Two closed organ pipes, when sounded simultaneously gave 4 beats per sec. If longer pipe has a length of 1 m. Then length of shorter pipe will be (v=300 m//s`

Options

A. `185.5cm`
B. `94.9 cm`
C. `90 cm`
D. `80 cm`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between frequency and length of the pipes
The frequency \( f \) of a closed organ pipe is given by the formula:
\[ f = \frac{v}{4L} \]
where \( v \) is the speed of sound in air and \( L \) is the length of the pipe.

### Step 2: Define the frequencies of the two pipes
Let:
- \( L_1 = 1 \, \text{m} \) (length of the longer pipe)
- \( L_2 \) = length of the shorter pipe
- \( f_1 \) = frequency of the longer pipe
- \( f_2 \) = frequency of the shorter pipe

Using the formula for frequency, we have:
\[ f_1 = \frac{v}{4L_1} = \frac{300}{4 \times 1} = \frac{300}{4} = 75 \, \text{Hz} \]
\[ f_2 = \frac{v}{4L_2} = \frac{300}{4L_2} \]

### Step 3: Set up the equation for beat frequency
The beat frequency is given as 4 beats per second, which means:
\[ |f_2 - f_1| = 4 \]
Since \( f_2 > f_1 \), we can write:
\[ f_2 - f_1 = 4 \]
Substituting the expression for \( f_2 \):
\[ \frac{300}{4L_2} - 75 = 4 \]

### Step 4: Solve for \( L_2 \)
Rearranging the equation:
\[ \frac{300}{4L_2} = 75 + 4 \]
\[ \frac{300}{4L_2} = 79 \]
Now, multiply both sides by \( 4L_2 \):
\[ 300 = 79 \times 4L_2 \]
\[ 300 = 316L_2 \]
Now, solve for \( L_2 \):
\[ L_2 = \frac{300}{316} \approx 0.9494 \, \text{m} \]

### Step 5: Convert to centimeters
To express \( L_2 \) in centimeters:
\[ L_2 \approx 0.9494 \, \text{m} = 94.94 \, \text{cm} \]

### Final Answer
The length of the shorter pipe \( L_2 \) is approximately \( 0.9494 \, \text{m} \) or \( 94.94 \, \text{cm} \).

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