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 louder sound between the two given wave equations. Let's break it down step by step.

### Step 1: Identify the wave equations
The two wave equations given are:
1. \( y_1 = A \cos(0.5 \pi x - 100 \pi t) \)
2. \( y_2 = A \cos(0.46 \pi x - 92 \pi t) \)

### Step 2: Identify the parameters
In the standard wave equation \( y = A \cos(kx - \omega t) \):
- \( A \) is the amplitude (which affects loudness),
- \( k \) is the wave number,
- \( \omega \) is the angular frequency.

From the equations:
- For \( y_1 \):
- \( k_1 = 0.5 \pi \)
- \( \omega_1 = 100 \pi \)

- For \( y_2 \):
- \( k_2 = 0.46 \pi \)
- \( \omega_2 = 92 \pi \)

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

#### For wave \( y_1 \):
\[ v_1 = \frac{\omega_1}{k_1} = \frac{100 \pi}{0.5 \pi} = \frac{100}{0.5} = 200 \, \text{m/s} \]

#### For wave \( y_2 \):
\[ v_2 = \frac{\omega_2}{k_2} = \frac{92 \pi}{0.46 \pi} = \frac{92}{0.46} = 200 \, \text{m/s} \]

### Step 4: Compare the amplitudes
Since both waves have the same amplitude \( A \), they have equal loudness.

### Conclusion
Both waves have the same wave velocity of \( 200 \, \text{m/s} \). Therefore, the wave velocity of the louder sound is:
\[ \text{Wave velocity of louder sound} = 200 \, \text{m/s} \]