Step-by-Step Solution

1. Calculate the frequency of the air column

The problem states that a vibrating string resonates with a mode of an air column in a tube closed at one end. The length of the air column is $L = 75 \text{ cm} = 0.75 \text{ m}$. The velocity of sound in air is given as $v_{air} = 340 \text{ m/s}$.

The mode of vibration is the first overtone, which for a closed pipe corresponds to the third harmonic. The frequency of the $k^{th}$ harmonic for a closed pipe is given by the formula: $f_k = \frac{(2k-1)v_{air}}{4L}$ For the first overtone, $k=2$, so we are looking for the frequency of the third harmonic ($2k-1=3$): $f_3 = \frac{3 \times v_{air}}{4L}$ Substituting the given values: $f_3 = \frac{3 \times 340 \text{ m/s}}{4 \times 0.75 \text{ m}} = \frac{1020}{3} = 340 \text{ Hz}$ Since the string resonates with this mode, the initial frequency of the string, $f_s$, is equal to the frequency of the air column. $f_s = 340 \text{ Hz}$

2. Determine the possible frequencies of the tuning fork

The string, with frequency $f_s = 340 \text{ Hz}$, produces 4 beats per second with a tuning fork of frequency $n$. The beat frequency is given by: $f_{beat} = |f_s - n|$ Substituting the values: $4 = |340 - n|$ This gives two possible values for the tuning fork frequency $n$:

Case 1: $n = 340 + 4 = 344 \text{ Hz}$ Case 2: $n = 340 - 4 = 336 \text{ Hz}$

3. Analyze the effect of increasing the string's tension

The frequency of a vibrating string is given by $f = \frac{p}{2l}\sqrt{\frac{T}{\mu}}$, where $T$ is the tension and $\mu$ is the linear mass density. This shows that the frequency is directly proportional to the square root of the tension: $f_s \propto \sqrt{T}$ When the tension $T$ is slightly increased, the frequency of the string $f_s$ also increases. Let the new frequency be $f_s'$. Therefore, we have: $f_s' > f_s = 340 \text{ Hz}$

4. Use the new beat frequency to identify the correct tuning fork frequency

The problem states that after increasing the tension, the new number of beats per second is 2. So, the new beat frequency is: $f'_{beat} = |f_s' - n| = 2$ We now test the two possible values of $n$:

• Case 1: $n = 344 \text{ Hz}$ In this case, the initial frequency of the string ($f_s = 340 \text{ Hz}$) is lower than the tuning fork's frequency. When the tension is increased, $f_s$ increases, moving closer to $n=344 \text{ Hz}$. As the two frequencies get closer, the beat frequency should decrease. The problem states that the beat frequency decreases from 4 Hz to 2 Hz. This is consistent with our analysis. The new string frequency would be $f_s' = 344 - 2 = 342 \text{ Hz}$, which is greater than $340 \text{ Hz}$.

• Case 2: $n = 336 \text{ Hz}$ In this case, the initial frequency of the string ($f_s = 340 \text{ Hz}$) is higher than the tuning fork's frequency. When the tension is increased, $f_s$ increases further, moving away from $n=336 \text{ Hz}$. As the two frequencies move further apart, the beat frequency should increase. However, the problem states that the beat frequency decreases. This contradicts our analysis.

Therefore, Case 1 is the only one consistent with the observations. The frequency of the tuning fork must be $n = 344 \text{ Hz}$.