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 break it down into two parts as mentioned in the question.

### Part (a): Finding the tension in the string

1. **Identify the Frequencies**: The two frequencies at which the sonometer wire vibrates are given as \( f_1 = 840 \, \text{Hz} \) and \( f_2 = 1120 \, \text{Hz} \).

2. **Find the Ratio of Frequencies**:
\[ \frac{f_1}{f_2} = \frac{840}{1120} = \frac{3}{4} \]
This implies that \( f_1 = 3f_0 \) and \( f_2 = 4f_0 \), where \( f_0 \) is the fundamental frequency.

3. **Calculate the Fundamental Frequency**:
From \( f_1 = 3f_0 \):
\[ f_0 = \frac{840}{3} = 280 \, \text{Hz} \]

4. **Use the Formula for Frequency**: The fundamental frequency \( f_0 \) is given by the formula:
\[ f_0 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
Rearranging this to find tension \( T \):
\[ T = 4L^2 f_0^2 \mu \]

5. **Substitute Values**: The linear mass density \( \mu \) is given as \( 0.75 \, \text{g/m} = 0.75 \times 10^{-3} \, \text{kg/m} \). The length \( L \) of the wire is not provided in the question, but we can assume a length for calculation purposes. Let's assume \( L = 0.25 \, \text{m} \).

6. **Calculate Tension**:
\[ T = 4 \times (0.25)^2 \times (280)^2 \times (0.75 \times 10^{-3}) \]
\[ T = 4 \times 0.0625 \times 78400 \times 0.00075 \]
\[ T = 4 \times 0.0625 \times 58.8 \approx 14.7 \, \text{N} \]

### Part (b): Finding the frequencies of the first and fourth overtone

1. **First Overtone**: The first overtone is the second harmonic, which is given by:
\[ f_1 = 2f_0 \]
Substituting the value of \( f_0 \):
\[ f_1 = 2 \times 280 = 560 \, \text{Hz} \]

2. **Fourth Overtone**: The fourth overtone corresponds to the fifth harmonic, which is given by:
\[ f_4 = 5f_0 \]
Substituting the value of \( f_0 \):
\[ f_4 = 5 \times 280 = 1400 \, \text{Hz} \]

### Final Answers:
- Tension in the string \( T \approx 14.7 \, \text{N} \)
- Frequency of the first overtone \( = 560 \, \text{Hz} \)
- Frequency of the fourth overtone \( = 1400 \, \text{Hz} \)