An open organ pipe of length `l` is sounded together with another open organ pipe of length ` l + x` in their fundamental tones. Speed of sound in air is `v` . The beat frequency heard will be `(x lt lt l)`:

Options

A. `( vx)/( 4 l^(2))`
B. `(vl^(2))/( 2x)`
C. `( v x)/( 2 l^(2))`
D. `( v x^(2))/( 2l)`

Explanation

To find the beat frequency heard when two open organ pipes of lengths \( l \) and \( l + x \) are sounded together, we can follow these steps:

### Step 1: Determine the Fundamental Frequencies
The fundamental frequency \( f \) of an open organ pipe is given by the formula:
\[ f = \frac{v}{2L} \]
where \( v \) is the speed of sound in air and \( L \) is the length of the pipe.

For the first pipe of length \( l \):
\[ f_1 = \frac{v}{2l} \]

For the second pipe of length \( l + x \):
\[ f_2 = \frac{v}{2(l + x)} \]

### Step 2: Calculate the Beat Frequency
The beat frequency \( f_b \) is the absolute difference between the two frequencies:
\[ f_b = |f_1 - f_2| = \left| \frac{v}{2l} - \frac{v}{2(l + x)} \right| \]

### Step 3: Simplify the Expression
To simplify the expression, we can factor out \( \frac{v}{2} \):
\[ f_b = \frac{v}{2} \left| \frac{1}{l} - \frac{1}{l + x} \right| \]

Now, we can find a common denominator:
\[ \frac{1}{l} - \frac{1}{l + x} = \frac{(l + x) - l}{l(l + x)} = \frac{x}{l(l + x)} \]

Thus, we have:
\[ f_b = \frac{v}{2} \cdot \frac{x}{l(l + x)} \]

### Step 4: Approximate for Small \( x \) Compared to \( l \)
Since we are given that \( x \ll l \), we can approximate \( l + x \) as \( l \) in the denominator:
\[ f_b \approx \frac{v}{2} \cdot \frac{x}{l^2} \]

### Final Result
Therefore, the beat frequency heard will be:
\[ f_b \approx \frac{vx}{2l^2} \]

### Summary
The beat frequency heard when two open organ pipes of lengths \( l \) and \( l + x \) are sounded together is approximately:
\[ f_b \approx \frac{vx}{2l^2} \]