A composition string is made up by joining two strings of different masses per unit length 1g/ m and 4 g/m. The composite string is under the same tension. A transverse wave pulse : Y = (6 mm) sin(5t + 40x), where it is in seconds and 'x' in meters, is sent along the lighter string towards the joint. The joint is at x = 0. The equation of the wave pulse reflected from the joint is

Options

A. (2 mm) sin(5t – 40x)
B. (4 mm) sin(40x - 5t)
C. -(2 mm) sin(5t – 40x)
D. (2 mm) sin (5t - 10x)

Answer

To solve the problem, we need to find the equation of the wave pulse reflected from the joint of two strings with different masses per unit length. Here’s the step-by-step solution:

### Step 1: Identify the Parameters
We have two strings with the following properties:
- String 1 (lighter string): mass per unit length \( \mu_1 = 1 \, \text{g/m} = 0.001 \, \text{kg/m} \)
- String 2 (heavier string): mass per unit length \( \mu_2 = 4 \, \text{g/m} = 0.004 \, \text{kg/m} \)

The wave pulse traveling along the lighter string is given by:
\[ Y = 6 \sin(5t + 40x) \]

### Step 2: Determine the Wave Speed in Each String
The wave speed \( v \) in a string is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the string and \( \mu \) is the mass per unit length.

Since both strings are under the same tension \( T \), we can express the wave speeds as:
- For String 1:
\[ v_1 = \sqrt{\frac{T}{\mu_1}} = \sqrt{\frac{T}{0.001}} \]

- For String 2:
\[ v_2 = \sqrt{\frac{T}{\mu_2}} = \sqrt{\frac{T}{0.004}} \]

### Step 3: Calculate the Reflection Coefficient
When a wave travels from a lighter medium to a heavier medium, it reflects with an inversion. The amplitude of the reflected wave can be calculated using the formula:
\[ A_r = \frac{Z_1 - Z_2}{Z_1 + Z_2} A \]
where \( Z \) is the impedance given by \( Z = \mu v \).

Calculating the impedances:
- For String 1:
\[ Z_1 = \mu_1 v_1 = 0.001 \cdot \sqrt{\frac{T}{0.001}} = \sqrt{T \cdot 0.001} \]

- For String 2:
\[ Z_2 = \mu_2 v_2 = 0.004 \cdot \sqrt{\frac{T}{0.004}} = \sqrt{T \cdot 0.004} \]

### Step 4: Substitute into the Reflection Coefficient Formula
Substituting \( Z_1 \) and \( Z_2 \) into the reflection coefficient formula:
\[ A_r = \frac{\sqrt{T \cdot 0.001} - \sqrt{T \cdot 0.004}}{\sqrt{T \cdot 0.001} + \sqrt{T \cdot 0.004}} \cdot 6 \]

### Step 5: Simplify the Amplitude of the Reflected Wave
Calculating the values:
- The amplitude of the incident wave \( A = 6 \, \text{mm} \)
- The expression simplifies to:
\[ A_r = \frac{\sqrt{0.001} - \sqrt{0.004}}{\sqrt{0.001} + \sqrt{0.004}} \cdot 6 \]

Calculating \( \sqrt{0.001} = 0.03162 \) and \( \sqrt{0.004} = 0.06325 \):
\[ A_r = \frac{0.03162 - 0.06325}{0.03162 + 0.06325} \cdot 6 \]
\[ A_r = \frac{-0.03163}{0.09487} \cdot 6 \approx -2 \]

### Step 6: Write the Equation of the Reflected Wave
Since the wave reflects with inversion, the equation of the reflected wave becomes:
\[ Y_r = -2 \sin(5t - 40x) \]

### Final Answer
The equation of the wave pulse reflected from the joint is:
\[ Y_r = -2 \sin(5t - 40x) \]

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