A pipe closed at one end resonates with sound waves of frequency 77 Hz and with 99 Hz. What will be frequency of fundamental note if the pipe does not resonate with any intermediate frequency between the two ?

Options

A. 11 Hz
B. 22 Hz
C. 10 Hz
D. 12 Hz

Explanation

To find the fundamental frequency of a pipe closed at one end that resonates at 77 Hz and 99 Hz without any intermediate frequencies, we can follow these steps:

### Step 1: Understand the Harmonics in a Closed Pipe
In a pipe closed at one end, the frequencies of the harmonics are given by the formula:
\[ f_n = \frac{(2n - 1) V}{4L} \]
where \( n \) is the harmonic number (1, 2, 3,...), \( V \) is the speed of sound, and \( L \) is the length of the pipe. Only odd harmonics are present.

### Step 2: Identify the Frequencies
Given that the pipe resonates at 77 Hz and 99 Hz, we can denote these frequencies as:
- \( f_1 = 77 \, \text{Hz} \)
- \( f_2 = 99 \, \text{Hz} \)

### Step 3: Establish the Relationship Between Frequencies
Since these two frequencies are consecutive harmonics, we can express them in terms of the harmonic number \( n \):
- For \( f_1 \): \( f_1 = \frac{(2n - 1) V}{4L} \)
- For \( f_2 \): \( f_2 = \frac{(2(n + 1) - 1) V}{4L} = \frac{(2n + 1) V}{4L} \)

### Step 4: Set Up the Equations
From the equations above, we have:
1. \( f_1 = \frac{(2n - 1) V}{4L} = 77 \)
2. \( f_2 = \frac{(2n + 1) V}{4L} = 99 \)

### Step 5: Divide the Two Equations
Dividing the two equations gives:
\[ \frac{f_1}{f_2} = \frac{(2n - 1)}{(2n + 1)} \]
Substituting the values:
\[ \frac{77}{99} = \frac{(2n - 1)}{(2n + 1)} \]

### Step 6: Cross Multiply and Solve for \( n \)
Cross multiplying gives:
\[ 77(2n + 1) = 99(2n - 1) \]
Expanding both sides:
\[ 154n + 77 = 198n - 99 \]
Rearranging gives:
\[ 198n - 154n = 77 + 99 \]
\[ 44n = 176 \]
\[ n = 4 \]

### Step 7: Find the Fundamental Frequency
Now, we can find the fundamental frequency \( f_0 \):
Using \( f_1 = \frac{(2n - 1) V}{4L} \):
\[ f_0 = \frac{(2n - 1) V}{4L} \text{ for } n = 4 \]
\[ f_0 = \frac{(2 \times 4 - 1) V}{4L} = \frac{(8 - 1) V}{4L} = \frac{7V}{4L} \]
Substituting \( f_1 = 77 \):
\[ 77 = \frac{7V}{4L} \]
Thus, the fundamental frequency \( f_0 \) is:
\[ f_0 = \frac{77}{7} = 11 \, \text{Hz} \]

### Final Answer
The fundamental frequency of the pipe is **11 Hz**.

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