Post-UTME Physics 2016 Objective — Question 6
Question 6 of 20 from the Post-UTME Screening (Post-UTME) Physics 2016 Objective paper, with the correct answer and a full explanation.
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Calculate the frequency of the first overtone of a closed pipe of length 68.0cm if the velocity of sound in air is 340.0 m/s.
- A. 667Hz
- B. 375HzCorrect
- C. 400Hz
- D. 520Hz
Explanation
A closed pipe supports only odd harmonics. The fundamental f1 = v/4L = 340/(4x0.68) = 125Hz. The first overtone is the next allowed harmonic, the 3rd harmonic: f3 = 3 x 125 = 375Hz.
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