7. A vibrating string a certain length under a tension T resonates with a mode corresponding to the first overtone (third harmonic) of an air column of length 75 cm 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 increasing the number of beats reduces to 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(A) 344(B) 336(C) 117.3(D) 109.3

To solve this problem, we need to analyze the relationship between the vibrating string and the air column in the tube, as well as the concept of beats. Let's break it down step by step.

Understanding the Vibrating String and Air Column

The string is vibrating in a mode that corresponds to the first overtone, which is the third harmonic. For a string fixed at both ends, the frequency of the nth harmonic can be calculated using the formula:

f_n = n * (v / 2L)

Where:

• f_n = frequency of the nth harmonic
• v = velocity of the wave on the string
• L = length of the string

In this case, since we are dealing with the third harmonic (n = 3), we can express the frequency of the string as:

f_string = 3 * (v / 2L)

Frequency of the Air Column

Next, we need to find the frequency of the air column in the tube closed at one end. The fundamental frequency (first harmonic) for a tube closed at one end is given by:

f_air = v_sound / 4L

Where:

• v_sound = speed of sound in air (340 m/s)
• L = length of the air column (0.75 m)

Substituting the values:

f_air = 340 / (4 * 0.75) = 340 / 3 = 113.33 Hz

Finding the Frequency of the Tuning Fork

Initially, the string generates 4 beats per second when excited with a tuning fork of frequency n. The beat frequency is given by the absolute difference between the two frequencies:

Beat Frequency = |f_string - n|

Thus, we can write:

|f_string - n| = 4

Now, we also know that when the tension of the string is slightly increased, the beat frequency reduces to 2 beats per second:

|f_string' - n| = 2

Here, f_string' is the new frequency of the string after the tension is increased. As the tension increases, the frequency of the string also increases. Therefore, we can express this as:

f_string' = f_string + Δf

Setting Up the Equations

From the first condition, we have:

f_string - n = 4 or n - f_string = 4

f_string' - n = 2 or n - f_string' = 2

Substituting f_string' into the second equation gives us:

n - (f_string + Δf) = 2

Solving the Equations

Now we can solve these equations. Let's take the first case where:

n = f_string + 4

Substituting this into the second equation:

(f_string + 4) - (f_string + Δf) = 2

This simplifies to:

4 - Δf = 2

Thus, we find:

Δf = 2

Now, substituting Δf back into the first equation gives us:

n = f_string + 4 = (113.33 + 4) = 117.33 Hz

Final Calculation

Since we are looking for the frequency of the tuning fork n, we can round it to:

n ≈ 117.3 Hz

Thus, the correct answer is (C) 117.3 Hz.