Practice question
Question
A pipe closed at one end has a length of 0.6 m and resonates at its second harmonic with a speed of
sound of 360 m/s. What is the frequency?
Explanation
**Sound wave reflection** at rigid wall behaves like string fixed end, displacement inverted. Equation of reflected wave includes sign change and direction reversal, amplitude unchanged but sign may flip, explaining standing wave formation with incident wave. For closed pipe: v_n = (n + (1/2)) (v/2L) , n = 1 for second harmonic. v₁ = (1 + (1/2)) (360/2 × 0.6) = 1.5 × (360/1.2) = 1.5 × 300 = 450 Hz . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 450 Hz, illustrating frequency-length-speed interdependence and quantization by boundaries.
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