A vibrating string of certain length `l` under a tension `T` resonates with a mode corresponding to the first overtone (third harmonic) of an air column of length `75cm` inside a tube closed at one end. The string also generates `4` beats per second when excited along with a tuning fork of frequency `n`. Now when the tension of the string is slightly increased the number of beats reduces `2` per second. Assuming the velocity of sound in air to be `340 m//s`, the frequency `n` of the tuning fork in `Hz` is

Options

A. (a)344
B. (b)336
C. (c)117.3
D. (d)109.3

Explanation

To solve the problem step by step, we will follow the given information and apply the relevant formulas.

### Step 1: Determine the frequency of the air column
The air column is closed at one end and has a length of \( l = 75 \, \text{cm} = 0.75 \, \text{m} \). The fundamental frequency (first harmonic) for a closed pipe is given by:

\[ f_1 = \frac{v}{4l} \]

where \( v \) is the speed of sound in air. For the first overtone (third harmonic), the frequency can be expressed as:

\[ f_3 = 3 \cdot f_1 = 3 \cdot \frac{v}{4l} \]

Substituting the values:

\[ v = 340 \, \text{m/s}, \quad l = 0.75 \, \text{m} \]

Calculating \( f_3 \):

\[ f_3 = 3 \cdot \frac{340}{4 \cdot 0.75} = 3 \cdot \frac{340}{3} = 340 \, \text{Hz} \]

### Step 2: Determine the frequency of the string
Let \( f \) be the frequency of the vibrating string. The string produces beats with a tuning fork of frequency \( n \), and the number of beats is given as 4 beats per second. This means:

\[ |f - n| = 4 \]

This gives us two equations:

1. \( f = n + 4 \)
2. \( f = n - 4 \)

### Step 3: Increase in tension and reduction of beats
When the tension of the string is slightly increased, the frequency of the string increases. The number of beats reduces to 2 beats per second. Therefore:

\[ |f' - n| = 2 \]

where \( f' \) is the new frequency of the string after increasing the tension. Since increasing tension increases frequency, we can say:

\[ f' = f + \Delta f \]

where \( \Delta f \) is a small increase in frequency. Thus, we have:

1. \( f' = n + 2 \)
2. \( f' = n - 2 \)

### Step 4: Relating the frequencies
From the equations we have:

1. \( f = n + 4 \) leads to \( f' = n + 4 + \Delta f \)
2. \( f' = n + 2 \)

Setting these equal gives:

\[ n + 4 + \Delta f = n + 2 \]

This simplifies to:

\[ \Delta f = -2 \]

This indicates that the frequency of the string decreased by 2 Hz, which contradicts our assumption that increasing tension increases frequency. Therefore, we should consider the other case:

From \( f = n - 4 \):

1. \( f' = n - 4 + \Delta f \)
2. \( f' = n - 2 \)

Setting these equal gives:

\[ n - 4 + \Delta f = n - 2 \]

This simplifies to:

\[ \Delta f = 2 \]

### Step 5: Solving for \( n \)
Now we have:

1. \( f = n - 4 \)
2. \( f' = n - 2 \)

Since we know \( f = 340 \, \text{Hz} \):

\[ 340 = n - 4 \]

Solving for \( n \):

\[ n = 340 + 4 = 344 \, \text{Hz} \]

### Final Answer
The frequency \( n \) of the tuning fork is:

\[ \boxed{344 \, \text{Hz}} \]