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Sound in two-path tube networks: the smallest length l for which the detector hears a maximum, matched to List-II

Sound of wavelength 0.29 m splits between a straight tube of length l and a detour in four tube networks. The detector hears the first maximum when the path difference is one wavelength: l = 0.51, 0.29, 0.13 and 1.19 m for P, Q, R and S - option D.

Sub-topics covered

• Superposition of two waves from one source: amplitude |1 + e^(i 2 pi Delta/lambda)|
• Constructive interference: path difference Delta = m lambda
• Path lengths of semicircles, rectangles and triangles (sine rule) in units of l
• Smallest length: l_min = lambda/(k - 1) for a detour of k l

What it teaches

• The sound splits at S and recombines at D, so only the path difference matters: a maximum needs Delta = m lambda, the smallest non-zero l has Delta = lambda
• Every length in a figure scales with l, so the detour is k l and l_min = lambda/(k - 1)
• k = pi/2 (semicircle on l), 2 (rectangle of height l/2), 1 + pi/sqrt2 (side l plus a semicircle on the diagonal), (sin 30 + sin 45)/cos 15 (triangle by the sine rule)
• l = 0.508, 0.290, 0.131, 1.186 m -> (3), (4), (5), (2): option D; with the exact cos 15 the triangle gives 1.16 m, still nearest (2)

Simulation ID

• ADV-2026-P1-PHY-Q13 · revision 1

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