Two wires of different linear mass densities are soldered together end to end and then stretched under a tension `F`. The wave speed in the first wire is thrice that in the second. If a harmonic wave travelling in the first wire is incident on the junction of the wires and if the amplitude of the incident wave is `A = sqrt(13) cm`, find the amplitude of reflected wave.

Explanation

To solve the problem, we need to find the amplitude of the reflected wave when a harmonic wave traveling in the first wire is incident on the junction of two wires with different linear mass densities.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- The wave speed in the first wire (V1) is three times that in the second wire (V2).
- The amplitude of the incident wave (Ai) is given as \( A_i = \sqrt{13} \) cm.

2. **Express the Relationship Between Wave Speeds:**
- From the problem, we know:
\[ V_1 = 3V_2 \]

3. **Use the Reflection Amplitude Formula:**
- The formula for the amplitude of the reflected wave (Ar) when a wave travels from one medium to another is given by:
\[ A_r = \frac{V_1 - V_2}{V_1 + V_2} A_i \]

4. **Substitute the Values:**
- Substitute \( V_1 = 3V_2 \) into the reflection amplitude formula:
\[ A_r = \frac{3V_2 - V_2}{3V_2 + V_2} A_i \]
- Simplifying the expression:
\[ A_r = \frac{2V_2}{4V_2} A_i = \frac{1}{2} A_i \]

5. **Calculate the Amplitude of the Reflected Wave:**
- Now substitute \( A_i = \sqrt{13} \) cm into the equation:
\[ A_r = \frac{1}{2} \sqrt{13} \text{ cm} \]
- Therefore, the amplitude of the reflected wave is:
\[ A_r = \frac{\sqrt{13}}{2} \text{ cm} \]

### Final Answer:
The amplitude of the reflected wave is \( \frac{\sqrt{13}}{2} \) cm.

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