Question

Obtain an expression for practical determination of end correction -- (i) for a pipe open at both ends and (ii) for a pipe closed at one end.

Hint:

End correction is like adding a little extra "acoustic length" to the pipe. The key to finding it experimentally is to use two different resonance conditions (either different lengths or different frequencies) and solve the resulting simultaneous equations.

Answer 1

End correction (\(e\)) accounts for the fact that the antinode at the open end of a pipe is not formed exactly at the edge, but slightly outside it. The corrected length is \(L' = L + e\) for a closed pipe and \(L' = L + 2e\) for an open pipe.
Let \(n_1\) and \(n_2\) be the frequencies of two successive resonating modes.
(i) For a pipe open at both ends:
The fundamental frequency is \(n = \frac{v}{2L'} = \frac{v}{2(L+2e)}\).
Let the pipe resonate with a tuning fork of frequency \(n_1\) at length \(L_1\),
and with a tuning fork of frequency \(n_2\) at length \(L_2\).
Then, \(v = 2n_1(L_1 + 2e)\) and \(v = 2n_2(L_2 + 2e)\). \[ 2n_1(L_1 + 2e) = 2n_2(L_2 + 2e) \] \[ n_1L_1 + 2n_1e = n_2L_2 + 2n_2e \] \[ 2e(n_1 - n_2) = n_2L_2 - n_1L_1 \] \[ e = \frac{n_2L_2 - n_1L_1}{2(n_1 - n_2)} \] (ii) For a pipe closed at one end:
The fundamental frequency is \(n = \frac{v}{4L'} = \frac{v}{4(L+e)}\). Let the pipe resonate with a tuning fork of frequency \(n\) at two successive lengths, \(L_1\) (for the fundamental mode) and \(L_2\) (for the third harmonic). For the first resonance: \(v = 4n(L_1 + e)\).
For the second resonance (3rd harmonic): The corrected length is effectively \(\frac{3\lambda}{4}\), so \(v = \frac{4n(L_2+e)}{3}\) is not the way. Instead, we use the fact that the distance between two consecutive nodes is \(\lambda/2\).
Let the first resonating length be \(L_1\) and the second be \(L_2\).
Then \(L_1 + e = \frac{\lambda}{4}\) and \(L_2 + e = \frac{3\lambda}{4}\).
Subtracting the first equation from the second: \[ (L_2 + e) - (L_1 + e) = \frac{3\lambda}{4} - \frac{\lambda}{4} \] \[ L_2 - L_1 = \frac{2\lambda}{4} = \frac{\lambda}{2} \implies \lambda = 2(L_2 - L_1) \] Substitute \(\lambda\) back into the first equation: \[ L_1 + e = \frac{2(L_2 - L_1)}{4} = \frac{L_2 - L_1}{2} \] \[ e = \frac{L_2 - L_1}{2} - L_1 = \frac{L_2 - 3L_1}{2} \] A simpler experimental method uses two different frequencies \(n_1, n_2\) and one length L. Let's re-derive for two lengths.
From \(v = 4n_1(L_1+e)\) and \(v=4n_2(L_2+e)\): \[ n_1(L_1+e) = n_2(L_2+e) \implies n_1L_1 + n_1e = n_2L_2 + n_2e \] \[ e(n_1-n_2) = n_2L_2 - n_1L_1 \implies e = \frac{n_2L_2 - n_1L_1}{n_1 - n_2} \] The second derivation is correct for closed pipes as well.