Question

Prove that the fundamental frequancy of an open pipe is 2 times the fundamental frequancy of a closed pipe of the same length

Answer 1

For an open pipe,

$\displaystyle\longrightarrow\sf{L=\dfrac{\lambda}{2}}$

$\displaystyle\longrightarrow\sf{\lambda=2L}$

Then its fundamental frequency is,

$\displaystyle\longrightarrow\sf{\nu_{o}=\dfrac{v}{\lambda}}$

$\displaystyle\longrightarrow\sf{\nu_o=\dfrac{v}{2L}\quad\quad\dots(1)}$

where $\sf{v}$ is the velocity of sound.

For a closed pipe,

$\displaystyle\longrightarrow\sf{L=\dfrac{\lambda}{4}}$

$\displaystyle\longrightarrow\sf{\lambda=4L}$

Then its fundamental frequency is,

$\displaystyle\longrightarrow\sf{\nu_{c}=\dfrac{v}{\lambda}}$

$\displaystyle\longrightarrow\sf{\nu_c=\dfrac{v}{4L}\quad\quad\dots(2)}$

From (1) and (2) we see that, for the constant velocity of sound,

$\displaystyle\longrightarrow\sf{\underline{\underline{\nu_o=2\nu_c}}}$

$\displaystyle\longrightarrow\sf{\dfrac{v}{2L}=2\cdot\dfrac{v}{4L}}$

$\displaystyle\longrightarrow\sf{\dfrac{v}{2L}=\dfrac{v}{2L}}$

Hence the Proof!