AS Daily A Level Physics question
A cylindrical tube of fixed length L is driven by a speaker and resonates with both ends open at frequency f, corresponding to the second harmonic in that configuration. The tube is then closed at one end while L and f are unchanged (air temperature and end corrections are assumed unchanged/negligible). Which statement must be true about stationary-wave resonance in the closed tube?
Answer
The correct answer is A.
Correct: A — No exact resonance occurs, because matching f would require the fourth multiple of the closed‑tube fundamental, and only odd multiples are allowed. For the open–open tube, the second harmonic has f = v/L. A closed–open tube resonates at odd multiples of v/(4L); to match v/L would need 4×v/(4L), which is not permitted since even multiples are excluded by the node–antinode boundary conditions. B is wrong because the third closed‑tube harmonic is 3v/(4L) = 0.75 v/L, which is not equal to f = v/L. C is wrong because the closed‑tube fundamental is v/(4L), a quarter of v/L, and closing one end does not halve the wavelength for the same frequency; it enforces a node–antinode pair, changing the fundamental to λ = 4L. D is wrong because the fifth closed‑tube harmonic is 5v/(4L) = 1.25 v/L, which overshoots f; choosing a higher odd harmonic does not make an exact match here.