The two nearest harmonics of a tube closed at one end and open at other end are 220 Hz and 260 Hz. W

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Question

The two nearest harmonics of a tube closed at one end and open at other end are 220 Hz and 260 Hz. What is the fundamental frequency of the system?

Accepted Answer

Correct Answer: C — 20 HzStep-by-step Solution:A tube closed at one end and open at the other supports only odd harmonics:f_n = (2n - 1)f₀ for n = 1, 2, 3, ...Two consecutive (nearest) harmonics correspond to n and (n+1):f_n = (2n - 1)f₀ = 220 Hzf_{n+1} = (2n + 1)f₀ = 260 HzSubtracting: 2f₀ = 260 - 220 = 40 Hzf₀ = 20 HzVerification: 220 = (2n-1) × 20 → 2n-1 = 11 → n = 6 (11th harmonic). 260 = (2×7-1) × 20 = 13 × 20 = 260. Consistent.Why other options are incorrect:Option A (40 Hz): This is the difference between the two harmonics (260 - 220), which equals 2f₀, not f₀. Option B (10 Hz): This would give harmonics at multiples of 10 Hz, but the difference between consecutive odd harmonics would be 2 × 10 = 20 Hz, not 40 Hz. Option D (30 Hz): 2 × 30 = 60 ≠ 40 Hz.Key Concept: Closed organ pipe (one end closed, one end open): Only odd harmonics are present. Frequencies are f₀, 3f₀, 5f₀, 7f₀, ... The difference between any two consecutive harmonics (in terms of n) is always 2f₀. So f₀ = (difference between consecutive harmonics) / 2.

The two nearest harmonics of a tube closed at one end and open at other end are 220 Hz and 260 Hz. What is the fundamental frequency of the system?

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The two nearest harmonics of a tube closed at one end and open at other end are 220 Hz and 260 Hz. What is the fundamental frequency of the system?

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