Question

the frequency of fundamental note of a closed organ pipe and that of an open organ pipe are the same. what is the ratio of their lengths?

Answer 1

Answer:

$\boxed{ \sf{ \red{Closed \: Organ \: Pipe }}}$

Nodes will be formed at both the closed ends.

Let length of tube be l1

$\therefore \: \dfrac{ \lambda}{2} = l_{1}$

$\implies \: \lambda = 2 l_{1}$

So, frequency will be :

$f = \dfrac{velocity}{wavelength}$

$\boxed{\implies \: f = \dfrac{v}{2 l_{1} } \: ......(1)}$

$\boxed{ \orange{ \sf{Open \: Organ \: Pipe}}}$

Anti-node will be formed at open end and node will be formed at closed end.

Let length I tube be l2

$\therefore \: \dfrac{ \lambda}{4} = l_{2}$

$\implies \: \lambda = 4 l_{2}$

So, Frequency is :

$f = \dfrac{velocity}{wavelength}$

$\boxed{\implies \: f = \dfrac{v}{4 l_{2} } \: ......(2)}$

Now, as per the question , the frequencies are same :

$\therefore \: \dfrac{v}{2l_{1}} = \dfrac{v}{4l_{2} }$

$\implies \: l_{1} : l_{2} = 2 : 1$

So final answer is :

$\boxed{ \huge{ \blue{ \sf{ \bold{ \: l_{1} : l_{2} = 2 : 1}}}}}$