A pipe is open at both ends and has a 2nd overtone of 353 Hz and speed of 353 m/s. Determine (a) the temperature to which the wave is traveling, (b) length of the pipe (c) the fundamental frequency of the wave?
Answer (2)
Given:2nd overtone = 353 HzSpeed of sound = 353 m/sThe pipe is open at both endsUnderstanding the overtone (Open pipe):For a pipe open at both ends, the harmonics are:Fundamental (1st harmonic) = f₁1st overtone (2nd harmonic) = 2f₁2nd overtone (3rd harmonic) = 3f₁So,3f₁ = 353 Hz → f₁ = 353 / 3 = 117.67 Hz(a) Temperature of the air:Speed of sound in air is related to temperature by:v = 331 + 0.6TWhere:v = 353 m/s (given)T = temperature in °CSolving:353 = 331 + 0.6T0.6T = 22T = 36.67°C(b) Length of the pipe:Use the formula for frequency in an open pipe:f₁ = v / (2L)Rearranging:L = v / (2f₁) = 353 / (2 × 117.67) ≈ 1.5 meters(c) Fundamental frequency:Already solved:f₁ = 117.67 Hz✅ Final Answers:(a) Temperature = 36.67°C(b) Length of pipe = 1.5 meters(c) Fundamental frequency = 117.67 Hz
Final Answers:(a) Temperature: 36.67 ∘C(b) Length of the pipe: 1.5m(c) Fundamental frequency: 117.67Hz
