Question

The length of the air column for fundamental mode in the resonance tube is 16 cm and that for the second resonance further is 50.25 cm. Find the end correction and inner diameter of the tube. Data:-
$l_1$ = $16cm$
$l_2$ = $50.25cm$
$e$ = ? and $d$ =?

Answer 1

Hint: To calculate the exact resonance frequency pipe, the end correction is applied to actual length of pipe. Air columns behave as closed organ pipes.

Complete step by step solution:
Let end correction be e
Inner diameter be d
$\eqalign{
  & {\text{We know}} \cr
  & {\text{for length of air column , }}l = 3{l_1} + 2e{\text{ }} \cr
  & \Rightarrow {\text{e = }}\dfrac{{{l_1} - {3l_2}}}{2} \cr
  & \Rightarrow \dfrac{{50.25 - (3{\text{ }} \times {\text{ 16)}}}}{2}{\text{ = }}\dfrac{{50.25 - 48}}{2} \cr
  & \Rightarrow e{\text{ = }}\dfrac{{2.25}}{2} \cr
  & \Rightarrow e{\text{ = 1}}{\text{.125 cm}} \cr
  & {\text{Also we know e = 0}}{\text{.3d}} \cr
  & \Rightarrow d = \dfrac{e}{{0.3}} \cr
  & \Rightarrow d = \dfrac{{1.125}}{{0.3}} \cr
  & \therefore d = 3.75cm \cr} $

Note: A simple assumption is that the basic resonance of a pipe occurs when the resonant length is half or one-fourth of the sound wavelength. Although it is well assumed that the practical frequency is less than this, you must apply a final correction, the pipe appears to be somewhat acoustic compared to its physical length. When the end correction is included, an exact value of the resonance frequency of the pipe can be estimated.