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Two organ pipes closed at one end have the same diameters but different lengths. Show that the end correction at each end is e = n1l1-n2l2n2-n1, where the symbols have their usual meanings. -

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Question

Two organ pipes closed at one end have the same diameters but different lengths. Show that the end correction at each end is e = `(n_1l_1 - n_2l_2)/(n_2 - n_1)`, where the symbols have their usual meanings. Take `γ = 5/3`.

Numerical

Solution

Suppose two organ pipes, closed at one end and of the same inner diameter d, have lengths l1 and l2.

Then, the effective lengths of the air columns are respectively

L1 = l1 + e = l1 + 0.3 d and L2 = l2 + e = l2 + 0.3 d

where e = 0.3 d is the end correction for the open end. The fundamental frequencies of the corresponding air columns are

`n_1 = v/(4L_1) = v/(4(l_1 + e)` and `n_2 = v/(4L_2) = v/(4(l_2 + e)`

where v is the speed of sound in air.

∴ v = 4n1 (l1 + e) = 4n2 (l2 + e)

∴ n1l1 + n1e = n2l2 + n2e

∴ n1l1 − n2l2 = (n2 − n1)e

∴ e = `(n_1l_1 - n_2l_2)/(n_2 - n_1)`

which is the required expression.

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