Practice question
Question
A pipe open at both ends has a length of 0.25 m and a speed of sound of 340 m/s. What is the frequency
of its second harmonic?
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 open pipe: v_n = (n v/2L) . Second harmonic ( n = 2 ): v₂ = (2 × 340/2 × 0.25) = (680/0.5) = 1360 Hz . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 1360 Hz, illustrating frequency-length-speed interdependence and quantization by boundaries.
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