Practice question
Question
A pipe closed at one end has a length of 0.45 m and a speed of sound of 360 m/s. What is the frequency
of its fifth harmonic?
Explanation
**Resonance in pipes** occurs when length accommodates standing wave pattern. Closed pipe L = (2n-1)λ/4, so f₁ = v/(4L). Given f₁ and v, length follows L = v/(4f₁), enabling length calculation from measured resonance frequency and sound speed 330-340 m/s. For closed pipe: v_n = (n + (1/2)) (v/2L) , n = 4 for fifth harmonic. v₄ = (4 + (1/2)) (360/2 × 0.45) = 4.5 × (360/0.9) = 4.5 × 400 = 1800 Hz . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 1800 Hz, illustrating frequency-length-speed interdependence and quantization by boundaries.
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