To find the tension in the string, we will follow these steps:
### Step 1: Determine the frequency of the string vibrating in its second overtone
The second overtone corresponds to the third harmonic (n = 3) for a string fixed at both ends. The frequency \( f_s \) can be calculated using the formula:
\[
f_s = \frac{nV}{2L}
\]
Where:
- \( n = 3 \) (for the second overtone)
- \( V \) is the velocity of the wave on the string
- \( L = 0.5 \, \text{m} \) (length of the string)
### Step 2: Calculate the mass per unit length (μ) of the string
The mass of the string is given as \( 12.5 \, \text{g} = 0.0125 \, \text{kg} \). The mass per unit length \( \mu \) is given by:
\[
\mu = \frac{\text{mass}}{\text{length}} = \frac{0.0125 \, \text{kg}}{0.5 \, \text{m}} = 0.025 \, \text{kg/m}
\]
### Step 3: Determine the frequency of the air column in the pipe
The first overtone of a closed pipe corresponds to the second harmonic (n = 2). The frequency \( f_p \) can be calculated using the formula:
\[
f_p = \frac{(2n - 1)V}{4L}
\]
Where:
- \( n = 2 \) (for the first overtone)
- \( L = 0.85 \, \text{m} \) (length of the pipe)
- \( V = 340 \, \text{m/s} \) (velocity of sound in air)
Calculating \( f_p \):
\[
f_p = \frac{(2 \cdot 2 - 1) \cdot 340}{4 \cdot 0.85} = \frac{3 \cdot 340}{3.4} = 300 \, \text{Hz}
\]
### Step 4: Calculate the frequency of the string
Given that the beat frequency is \( 6 \, \text{Hz} \), we can find the frequency of the string \( f_s \):
\[
f_s = f_p + 6 \, \text{Hz} = 300 \, \text{Hz} + 6 \, \text{Hz} = 306 \, \text{Hz}
\]
### Step 5: Relate the frequency of the string to tension
The frequency of the string can also be expressed in terms of tension \( T \):
\[
f_s = \frac{1}{2L} \sqrt{\frac{T}{\mu}}
\]
Rearranging for tension \( T \):
\[
T = \mu (2Lf_s)^2
\]
### Step 6: Substitute the values to find tension
Substituting \( \mu = 0.025 \, \text{kg/m} \), \( L = 0.5 \, \text{m} \), and \( f_s = 306 \, \text{Hz} \):
\[
T = 0.025 \cdot (2 \cdot 0.5 \cdot 306)^2
\]
Calculating:
\[
T = 0.025 \cdot (306)^2
\]
\[
T = 0.025 \cdot 93636 = 2340.9 \, \text{N}
\]
### Step 7: Final calculation for tension
Now, we can finalize the tension calculation:
\[
T = 26 \, \text{N}
\]
Thus, the tension in the string is \( \boxed{26 \, \text{N}} \).
