Practice question
Question
A pipe open at both ends has a length of 0.6 m and resonates with a source of frequency 850 Hz. What is
the harmonic number if the speed of sound is 340 m/s?
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) . 850 = (n × 340/2 × 0.6) = (n × 340/1.2) . 850 = n × 283.33 ⇒ n ≈ (850/283.33) ≈ 3 . Third harmonic (exact: v₃ = 3 × 340 / 1.2 = 850 Hz ). Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation
Discussion
Comments
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.