Question

The closed and open organ pipe have same length and when they are vibrating simultaneously in first overtone produce \(3\) beats. The length of open pipe is made \((\frac{1}{3})^{rd}\) and closed pipe is made \(3\) times the original, the number of beats produced will be {neglect end correction}

• A. \(8\)
• B. \(10\)
• C. \(17\)
• D. \(33\)

Hint:

Write the first overtone of the closed pipe as \(3v/4L\) and of the open pipe as \(v/L\), use the 3 beats to find \(v/L\), then rescale each frequency for its new length.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A closed pipe has a node at the closed end and an antinode at the open end. An open pipe has antinodes at both ends. Beats are the difference in frequency of two notes sounding together. Let \(v\) be the speed of sound and \(L\) the original length.

Step 2: Key Formula or Approach:
1. First overtone of a closed pipe is the third harmonic: \(f_c = \frac{3v}{4L}\).
2. First overtone of an open pipe is the second harmonic: \(f_o = \frac{2v}{2L} = \frac{v}{L}\).

Step 3: Use the first condition.
Both pipes have length \(L\). The open pipe note is higher, since \(\frac{v}{L} > \frac{3v}{4L}\). The beats are
\[ f_o - f_c = \frac{v}{L} - \frac{3v}{4L} = \frac{v}{4L} = 3 \]
So \(\frac{v}{4L} = 3\), which gives \(\frac{v}{L} = 12\) Hz.

Step 4: Change the lengths.
The open pipe is now \(\frac{L}{3}\) long, so its first overtone is
\[ f_o' = \frac{v}{L/3} = \frac{3v}{L} = 36\ \text{Hz} \]
The closed pipe is now \(3L\) long, so its first overtone is
\[ f_c' = \frac{3v}{4(3L)} = \frac{v}{4L} = 3\ \text{Hz} \]

Step 5: Find the new beats.
Beats \(= f_o' - f_c' = 36 - 3 = 33\).

Final Answer:
The number of beats becomes 33, which is option (D). \[ \boxed{33} \]