundamental wavelength of a standing wave in a 1 meter long pipe, if the pipe has one open end and one closed end is:
A. 4m
B. 1m
C. 2m
D. 4/3m
Click for Explanation
Standing waves consist of wave frequencies at which wave nodes occur at closed pipe ends and antinodes occur at open pipe ends.
For a pipe closed at one end, the equation relating the length of the pipe, L, to the λ of a standing wave in the pipe is λ = 4L / n, where n is the harmonic number, and can be any odd integer.
When n = 1, the standing wave is the “fundamental” wavelength. This is also called the “first harmonic”.
A. 4m, correct, the fundamental wavelength for a pipe with one closed end is λ1= 4L.
B. 1m, incorrect , the pipe length does not correspond to the fundamental wavelength.
C. 2m, incorrect, 2m corresponds to the fundamental wavelength of a pipe with both ends open λ1= 2L.
D. 4/3m, incorrect, 4/3m corresponds to the second resonant wavelength of a pipe with one end open, λ2= 4L/3.
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