Practice question
Question
A pipe closed at one end has a length of 0.85 m and resonates at its second harmonic with a speed of
sound of 340 m/s. What is the frequency?
Explanation
**Reflection at boundaries** follows phase change rules: rigid boundary (fixed end) introduces π phase shift, inverting displacement y → -y, while free boundary reflects without phase change. Reflected wave derived by reversing propagation direction kx → -kx and applying phase shift, preserving k = 2π/λ and ω = 2πf. For closed pipe: v_n = (n + (1/2)) (v/2L) , n = 1 for second harmonic. v₁ = (1 + (1/2)) (340/2 × 0.85) = 1.5 × (340/1.7) = 300 Hz . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 300 Hz, illustrating frequency-length-speed interdependence and quantization by boundaries.
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