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 will analyze the situation of two wires with different wave speeds and determine the amplitudes of the reflected and transmitted waves, as well as the fractions of power reflected and transmitted.

### Step-by-Step Solution:

**Given:**
- Speed of wave in the first wire (V1) = V
- Speed of wave in the second wire (V2) = 2V
- Amplitude of incident wave (AI) = A

**(a) Finding the amplitudes of reflected and transmitted waves:**

1. **Determine the amplitude of the reflected wave (AR):**
The formula for the amplitude of the reflected wave when there is a change in medium is given by:
\[ AR = \frac{V2 - V1}{V2 + V1} \cdot AI \]
Substituting the values:
\[ AR = \frac{2V - V}{2V + V} \cdot A = \frac{V}{3V} \cdot A = \frac{1}{3} A \]
Since the reflected wave travels in the opposite direction, we can denote it as:
\[ AR = -\frac{1}{3} A \]

2. **Determine the amplitude of the transmitted wave (AT):**
The formula for the amplitude of the transmitted wave is:
\[ AT = \frac{2V2}{V2 + V1} \cdot AI \]
Substituting the values:
\[ AT = \frac{2 \cdot 2V}{2V + V} \cdot A = \frac{4V}{3V} \cdot A = \frac{4}{3} A \]

**Final Answers for Part (a):**
- Amplitude of reflected wave (AR) = \(-\frac{1}{3} A\)
- Amplitude of transmitted wave (AT) = \(\frac{4}{3} A\)

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**(b) Finding the fraction of the incident power that is reflected and transmitted:**

1. **Calculate the power associated with the amplitudes:**
The power (P) associated with a wave is proportional to the square of its amplitude:
\[ P \propto A^2 \]

2. **Calculate the reflected power (PR):**
\[ PR \propto AR^2 = \left(-\frac{1}{3} A\right)^2 = \frac{1}{9} A^2 \]

3. **Calculate the transmitted power (PT):**
\[ PT \propto AT^2 = \left(\frac{4}{3} A\right)^2 = \frac{16}{9} A^2 \]

4. **Total incident power (PI):**
\[ PI \propto AI^2 = A^2 \]

5. **Calculate the fractions of power:**
- Fraction of power reflected:
\[ \text{Fraction reflected} = \frac{PR}{PI} = \frac{\frac{1}{9} A^2}{A^2} = \frac{1}{9} \]

- Fraction of power transmitted:
\[ \text{Fraction transmitted} = \frac{PT}{PI} = \frac{\frac{16}{9} A^2}{A^2} = \frac{16}{9} \]

**Final Answers for Part (b):**
- Fraction of incident power reflected = \(\frac{1}{9}\)
- Fraction of incident power transmitted = \(\frac{8}{9}\) (since total power must equal 1, \(1 - \frac{1}{9} = \frac{8}{9}\))

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