Two waves `y_(1) = A cos (0.5 pi x - 100 pi t)` and `y_(2) = A cos (0.46 pi x - 92 pi t)` are travelling in a pipe placed along the `x ` - axis.
Find wave velocity of louder sound

Options

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

Explanation

To find the wave velocity of the louder sound from the given waves \( y_1 = A \cos(0.5 \pi x - 100 \pi t) \) and \( y_2 = A \cos(0.46 \pi x - 92 \pi t) \), we will follow these steps:

### Step 1: Identify the angular frequency (\(\omega\)) and wave number (\(k\)) for both waves.

For wave \( y_1 \):
- The angular frequency \(\omega_1 = 100 \pi\) (from the term \(-100 \pi t\))
- The wave number \(k_1 = 0.5 \pi\) (from the term \(0.5 \pi x\))

For wave \( y_2 \):
- The angular frequency \(\omega_2 = 92 \pi\) (from the term \(-92 \pi t\))
- The wave number \(k_2 = 0.46 \pi\) (from the term \(0.46 \pi x\))

### Step 2: Calculate the wave velocity (\(v\)) for both waves 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 3: Compare the wave velocities.

Both waves have the same velocity:
\[ v_1 = v_2 = 200 \, \text{m/s} \]

### Step 4: Determine the louder sound.

Since both waves have the same velocity, we need to determine which wave is louder. The amplitude \(A\) is the same for both waves, so we need to consider the wave with the higher angular frequency \(\omega\) as it typically corresponds to a higher energy level, which can be interpreted as louder sound.

In this case, \(y_1\) has a higher angular frequency (\(100 \pi\)) than \(y_2\) (\(92 \pi\)), thus \(y_1\) is the louder sound.

### Final Answer:
The wave velocity of the louder sound is \(200 \, \text{m/s}\).

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