A string `25cm` long and having a mass of `2.5 gm` is under tension. A pipe closed at one end is `40cm` long. When the string is set vibrating in its first overtone and the air in the pipe in its fundamental frequency, `8` beats per second are heard. It is observed that decreasing the tension in the string decreases beat frequency. If the speed of sound in air is `320m//s`, find the tension in the string.

Explanation

To find the tension in the string, we will follow these steps:

### Step 1: Calculate the linear mass density (μ) of the string
The linear mass density (μ) is given by the formula:
\[ \mu = \frac{m}{L} \]
where \( m \) is the mass of the string and \( L \) is its length.

Given:
- Mass \( m = 2.5 \, \text{g} = 2.5 \times 10^{-3} \, \text{kg} \)
- Length \( L = 25 \, \text{cm} = 25 \times 10^{-2} \, \text{m} \)

Calculating μ:
\[ \mu = \frac{2.5 \times 10^{-3} \, \text{kg}}{25 \times 10^{-2} \, \text{m}} = 10^{-2} \, \text{kg/m} \]

### Step 2: Determine the wavelength of the first overtone of the string
The first overtone for a string fixed at both ends has a wavelength given by:
\[ \lambda_s = \frac{L}{n} \]
where \( n \) is the mode number. For the first overtone, \( n = 2 \).

Thus,
\[ \lambda_s = \frac{25 \times 10^{-2} \, \text{m}}{2} = 0.125 \, \text{m} \]

### Step 3: Calculate the frequency of the string (fs)
The frequency of the string can be calculated using the formula:
\[ f_s = \frac{1}{\lambda_s} \sqrt{\frac{T}{\mu}} \]

Substituting the values:
\[ f_s = \frac{1}{0.125} \sqrt{\frac{T}{10^{-2}}} \]
\[ f_s = 8 \sqrt{\frac{T}{10^{-2}}} \]

### Step 4: Determine the wavelength of the fundamental frequency of the pipe
For a pipe closed at one end, the fundamental frequency has a wavelength given by:
\[ \lambda_p = 4L \]
Given the length of the pipe \( L = 40 \, \text{cm} = 0.4 \, \text{m} \):
\[ \lambda_p = 4 \times 0.4 = 1.6 \, \text{m} \]

### Step 5: Calculate the frequency of the pipe (fp)
Using the speed of sound in air:
\[ v = 320 \, \text{m/s} \]
The frequency of the pipe can be calculated as:
\[ f_p = \frac{v}{\lambda_p} = \frac{320}{1.6} = 200 \, \text{Hz} \]

### Step 6: Set up the equation for beat frequency
The beat frequency is given by:
\[ f_s - f_p = 8 \, \text{Hz} \]
Substituting the expression for \( f_s \):
\[ 8 \sqrt{\frac{T}{10^{-2}}} - 200 = 8 \]
Rearranging gives:
\[ 8 \sqrt{\frac{T}{10^{-2}}} = 208 \]
Dividing by 8:
\[ \sqrt{\frac{T}{10^{-2}}} = 26 \]

### Step 7: Solve for Tension (T)
Squaring both sides:
\[ \frac{T}{10^{-2}} = 676 \]
Thus,
\[ T = 676 \times 10^{-2} = 6.76 \, \text{N} \]

### Step 8: Final Calculation
We need to verify if the tension is consistent with the beat frequency decreasing when tension decreases.

### Conclusion
The tension in the string is approximately:
\[ T \approx 6.76 \, \text{N} \]