The mathmaticaly form of three travelling waves are given by
`Y_(1)=(2 cm) sin (3x-6t)`
`Y_(2)=(3 cm) sin (4x-12t)`
And `Y_(3)=94 cm) sin(5x-11t)`
of these waves,

Options

A. Wave 1 has greatest wave speed and greatest maximum transverse string speed
B. wave 2 has greatest wave speed and wave 1 has greatest maximum transverse string speed
C. wave 3 has greatest wave speed and wave 1 has greatest maximum transverse string speed
D. wave 2 has greatest wave speed and wave 3 has greatest maximum transverse string speed

Explanation

To solve the problem, we need to analyze the three given traveling waves and determine which has the greatest wave speed and which has the greatest maximum transverse string speed.

### Step 1: Identify the wave equations
The three traveling waves are given as:
1. \( Y_1 = (2 \, \text{cm}) \sin(3x - 6t) \)
2. \( Y_2 = (3 \, \text{cm}) \sin(4x - 12t) \)
3. \( Y_3 = (94 \, \text{cm}) \sin(5x - 11t) \)

### Step 2: Calculate wave speed for each wave
The wave speed \( v \) can be calculated using the formula:
\[ v = \frac{\omega}{k} \]
where \( \omega \) is the coefficient of \( t \) and \( k \) is the coefficient of \( x \).

- For \( Y_1 \):
- \( \omega_1 = 6 \) (from \( -6t \))
- \( k_1 = 3 \) (from \( 3x \))
- Wave speed \( v_1 = \frac{6}{3} = 2 \, \text{cm/s} \)

- For \( Y_2 \):
- \( \omega_2 = 12 \) (from \( -12t \))
- \( k_2 = 4 \) (from \( 4x \))
- Wave speed \( v_2 = \frac{12}{4} = 3 \, \text{cm/s} \)

- For \( Y_3 \):
- \( \omega_3 = 11 \) (from \( -11t \))
- \( k_3 = 5 \) (from \( 5x \))
- Wave speed \( v_3 = \frac{11}{5} = 2.2 \, \text{cm/s} \)

### Step 3: Compare wave speeds
From the calculations:
- \( v_1 = 2 \, \text{cm/s} \)
- \( v_2 = 3 \, \text{cm/s} \)
- \( v_3 = 2.2 \, \text{cm/s} \)

The wave with the greatest speed is \( Y_2 \) with \( v_2 = 3 \, \text{cm/s} \).

### Step 4: Calculate maximum transverse string speed for each wave
The maximum transverse string speed (particle velocity) can be calculated using the formula:
\[ v_{max} = \omega \cdot A \]
where \( A \) is the amplitude.

- For \( Y_1 \):
- Amplitude \( A_1 = 2 \, \text{cm} \)
- Maximum speed \( v_{max1} = 6 \cdot 2 = 12 \, \text{cm/s} \)

- For \( Y_2 \):
- Amplitude \( A_2 = 3 \, \text{cm} \)
- Maximum speed \( v_{max2} = 12 \cdot 3 = 36 \, \text{cm/s} \)

- For \( Y_3 \):
- Amplitude \( A_3 = 94 \, \text{cm} \)
- Maximum speed \( v_{max3} = 11 \cdot 94 = 1034 \, \text{cm/s} \)

### Step 5: Compare maximum transverse string speeds
From the calculations:
- \( v_{max1} = 12 \, \text{cm/s} \)
- \( v_{max2} = 36 \, \text{cm/s} \)
- \( v_{max3} = 1034 \, \text{cm/s} \)

The wave with the greatest maximum transverse string speed is \( Y_3 \) with \( v_{max3} = 1034 \, \text{cm/s} \).

### Final Conclusion
- The wave with the greatest wave speed is \( Y_2 \).
- The wave with the greatest maximum transverse string speed is \( Y_3 \).