The first overtone frequency of an open organ pipe is equal to the fundamental frequency of a closed organ pipe. If the length of the closed organ pipe is 20 cm. The length of the open organ pipe is ________ cm.

CONCEPT:

When an organ pipe is opened from both ends, its frequency is written as;

f = \(\frac{nv}{2L}\)

and when it is closed then its frequency is written as;

f' = \(\frac{(2n-1)v}{4L}\)

CALCULATION:

Given: Length, L' = 20 cm 

Let us suppose P₁ is the open pipe and its frequency is written as;

f₁ = \(\frac{v}{2L}\)

and P₂ is the closed organ pipe and its frequency is written as;

f₂ = \(\frac{3v}{4L'}\)

It is given that the first overtone frequency of an open organ pipe is equal to the fundamental frequency of a closed organ pipe, therefore we have,

f₁ = f'₂

⇒ \(\frac{v}{2L}\) = \(\frac{3v}{4L'}\)

⇒ \(\frac{1}{L}\) = \(\frac{3}{4L'}\)

Now, on putting the value of length we have;

L = \(\frac{4L'}{3}\)

⇒ L = \(\frac{4}{3}\times 20\)

⇒ L = 26.66 cm

Hence, the length of an open organ pipe, L = 26.66 cm.