Question

An open organ pipe is closed suddenly with the result, that the second overtone of the closed pipe is found to be higher in frequency by $100$ than the first overtone of the original pipe. Then, the fundamental frequency of the open pipe is

• A. $200s^{-1}$
• B. $100s^{-1}$
• C. $250s^{-1}$
• D. $300s^{-1}$
• E. $150s^{-1}$

Answer 1

The correct option is A

$200s^{-1}$

Step 1. Given data

The second overtone of the closed pipe is found to be higher in frequency by $100$ than the first overtone of the original pipe.

Step 2. Finding the fundamental frequency of the open pipe, $f_1$

Let $l$ be the length of pipe.

Frequency of first overtone (second harmonic) of the open pipe,$f_2=\frac{2v}{2l}=\frac{v}{l}$ [$v$ is velocity]

Frequency of second overtone (fifth harmonic) of closed pipe,${f^{\text{'}}}_3=\frac{5v}{4l}$

Subtracting both we get,

$$\begin{gathered} {f^{\text{'}}}_3–f_2=(5v/4l)–(2v/2l) \\ 100=\frac{v}{4l} \\ l=\frac{v}{400} \end{gathered}$$………..(Given that ${f^{\text{'}}}_3–f_2=100$)

Thus, fundamental frequency of the open pipe will be

$$\begin{gathered} f_1=\frac{v}{2l} \\ =\frac{v}{2\left(v/400\right)} \\ =200s^{-1} \end{gathered}$$

Hence, Option $\left(A\right)$ is correct.