Question

The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is 60 cm, the length of the closed pipe will be :

• A. 60 cm
• B. 45 cm
• C. 30 cm
• D. 15 cm

Answer 1

Answer: (D)

15 cm

Answer 2

The relationship for the fundamental frequency of a closed organ pipe (length $L_1$) and the first overtone of an open organ pipe (length $L_2$) is given by:

$\frac{\lambda}{4} = L_1 \quad \text{and} \quad 2 \left( \frac{\lambda}{2} \right) = \lambda$

The velocity of sound is $v$, thus:

$v = f \lambda$

For the closed pipe:

$v = f_1 (4L_1)$

For the open pipe:

$f_2 = \frac{v}{2L_2}$

Equating the fundamental frequency of the closed pipe to the first overtone of the open pipe:

$f_1 = f_2$ $\frac{v}{4L_1} = \frac{v}{2L_2}$ $L_2 = 4L_1$

Given $L_2 = 60 \, \text{cm}$:

$60 = 4 \times L_1$ $L_1 = 15 \, \text{cm}$