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Question

A pipe open at both ends has a length of 0.45 m and resonates with a source of frequency 1133 Hz. What
is the harmonic number if the speed of sound is 340 m/s?

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Choose one · Correct answer highlighted

Explanation

**Air column vibrations** depend on end conditions. Pipe closed at one end has displacement node at closed end and antinode at open, allowing only odd harmonics, fundamental f₁ = v/(4L). Open pipe has antinodes at both ends, fₙ = n·v/(2L), all harmonics present, v sound speed. v_n = (n v/2L) . 1133 = (n × 340/2 × 0.45) = (n × 340/0.9) . 1133 = n × 377.78 ⇒ n ≈ (1133/377.78) ≈ 3 . Third harmonic (exact: v₃ = 3 × 340 / 0.9 = 1133.33 Hz ). Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation

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