The first two resonance lengths of a resonance tube are 15.0 cm and 46.2 cm. The end correction is related to the tube radius r by e = 0.6r. If the tube radius is N cm, find N.
Physics · Wave Motion · organ pipe and resonance tube
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Question

The first two resonance lengths of a resonance tube are 15.0 cm and 46.2 cm. The end correction is related to the tube radius r by e = 0.6r. If the tube radius is N cm, find N.

G1 Physics · Accepted Answer

For the first resonance, l₁+e = λ/4. For the second resonance, l₂+e = 3λ/4. Multiply the first equation by 3: 3l₁+3e = 3λ/4. Subtract the second-resonance equation: 3l₁+3e−(l₂+e) = 0. Therefore, 2e = l₂−3l₁. Hence, e = (l₂−3l₁)/2. Substitute: e = (46.2−45.0)/2 = 1.2/2 = 0.6 cm. Given e = 0.6r, 0.6 = 0.6r. Thus, r = 1 cm. Hence, N = 1.