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Question

A pipe closed at one end has a length of 0.25 m and resonates at its fundamental frequency with a speed
of sound of 340 m/s. What is the frequency?

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Choose one · Correct answer highlighted

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. Fundamental: v₁ = (v/4L) . v₁ = (340/4 × 0.25) = (340/1) = 340 Hz . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 340 Hz, illustrating frequency-length-speed interdependence and quantization by boundaries.

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