Two wires of different densities are soldered together end to end then stretched under tension T. The waves speed in the first wire is twice that in the second wire.
(a) If the amplitude of incident wave is A, what are amplitudes of reflected and transmitted waves?
(b) Assuming no energy loss in the wire, find the fraction of the incident power that is reflected at the junction and fraction of the same that is transmitted.

Explanation

To solve the problem step by step, we can break it down into two parts: (a) finding the amplitudes of the reflected and transmitted waves, and (b) finding the fractions of incident power that is reflected and transmitted.

### Part (a): Amplitudes of Reflected and Transmitted Waves

1. **Identify the Wave Speeds:**
Let the speed of the wave in the first wire be \( v_1 \) and in the second wire be \( v_2 \). According to the problem, we know that:
\[ v_1 = 2v_2 \]

2. **Use the Reflection and Transmission Amplitude Formulas:**
The amplitude of the reflected wave \( A_r \) and the transmitted wave \( A_t \) can be calculated using the following formulas:
\[ \frac{A_r}{A_i} = \frac{v_2 - v_1}{v_2 + v_1} \]
\[ \frac{A_t}{A_i} = \frac{2v_2}{v_1 + v_2} \]
where \( A_i \) is the amplitude of the incident wave, which is given as \( A \).

3. **Substituting the Values:**
Substitute \( v_1 = 2v_2 \) into the reflection formula:
\[ \frac{A_r}{A} = \frac{v_2 - 2v_2}{v_2 + 2v_2} = \frac{-v_2}{3v_2} = -\frac{1}{3} \]
Therefore, the amplitude of the reflected wave is:
\[ A_r = -\frac{1}{3} A \]

4. **Calculate the Transmitted Amplitude:**
Now substitute into the transmission formula:
\[ \frac{A_t}{A} = \frac{2v_2}{2v_2 + v_2} = \frac{2v_2}{3v_2} = \frac{2}{3} \]
Therefore, the amplitude of the transmitted wave is:
\[ A_t = \frac{2}{3} A \]

### Summary of Part (a):
- Amplitude of reflected wave \( A_r = -\frac{1}{3} A \)
- Amplitude of transmitted wave \( A_t = \frac{2}{3} A \)

### Part (b): Fraction of Incident Power Reflected and Transmitted

1. **Power Proportional to Amplitude Squared:**
The power \( P \) is proportional to the square of the amplitude:
\[ P \propto A^2 \]

2. **Calculate the Fraction of Reflected Power:**
The fraction of the reflected power \( P_r \) to the incident power \( P_i \) is given by:
\[ \frac{P_r}{P_i} = \left(\frac{A_r}{A_i}\right)^2 = \left(-\frac{1}{3}\right)^2 = \frac{1}{9} \]

3. **Calculate the Fraction of Transmitted Power:**
The fraction of the transmitted power \( P_t \) to the incident power \( P_i \) is given by:
\[ \frac{P_t}{P_i} = \left(\frac{A_t}{A_i}\right)^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \]

4. **Finding the Remaining Power:**
Since the total power must equal the sum of reflected and transmitted power:
\[ P_r + P_t = P_i \]
Therefore, the fraction of power that is transmitted is:
\[ P_t = P_i - P_r = P_i - \frac{1}{9} P_i = \frac{8}{9} P_i \]

### Summary of Part (b):
- Fraction of reflected power \( \frac{P_r}{P_i} = \frac{1}{9} \)
- Fraction of transmitted power \( \frac{P_t}{P_i} = \frac{8}{9} \)