An open pipe of physical length 0.840 m resonates at consecutive frequencies 200 Hz, 400 Hz and 600 Hz. Explain the mode pattern, infer wave speed if end correction is negligible, and calculate the end correction per end if an independent temperature measurement gives sound speed 343 m s⁻¹. Predict the first three resonances if one end is then closed but the same effective length applies. Discuss a limitation of that assumption.

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The open-pipe modes are n = 1, 2, 3 and imply 336 m s⁻¹ without correction. Effective length for 343 m s⁻¹ is 0.8575 m, so correction ≈ 8.75 mm per end. Closed-pipe resonances ≈ 100, 300 and 500 Hz under the same-length assumption. An open pipe has displacement antinodes at both ends and frequencies f_n = nv/(2L_eff). Integer multiples 200, 400 and 600 Hz identify consecutive n = 1, 2 and 3 modes. Ignoring end correction, v = 2Lf₁ = 2(0.840)(200) = 336 m s⁻¹. Using independent v = 343 m s⁻¹ gives L_eff = v/(2f₁) = 343/400 = 0.8575 m. The excess 0.0175 m shared by two open ends is 8.75 mm per end. With one end closed, only odd harmonics occur: f = v/(4L_eff), 3v/(4L_eff), 5v/(4L_eff). The assumed same effective length gives 100, 300 and 500 Hz. Closing an end changes the boundary geometry and its end correction; using the open-pipe effective length is therefore an approximation. Directly measure the new resonance series and check its odd-integer spacing.

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