A string of mass per unit length `"0.2 kg m"^(-1)` and length 0.6 m is fixed at both ends such that it has a tension of 80 N. If the string is vibrating in its second overtone mode, then the frequency of the string is

Options

A. 40 Hz
B. 50 Hz
C. 60 Hz
D. 70 Hz

Answer

To find the frequency of the string vibrating in its second overtone mode, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the parameters given in the problem:**
- Mass per unit length (μ) = 0.2 kg/m
- Length of the string (L) = 0.6 m
- Tension in the string (T) = 80 N

2. **Determine the mode of vibration:**
- The second overtone corresponds to the third harmonic (n = 3). This is because the nth overtone is given by (n - 1), where n is the harmonic number.

3. **Use the formula for the frequency of a vibrating string:**
- The frequency (f) of a string fixed at both ends is given by the formula:
\[ f_n = \frac{nV}{2L} \]
- Where:
- \( n \) = harmonic number
- \( V \) = wave speed
- \( L \) = length of the string

4. **Calculate the wave speed (V):**
- The wave speed (V) on a string is given by:
\[ V = \sqrt{\frac{T}{\mu}} \]
- Substituting the values:
\[ V = \sqrt{\frac{80 \, \text{N}}{0.2 \, \text{kg/m}}} = \sqrt{400} = 20 \, \text{m/s} \]

5. **Substitute the values into the frequency formula:**
- Now, substituting \( n = 3 \), \( V = 20 \, \text{m/s} \), and \( L = 0.6 \, \text{m} \):
\[ f_3 = \frac{3 \times 20}{2 \times 0.6} \]
- Simplifying this:
\[ f_3 = \frac{60}{1.2} = 50 \, \text{Hz} \]

6. **Final answer:**
- The frequency of the string vibrating in its second overtone mode is **50 Hz**.