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

Explanation

To solve the problem, we need to find the length of the shorter pipe given that the longer pipe has a length of 1 meter and they produce 4 beats per second when sounded together. We will use the formula for the frequency of a closed organ pipe and the concept of beat frequency.

### Step-by-Step Solution:

1. **Understand the Frequency Formula**:
For a closed organ pipe, the frequency (f) is given by the formula:
\[ f = \frac{v}{4L} \]
where \( v \) is the speed of sound in air (300 m/s) and \( L \) is the length of the pipe.

2. **Calculate the Frequency of the Longer Pipe**:
The length of the longer pipe \( L_1 \) is given as 1 m. Therefore, the frequency \( f_1 \) of the longer pipe is:
\[ f_1 = \frac{300}{4 \times 1} = \frac{300}{4} = 75 \text{ Hz} \]

3. **Set Up the Beat Frequency Equation**:
The beat frequency is the difference in frequencies of the two pipes. Given that the beat frequency is 4 beats per second, we can express this as:
\[ |f_2 - f_1| = 4 \]
where \( f_2 \) is the frequency of the shorter pipe.

4. **Express the Frequency of the Shorter Pipe**:
The frequency \( f_2 \) of the shorter pipe with length \( L_2 \) can be expressed as:
\[ f_2 = \frac{300}{4L_2} \]

5. **Substitute into the Beat Frequency Equation**:
We can substitute \( f_2 \) into the beat frequency equation:
\[ \left| \frac{300}{4L_2} - 75 \right| = 4 \]

6. **Solve the Equation**:
This gives us two cases to consider:
- Case 1:
\[ \frac{300}{4L_2} - 75 = 4 \]
Simplifying this:
\[ \frac{300}{4L_2} = 79 \implies 300 = 316L_2 \implies L_2 = \frac{300}{316} \approx 0.949 \text{ m} \]
- Case 2:
\[ 75 - \frac{300}{4L_2} = 4 \]
Simplifying this:
\[ \frac{300}{4L_2} = 71 \implies 300 = 284L_2 \implies L_2 = \frac{300}{284} \approx 1.056 \text{ m} \]
Since \( L_2 \) must be less than \( L_1 \), we discard Case 2.

7. **Final Result**:
The length of the shorter pipe \( L_2 \) is approximately:
\[ L_2 \approx 0.949 \text{ m} \text{ or } 94.9 \text{ cm} \]