An open pipe is in resonance in 2nd harmonic with frequency f1. Now one end of the tube is closed and frequency is increased to f2 such that the resonance again occurs in the nth harmonic. Choose the correct option:
Correct answer: C. n = 5, f2 = 5/4 f1
- A. n = 3, f2 = 3/4 f1
- B. n = 3, f2 = 5/8 f1
- C. n = 5, f2 = 5/4 f1
- D. n = 5, f2 = 3/4 f1
Explanation
In an open pipe, the harmonics occur at integer multiples of the fundamental frequency. When the pipe is closed, only odd harmonics are present. In this problem, the open pipe is in resonance at the 2nd harmonic with frequency f1. When one end is closed, the next possible resonance occurs at the 5th harmonic, because the closed pipe supports only odd harmonics. Therefore, the new frequency f2 is 5/4 of the original frequency f1. The other options incorrectly calculate this relationship based on incorrect assumptions about the harmonics in a closed pipe.
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