Answer to Question #103197 in Physics for kiiki

Question

Please give intutive expalnation for harmonucs in closed organ pipe.If there is no phase change at closed end then why standing waves are set up?perhaps there is phase change at open end ,but I cant imagine it pretty well.so please explain with diagram

Answer 1

A closed organ pipe is a cylindrical tube having an air column with one end is closed. Sound waves are sent into the tube by a vibrating source. An in going pressure get reflected from the fixed end. This inverted wave gets reflected at the open end, after the two reflection it moves towards the fixed end and interfere with the new waves.

Let L is the length of the tube,

As per the given diagram, in the fundamental mode, there is a node at the closed end and anti-node at the open end.

From the above we can see that , $L=\dfrac{\lambda_o}{4}$

Fundamental wavelength of the closed tube$(\lambda_o)=\dfrac{L}{4}$

If $f_o$ is the fundamental frequency, and velocity of the wave($v)$

So, frequency of the wave $f_o=\dfrac{v}{\lambda_o}=\dfrac{4v}{L}$

After the reflection of the then change in phase=$\dfrac{2\pi}{\lambda}\times 2L=\dfrac{4\pi L}{L}$

The twice reflected wave will suffer a phase of $\pi$

Then the phase difference between the two wave =$\dfrac{4\pi L}{L}+\pi$

The waves interfere constructively, if $\dfrac{4\pi L}{L}+\pi=2n\pi$

here $n\epsilon N$ (natural number)

So, $L=\dfrac{(2n+1)\lambda}{4}$

frequency $f=\dfrac{v}{\lambda}=\dfrac{(2n+1)v}{4L}$