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//^(2))`
B. `(vl^(2))/(2x)`
C. `(vl)/(2//^(2))`
D. `(vl^(2))/(2l)`

Answer

To solve the problem of finding 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: Understand the fundamental frequency of an open organ pipe
For an open organ pipe, the fundamental frequency \( f \) is given by the formula:
\[ f = \frac{v}{\lambda} \]
where \( v \) is the speed of sound in air and \( \lambda \) is the wavelength.

### Step 2: Determine the wavelength for the fundamental mode
For the fundamental mode of an open organ pipe, the relationship between the length \( L \) of the pipe and the wavelength \( \lambda \) is:
\[ \lambda = 2L \]
Thus, for the first pipe of length \( l \):
\[ \lambda_1 = 2l \]
And for the second pipe of length \( l + x \):
\[ \lambda_2 = 2(l + x) \]

### Step 3: Calculate the fundamental frequencies of both pipes
Using the wavelength, we can find the fundamental frequencies:
\[ f_1 = \frac{v}{\lambda_1} = \frac{v}{2l} \]
\[ f_2 = \frac{v}{\lambda_2} = \frac{v}{2(l + x)} \]

### Step 4: Find the beat frequency
The beat frequency \( f_b \) is given by the absolute difference of the two frequencies:
\[ f_b = |f_1 - f_2| = \left| \frac{v}{2l} - \frac{v}{2(l + x)} \right| \]

### Step 5: Simplify the expression for beat frequency
Factoring out \( \frac{v}{2} \):
\[ f_b = \frac{v}{2} \left| \frac{1}{l} - \frac{1}{l + x} \right| \]
Now, simplifying the expression inside the absolute value:
\[ \frac{1}{l} - \frac{1}{l + x} = \frac{(l + x) - l}{l(l + x)} = \frac{x}{l(l + x)} \]
Thus, the beat frequency becomes:
\[ f_b = \frac{v}{2} \cdot \frac{x}{l(l + x)} = \frac{vx}{2l(l + x)} \]

### Step 6: Approximate for small \( x \)
Since \( x \) is very small compared to \( l \) (i.e., \( x \ll l \)), we can approximate \( l + x \approx l \):
\[ f_b \approx \frac{vx}{2l^2} \]

### Final Result
Thus, the beat frequency heard when the two pipes are sounded together is:
\[ f_b \approx \frac{vx}{4l^2} \]