Question

The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If the length of the open pipe is $60\text{cm}$, what is the length of the closed pipe?

• A. $30\text{cm}$
• B. $15\text{cm}$
• C. $120\text{cm}$
• D. $240\text{cm}$

Answer 1

The correct option is B $15\text{cm}$

Given that, the length of the open pipe, $l_o=60\text{cm}$

Let the length of the closed pipe be $l_c$.

Now, fundamental frequency of closed organ pipe is

$f_1=\frac{V}{4l_c}$

The fundamental frequency of open organ pipe is $\frac{V}{2l_o}$ and first over tone is

$f_2=\frac{2V}{2l_o}=\frac{V}{l_o}$

From the question : $f_1=f_2$

$\Rightarrow \frac{V}{4l_c}=\frac{V}{l_o}$

$\Rightarrow l_c=\frac{l_o}{4}=\frac{60}{4}=15\text{cm}$

Hence, option $\left(B\right)$ is correct.

Why this question ? Concept : Frequency of the open organ pipe is same as frequency of vibrating string fixed at both end.