The harmonic wave `y_i = (2.0 xx 10^(-3)) cos pi (2.0x - 50t)` travels along a string towards a boundary at x=0 with a second string. The wave speed on the second string is `50 m//s`. Write expressions for reflected and transmitted waves. Assume SI units.

Explanation

To solve the problem, we need to find the expressions for the reflected and transmitted waves based on the given incident wave and the properties of the two strings. Let's break it down step by step.

### Step 1: Identify the Incident Wave
The incident wave is given by the equation:
\[ y_i = (2.0 \times 10^{-3}) \cos(\pi (2.0x - 50t)) \]

### Step 2: Determine the Wave Speed of the Incident Wave
The wave speed \( v_1 \) of the incident wave can be calculated using the coefficients of \( t \) and \( x \) in the wave equation. The general form of a wave is:
\[ y = A \cos(kx - \omega t) \]
where:
- \( \omega \) is the angular frequency,
- \( k \) is the wave number.

From the given equation:
- Coefficient of \( t \) (angular frequency \( \omega \)) = 50
- Coefficient of \( x \) (wave number \( k \)) = \( \pi \times 2 \)

Using the formula for wave speed:
\[ v_1 = \frac{\omega}{k} = \frac{50}{2\pi} \]
Calculating this gives:
\[ v_1 = \frac{50}{2} = 25 \, \text{m/s} \]

### Step 3: Determine the Wave Speed of the Second String
The wave speed of the second string is given as:
\[ v_2 = 50 \, \text{m/s} \]

### Step 4: Analyze the Mediums
Since \( v_2 > v_1 \), we conclude that the second medium is a rarer medium. In a rarer medium, the reflected wave will have a positive phase shift.

### Step 5: Calculate the Amplitude of the Reflected Wave
The amplitude of the reflected wave \( A_R \) can be calculated using the formula:
\[ A_R = \frac{v_2 - v_1}{v_2 + v_1} A_I \]
where \( A_I = 2.0 \times 10^{-3} \).

Substituting the values:
\[ A_R = \frac{50 - 25}{50 + 25} \times (2.0 \times 10^{-3}) \]
\[ A_R = \frac{25}{75} \times (2.0 \times 10^{-3}) = \frac{1}{3} \times (2.0 \times 10^{-3}) = \frac{2.0}{3} \times 10^{-3} \, \text{m} \]

### Step 6: Calculate the Amplitude of the Transmitted Wave
The amplitude of the transmitted wave \( A_T \) can be calculated using the formula:
\[ A_T = \frac{2v_2}{v_2 + v_1} A_I \]

Substituting the values:
\[ A_T = \frac{2 \times 50}{50 + 25} \times (2.0 \times 10^{-3}) \]
\[ A_T = \frac{100}{75} \times (2.0 \times 10^{-3}) = \frac{4}{3} \times (2.0 \times 10^{-3}) = \frac{8.0}{3} \times 10^{-3} \, \text{m} \]

### Step 7: Write the Expressions for the Reflected and Transmitted Waves
The general form of the reflected wave is:
\[ y_R = A_R \cos(kx - \omega t) \]
Since the phase is positive for the reflected wave:
\[ y_R = \frac{2.0}{3} \times 10^{-3} \cos(\pi(2.0x - 50t)) \]

The transmitted wave will have a positive phase shift:
\[ y_T = A_T \cos(kx + \omega t) \]
Thus:
\[ y_T = \frac{8.0}{3} \times 10^{-3} \cos(\pi(2.0x + 50t)) \]

### Final Expressions
- Reflected Wave:
\[ y_R = \frac{2.0}{3} \times 10^{-3} \cos(\pi(2.0x - 50t)) \]
- Transmitted Wave:
\[ y_T = \frac{8.0}{3} \times 10^{-3} \cos(\pi(2.0x + 50t)) \]