Question

Two open organ pipes of fundamental frequencies \( n_1 \) and \( n_2 \) are joined in series. The fundamental frequency of the new pipe is

• A. \( n_1 - n_2 \)
• B. \( \dfrac{n_1 n_2}{n_1 + n_2} \)
• C. \( \dfrac{1}{n_1 n_2} \)
• D. \( \dfrac{n_1 + n_2}{n_1 n_2} \)

Hint:

When organ pipes are joined in series, always add their lengths first and then calculate the new frequency.

Answer 1

The Correct Option is B

Step 1: Write the formula for fundamental frequency of an open pipe.
For an open organ pipe, \[ n = \frac{v}{2L} \] where \( v \) is the velocity of sound and \( L \) is the length of the pipe.
Step 2: Express lengths of individual pipes.
\[ L_1 = \frac{v}{2n_1}, \quad L_2 = \frac{v}{2n_2} \]
Step 3: Find the total length of the combined pipe.
\[ L = L_1 + L_2 = \frac{v}{2}\left(\frac{1}{n_1} + \frac{1}{n_2}\right) \]
Step 4: Calculate the new fundamental frequency.
\[ n = \frac{v}{2L} = \frac{v}{2 \cdot \frac{v}{2}\left(\frac{1}{n_1} + \frac{1}{n_2}\right)} \] \[ = \frac{1}{\frac{1}{n_1} + \frac{1}{n_2}} = \frac{n_1 n_2}{n_1 + n_2} \]
Step 5: Conclusion.
The fundamental frequency of the new pipe is \( \dfrac{n_1 n_2}{n_1 + n_2} \).