A tuning fork vibrats with frequnecy `256 Hz` and gives one beat per seond with the third normal mode of vibration of an open pipe. What is the length of the pipe (Speed of sound in air is `340 ms^(-1)`)

Options

A. 190 cm
B. 200 cm
C. 220 cm
D. 180 cm

Answer

To solve the problem, we need to find the length of an open pipe that produces a specific frequency of sound. Here are the steps to arrive at the solution:

### Step 1: Understand the Frequency of the Tuning Fork and the Beats
The tuning fork vibrates at a frequency of \( f_t = 256 \, \text{Hz} \). It produces 1 beat per second with the frequency of the third normal mode of vibration of the open pipe. This means the frequency of the pipe can either be \( f_p = 256 + 1 = 257 \, \text{Hz} \) or \( f_p = 256 - 1 = 255 \, \text{Hz} \).

### Step 2: Identify the Frequency of the Third Normal Mode of Vibration
For an open pipe, the frequency of the \( n \)-th normal mode of vibration is given by the formula:
\[ f_n = \frac{nV}{2L} \]
where:
- \( n \) is the mode number (in this case, \( n = 3 \)),
- \( V \) is the speed of sound in air (\( V = 340 \, \text{m/s} \)),
- \( L \) is the length of the pipe.

### Step 3: Set Up the Equation for the Third Normal Mode
For the third normal mode (\( n = 3 \)):
\[ f_3 = \frac{3V}{2L} \]
We can set this equal to the frequency of the pipe. Let's use \( f_p = 255 \, \text{Hz} \):
\[ 255 = \frac{3 \times 340}{2L} \]

### Step 4: Solve for the Length \( L \)
Rearranging the equation to solve for \( L \):
\[ 255 \cdot 2L = 3 \cdot 340 \]
\[ 510L = 1020 \]
\[ L = \frac{1020}{510} = 2 \, \text{m} \]

### Step 5: Convert Length to Centimeters
Since the problem asks for the length in centimeters:
\[ L = 2 \, \text{m} = 200 \, \text{cm} \]

### Final Answer
The length of the open pipe is \( \boxed{200 \, \text{cm}} \).

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