A cord 80 cm long is stretched by a load of 8.0 kg f.The mass per unit length of the cord is `4.0xx10^(-5) kg m^(-1)`.Find (i)speed of the transverse wave in the cord and (ii)frequency of the fundamental and that of the second overtone.

Answer

To solve the given problem, we will break it down into two parts: (i) finding the speed of the transverse wave in the cord and (ii) finding the frequency of the fundamental and the second overtone.

### Step 1: Finding the Speed of the Transverse Wave

1. **Identify the given values:**
- Length of the cord, \( L = 80 \, \text{cm} = 0.8 \, \text{m} \)
- Load (mass), \( m = 8.0 \, \text{kg} \)
- Mass per unit length, \( \mu = 4.0 \times 10^{-5} \, \text{kg/m} \)

2. **Convert the load into force (tension):**
- The weight (force) due to the load is given by:
\[ T = m \cdot g = 8.0 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 78.4 \, \text{N} \]

3. **Use the formula for the speed of a wave on a string:**
- The speed \( v \) of the transverse wave is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
- Substituting the values:
\[ v = \sqrt{\frac{78.4 \, \text{N}}{4.0 \times 10^{-5} \, \text{kg/m}}} \]

4. **Calculate the speed:**
- First, calculate \( \frac{78.4}{4.0 \times 10^{-5}} \):
\[ \frac{78.4}{4.0 \times 10^{-5}} = 1960000 \, \text{m}^2/\text{s}^2 \]
- Now take the square root:
\[ v = \sqrt{1960000} \approx 1400 \, \text{m/s} \]

### Step 2: Finding the Frequency of the Fundamental and Second Overtone

1. **Fundamental frequency formula:**
- The fundamental frequency \( f_1 \) is given by:
\[ f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
- Substitute \( L = 0.8 \, \text{m} \) and \( T = 78.4 \, \text{N} \):
\[ f_1 = \frac{1}{2 \times 0.8} \sqrt{\frac{78.4}{4.0 \times 10^{-5}}} \]
- We already calculated \( \sqrt{\frac{78.4}{4.0 \times 10^{-5}}} \approx 1400 \):
\[ f_1 = \frac{1}{1.6} \times 1400 \approx 875 \, \text{Hz} \]

2. **Finding the second overtone:**
- The second overtone corresponds to the third harmonic, which is given by:
\[ f_3 = 3 \cdot f_1 \]
- Therefore:
\[ f_3 = 3 \cdot 875 = 2625 \, \text{Hz} \]

### Final Answers:
- (i) The speed of the transverse wave in the cord is \( 1400 \, \text{m/s} \).
- (ii) The frequency of the fundamental is \( 875 \, \text{Hz} \) and the frequency of the second overtone is \( 2625 \, \text{Hz} \).