Frequency of Tuning Fork Resonating with Closed Air Column

To find the frequency $n$ of the tuning fork, we proceed step-by-step:

1. Frequency of the Air Column:
   For a tube closed at one end, the fundamental frequency is given by $f_1 = \frac{v}{4L}$, where $v$ is the speed of sound in air and $L$ is the length of the air column.
   The first overtone corresponds to the third harmonic ($m = 3$):
   $f_{\text{pipe}} = \frac{3v}{4L}$
   Given $v = 340 \text{ m/s}$ and $L = 75 \text{ cm} = 0.75 \text{ m}$:
   $f_{\text{pipe}} = \frac{3 \times 340}{4 \times 0.75} = \frac{1020}{3.00} = 340 \text{ Hz}$

2. Frequency of the Vibrating String:
   Since the string resonates with the first overtone of the closed air column, the frequency of vibration of the string is:
   $f_{\text{string}} = 340 \text{ Hz}$

3. Determining the Frequency of the Tuning Fork ($n$):
   The string produces $4\text{ beats per second}$ with a tuning fork of frequency $n$:
   $|f_{\text{string}} - n| = 4$
   This gives two possible values for $n$:
   $n = 340 + 4 = 344 \text{ Hz} \quad \text{or} \quad n = 340 - 4 = 336 \text{ Hz}$

4. Effect of Increasing Tension:
   The frequency of a vibrating string is directly proportional to the square root of tension ($f \propto \sqrt{T}$). When the tension $T$ is slightly increased, the frequency of the string increases ($f_{\text{string}}' > 340 \text{ Hz}$).

   The problem states that the number of beats decreases from $4$ to $2$ per second:

   • If $n = 344 \text{ Hz}$, an increase in $f_{\text{string}}$ (e.g., to $342 \text{ Hz}$) brings it closer to $n$, reducing the beat frequency to $|342 - 344| = 2 \text{ Hz}$.
   • If $n = 336 \text{ Hz}$, an increase in $f_{\text{string}}$ would move it further away from $n$, increasing the beat frequency (e.g., to $|342 - 336| = 6 \text{ Hz}$).

Thus, the frequency of the tuning fork must be $n = 344 \text{ Hz}$.

Correct Option: A