Here are the equations of three waves:
(1) (y) (x,t) =2 sin (4x-2t), (2). (y)(x,t) =sin (3x-4t), (3) (y) (x,t)=2 sin (3x-3t). Rank the waves according to their (a) wave speed and (b) maximum speed perpendicular to the wave's direction of travel (the transverse speed), greatest first.

Answer

To solve the problem, we need to analyze the given wave equations and find the wave speed and the maximum transverse speed for each wave. Let's break it down step by step.

### Step 1: Identify the wave equations
The given wave equations are:
1. \( y_1(x,t) = 2 \sin(4x - 2t) \)
2. \( y_2(x,t) = \sin(3x - 4t) \)
3. \( y_3(x,t) = 2 \sin(3x - 3t) \)

### Step 2: Extract parameters from the wave equations
From the general wave equation \( y(x,t) = A \sin(kx - \omega t) \), we can identify:
- Amplitude \( A \)
- Wave number \( k \)
- Angular frequency \( \omega \)

For each wave:
- For wave 1: \( A_1 = 2, k_1 = 4, \omega_1 = 2 \)
- For wave 2: \( A_2 = 1, k_2 = 3, \omega_2 = 4 \)
- For wave 3: \( A_3 = 2, k_3 = 3, \omega_3 = 3 \)

### Step 3: Calculate wave speed
The wave speed \( v \) can be calculated using the formula:
\[ v = \frac{\omega}{k} \]

Calculating for each wave:
1. For wave 1:
\[ v_1 = \frac{\omega_1}{k_1} = \frac{2}{4} = 0.5 \]
2. For wave 2:
\[ v_2 = \frac{\omega_2}{k_2} = \frac{4}{3} \approx 1.33 \]
3. For wave 3:
\[ v_3 = \frac{\omega_3}{k_3} = \frac{3}{3} = 1 \]

### Step 4: Rank the waves according to wave speed
From the calculated speeds:
- \( v_2 \approx 1.33 \) (wave 2)
- \( v_3 = 1 \) (wave 3)
- \( v_1 = 0.5 \) (wave 1)

**Ranking according to wave speed:**
1. Wave 2
2. Wave 3
3. Wave 1

### Step 5: Calculate maximum transverse speed
The maximum transverse speed \( v_t \) can be calculated using:
\[ v_t = A \cdot \omega \]

Calculating for each wave:
1. For wave 1:
\[ v_{t1} = A_1 \cdot \omega_1 = 2 \cdot 2 = 4 \]
2. For wave 2:
\[ v_{t2} = A_2 \cdot \omega_2 = 1 \cdot 4 = 4 \]
3. For wave 3:
\[ v_{t3} = A_3 \cdot \omega_3 = 2 \cdot 3 = 6 \]

### Step 6: Rank the waves according to maximum transverse speed
From the calculated transverse speeds:
- \( v_{t3} = 6 \) (wave 3)
- \( v_{t1} = 4 \) (wave 1)
- \( v_{t2} = 4 \) (wave 2)

**Ranking according to maximum transverse speed:**
1. Wave 3
2. Wave 1 and Wave 2 (equal)

### Final Answers
(a) Ranking according to wave speed:
1. Wave 2
2. Wave 3
3. Wave 1

(b) Ranking according to maximum transverse speed:
1. Wave 3
2. Wave 1 and Wave 2 (equal)