An audio oscillator capable of producing notes of frequencies ranging from `500 Hz "to" 1500 Hz` is placed constant tension `T`. The linear mass density of the wire is `0.75 g//m`. It is observed that by varying the frequency of the oscillator over the given permissible rang the sonometer wire sets into vibration at frequencies `840 Hz and 1120 Hz`.
a. Find the tension in the string .
b. What are the frequencies of the first and fourth overtone produced by the vibrating string?

Explanation

To solve the problem step by step, we will first find the tension in the string and then calculate the frequencies of the first and fourth overtones.

### Part (a): Finding the Tension in the String

1. **Identify the given values:**
- Frequencies where the wire vibrates: \( f_1 = 840 \, \text{Hz} \) and \( f_2 = 1120 \, \text{Hz} \)
- Linear mass density \( \mu = 0.75 \, \text{g/m} = 0.75 \times 10^{-3} \, \text{kg/m} \)

2. **Determine the ratio of the frequencies:**
\[ \frac{f_1}{f_2} = \frac{840}{1120} = \frac{3}{4} \]

3. **Use the fundamental frequency formula:**
The fundamental frequency \( f_0 \) can be expressed in terms of the tension \( T \) and the linear mass density \( \mu \):
\[ f_0 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
Since \( f_1 \) and \( f_2 \) are harmonics, we can express \( f_1 \) as:
\[ f_1 = \frac{3}{4} f_2 \]
This implies that \( f_1 \) is the third harmonic and \( f_2 \) is the fourth harmonic.

4. **Relate the harmonics to the fundamental frequency:**
The fundamental frequency \( f_0 \) can be calculated as:
\[ f_0 = \frac{f_1}{3} = \frac{840}{3} = 280 \, \text{Hz} \]

5. **Calculate the tension using the fundamental frequency:**
Rearranging the formula for tension:
\[ T = 4L^2 \mu f_0^2 \]
Here, we need to express \( L \) in terms of the frequency. Since we don't have \( L \), we can assume \( L = 1 \) m for simplicity in calculations (as it will cancel out later). Thus:
\[ T = 4(1^2)(0.75 \times 10^{-3}) (280^2) \]

6. **Calculate \( T \):**
\[ T = 4 \times 0.75 \times 10^{-3} \times 78400 \]
\[ T = 4 \times 0.75 \times 78.4 = 234.6 \, \text{N} \]

### Part (b): Finding the Frequencies of the First and Fourth Overtone

1. **Calculate the first overtone (2nd harmonic):**
\[ f_2 = 2 f_0 = 2 \times 280 = 560 \, \text{Hz} \]

2. **Calculate the fourth overtone (5th harmonic):**
\[ f_5 = 5 f_0 = 5 \times 280 = 1400 \, \text{Hz} \]

### Final Answers:
- **Tension in the string:** \( T \approx 234.6 \, \text{N} \)
- **Frequency of the first overtone:** \( 560 \, \text{Hz} \)
- **Frequency of the fourth overtone:** \( 1400 \, \text{Hz} \)