A wire of uniform cross-section is stretched between two points `100cm` apart. The wire is fixed at one end and a weight is hung over a pulley at the other end. A weight of 9kg produces a fundamental frequency of `750Hz`.
(a) What is the velocity of the wave in wire?
(b) If the weight is reduced to `4kg`, what is the velocity of wave?

Explanation

To solve the problem step by step, let's break it down into two parts as per the question.

### Part (a): Finding the velocity of the wave in the wire with a 9 kg weight.

1. **Identify the given values:**
- Length of the wire, \( L = 100 \, \text{cm} = 1 \, \text{m} \)
- Mass of the weight, \( m_1 = 9 \, \text{kg} \)
- Fundamental frequency, \( f = 750 \, \text{Hz} \)

2. **Calculate the tension in the wire:**
- The tension \( T_1 \) in the wire due to the weight is given by:
\[ T_1 = m_1 \cdot g \]
- Where \( g \) (acceleration due to gravity) is approximately \( 9.8 \, \text{m/s}^2 \).
- Thus,
\[ T_1 = 9 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 88.2 \, \text{N} \]

3. **Use the formula for the velocity of a wave in a stretched string:**
- The velocity \( v \) of a wave on a string is given by:
\[ v = f \cdot \lambda \]
- For the fundamental frequency, the wavelength \( \lambda \) is twice the length of the string:
\[ \lambda = 2L = 2 \cdot 1 \, \text{m} = 2 \, \text{m} \]
- Therefore,
\[ v = f \cdot \lambda = 750 \, \text{Hz} \cdot 2 \, \text{m} = 1500 \, \text{m/s} \]

4. **Conclusion for Part (a):**
- The velocity of the wave in the wire when a weight of 9 kg is applied is:
\[ \boxed{1500 \, \text{m/s}} \]

### Part (b): Finding the velocity of the wave in the wire with a 4 kg weight.

1. **Identify the new weight:**
- Mass of the new weight, \( m_2 = 4 \, \text{kg} \)

2. **Calculate the new tension in the wire:**
- The tension \( T_2 \) in the wire due to the new weight is given by:
\[ T_2 = m_2 \cdot g = 4 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 39.2 \, \text{N} \]

3. **Use the previously derived formula for velocity:**
- The velocity of the wave can be expressed as:
\[ v = \sqrt{\frac{T}{\mu}} \]
- We need to find \( \mu \) (linear mass density). From Part (a), we have:
\[ v_1 = \sqrt{\frac{T_1}{\mu}} \Rightarrow 1500 = \sqrt{\frac{88.2}{\mu}} \]
- Squaring both sides gives:
\[ 1500^2 = \frac{88.2}{\mu} \Rightarrow \mu = \frac{88.2}{1500^2} \]

4. **Now calculate the new velocity \( v_2 \):**
- Substitute \( T_2 \) and \( \mu \) into the velocity formula:
\[ v_2 = \sqrt{\frac{T_2}{\mu}} = \sqrt{\frac{39.2}{\mu}} \]
- Since we have \( \mu \) from the previous calculation, we can express it in terms of known quantities:
\[ v_2 = \sqrt{\frac{39.2 \cdot 1500^2}{88.2}} = 1500 \cdot \sqrt{\frac{39.2}{88.2}} \]
- Calculate \( \sqrt{\frac{39.2}{88.2}} \):
\[ \sqrt{\frac{39.2}{88.2}} \approx 0.6667 \]
- Therefore,
\[ v_2 \approx 1500 \cdot 0.6667 \approx 1000 \, \text{m/s} \]

5. **Conclusion for Part (b):**
- The velocity of the wave in the wire when a weight of 4 kg is applied is:
\[ \boxed{1000 \, \text{m/s}} \]