Waves `y_(1) = Acos(0.5pix - 100pit)` and `y_(2)=Acos(0.46pix - 92pit)` are travelling along x-axis. (Here `x` is in `m` and `t` is in second)
(2) The wave velocity of louder sound is

Options

A. 100 m//s
B. 192 m//s
C. 200 m//s
D. 96 m//s

Explanation

To solve the problem, we need to find the wave velocity of the two given waves and determine which one corresponds to the louder sound. However, the question states that both waves are traveling in the same medium, so they will have the same velocity.

### Step-by-Step Solution:

1. **Identify the wave equations**:
- The first wave is given by:
\[ y_1 = A \cos(0.5 \pi x - 100 \pi t) \]
- The second wave is given by:
\[ y_2 = A \cos(0.46 \pi x - 92 \pi t) \]

2. **Extract the wave numbers (k) and angular frequencies (ω)**:
- For the first wave \( y_1 \):
- Wave number \( k_1 = 0.5 \pi \) (from the term \( 0.5 \pi x \))
- Angular frequency \( \omega_1 = 100 \pi \) (from the term \( 100 \pi t \))
- For the second wave \( y_2 \):
- Wave number \( k_2 = 0.46 \pi \)
- Angular frequency \( \omega_2 = 92 \pi \)

3. **Use the wave velocity formula**:
- The wave velocity \( v \) is given by the formula:
\[ v = \frac{\omega}{k} \]

4. **Calculate the wave velocity for the first wave**:
- For the first wave:
\[ v_1 = \frac{\omega_1}{k_1} = \frac{100 \pi}{0.5 \pi} = \frac{100}{0.5} = 200 \text{ m/s} \]

5. **Calculate the wave velocity for the second wave**:
- For the second wave:
\[ v_2 = \frac{\omega_2}{k_2} = \frac{92 \pi}{0.46 \pi} = \frac{92}{0.46} = 200 \text{ m/s} \]

6. **Conclusion**:
- Both waves have the same velocity of \( 200 \text{ m/s} \). Since the problem states that the wave velocity of the louder sound is to be found, but both waves have the same velocity, the answer remains:
\[ \text{Wave velocity of the louder sound} = 200 \text{ m/s} \]