A harmonic wave is travelling on string 1. At a junction with string 2 it is partly reflected and partly transmitted. The linear mass density of the second string is four times that of the first string, and that the boundary between the two strings is at x=0. If the expression for the incident wave is
`y_i = A_i cos (k_1 x-omega_1t)`
(a) What are the expressions for the transmitted and the reflected waves in terms of `A_i, k_1` and `omega_1`?
(b) Show that the average power carried by the incident wave is equal to the sum of the average power carried by the transmitted and reflected waves.

Explanation

To solve the problem step by step, let's break it down into parts (a) and (b) as specified in the question.

### Part (a): Expressions for the Transmitted and Reflected Waves

1. **Identify the Incident Wave**:
The incident wave is given by:
\[ y_i = A_i \cos(k_1 x - \omega_1 t) \]

2. **Understand the Relationship Between the Strings**:
- The linear mass density of string 2 is four times that of string 1:
\[ \mu_2 = 4\mu_1 \]
- The tension in both strings is equal since they are connected.

3. **Determine the Velocities**:
- The velocity of the wave in string 1 (incident wave) is:
\[ v_1 = \sqrt{\frac{T}{\mu_1}} \]
- The velocity of the wave in string 2 (transmitted wave) is:
\[ v_2 = \sqrt{\frac{T}{\mu_2}} = \sqrt{\frac{T}{4\mu_1}} = \frac{1}{2}\sqrt{\frac{T}{\mu_1}} = \frac{v_1}{2} \]

4. **Determine the Wave Numbers**:
- The wave number in string 1 is:
\[ k_1 = \frac{\omega_1}{v_1} \]
- The wave number in string 2 is:
\[ k_2 = \frac{\omega_2}{v_2} = \frac{\omega_1}{v_1/2} = \frac{2\omega_1}{v_1} = 2k_1 \]

5. **Calculate the Amplitudes of Reflected and Transmitted Waves**:
- The amplitude of the transmitted wave \(A_t\) is given by:
\[ A_t = \frac{2v_2}{v_1 + v_2} A_i = \frac{2 \cdot \frac{v_1}{2}}{v_1 + \frac{v_1}{2}} A_i = \frac{2}{3} A_i \]
- The amplitude of the reflected wave \(A_r\) is given by:
\[ A_r = \frac{v_2 - v_1}{v_1 + v_2} A_i = \frac{\frac{v_1}{2} - v_1}{v_1 + \frac{v_1}{2}} A_i = -\frac{1}{3} A_i \]

6. **Write the Expressions for the Transmitted and Reflected Waves**:
- The transmitted wave \(y_t\) is:
\[ y_t = A_t \cos(k_2 x - \omega_1 t) = \frac{2}{3} A_i \cos(2k_1 x - \omega_1 t) \]
- The reflected wave \(y_r\) is:
\[ y_r = A_r \cos(k_1 x + \omega_1 t) = -\frac{1}{3} A_i \cos(k_1 x + \omega_1 t) \]

### Part (b): Show that the Average Power is Conserved

1. **Calculate the Average Power of the Incident Wave**:
The average power \(P_i\) carried by the incident wave is given by:
\[ P_i = \frac{1}{2} \mu_1 A_i^2 \omega_1^2 v_1 \]

2. **Calculate the Average Power of the Reflected Wave**:
The average power \(P_r\) carried by the reflected wave is:
\[ P_r = \frac{1}{2} \mu_1 A_r^2 \omega_1^2 v_1 = \frac{1}{2} \mu_1 \left(-\frac{1}{3} A_i\right)^2 \omega_1^2 v_1 = \frac{1}{2} \mu_1 \frac{1}{9} A_i^2 \omega_1^2 v_1 = \frac{1}{18} \mu_1 A_i^2 \omega_1^2 v_1 \]

3. **Calculate the Average Power of the Transmitted Wave**:
The average power \(P_t\) carried by the transmitted wave is:
\[ P_t = \frac{1}{2} \mu_2 A_t^2 \omega_1^2 v_2 = \frac{1}{2} (4\mu_1) \left(\frac{2}{3} A_i\right)^2 \omega_1^2 \left(\frac{v_1}{2}\right) = \frac{1}{2} (4\mu_1) \frac{4}{9} A_i^2 \omega_1^2 \frac{v_1}{2} = \frac{8}{18} \mu_1 A_i^2 \omega_1^2 v_1 \]

4. **Sum of Reflected and Transmitted Powers**:
\[ P_r + P_t = \frac{1}{18} \mu_1 A_i^2 \omega_1^2 v_1 + \frac{8}{18} \mu_1 A_i^2 \omega_1^2 v_1 = \frac{9}{18} \mu_1 A_i^2 \omega_1^2 v_1 = \frac{1}{2} \mu_1 A_i^2 \omega_1^2 v_1 = P_i \]

### Conclusion
Thus, we have shown that the average power carried by the incident wave is equal to the sum of the average power carried by the transmitted and reflected waves.