`y_(t)=2sin3pit`
`y_(2)=2sin(3pit-(pi)/(8))`
The wave velocity is , if the path difference is 1cm .

Options

A. `16cm//sec`
B. `24cm//sec`
C. `12cm//sec`
D. `8cm//sec`

Answer

To solve the problem step by step, let's break it down:

### Step 1: Identify the given information
We have two wave equations:
1. \( y_1(t) = 2 \sin(3 \pi t) \)
2. \( y_2(t) = 2 \sin(3 \pi t - \frac{\pi}{8}) \)

We also know that the path difference \( \Delta x = 1 \) cm.

### Step 2: Determine the phase difference
From the equations, we can see that the phase difference \( \Delta \phi \) between the two waves is:
\[ \Delta \phi = \frac{\pi}{8} \]

### Step 3: Relate phase difference to path difference
The relationship between phase difference and path difference is given by:
\[ \Delta \phi = \frac{2 \pi}{\lambda} \Delta x \]
Substituting the known values:
\[ \frac{\pi}{8} = \frac{2 \pi}{\lambda} \cdot 1 \text{ cm} \]

### Step 4: Solve for wavelength \( \lambda \)
To isolate \( \lambda \), we can rearrange the equation:
\[ \lambda = \frac{2 \pi \cdot 1 \text{ cm}}{\frac{\pi}{8}} \]
This simplifies to:
\[ \lambda = \frac{2 \pi \cdot 1 \text{ cm} \cdot 8}{\pi} = 16 \text{ cm} \]

### Step 5: Find the frequency \( f \)
We need to find the frequency \( f \). From the wave equation, we know:
\[ y = A \sin(\omega t - kx) \]
Comparing \( y_1(t) = 2 \sin(3 \pi t) \) with the standard form, we see:
\[ \omega = 3 \pi \]
Using the relationship \( \omega = 2 \pi f \):
\[ 3 \pi = 2 \pi f \]
Solving for \( f \):
\[ f = \frac{3 \pi}{2 \pi} = \frac{3}{2} \text{ Hz} \]

### Step 6: Calculate wave velocity \( v \)
The wave velocity \( v \) is given by:
\[ v = f \cdot \lambda \]
Substituting the values we found:
\[ v = \left(\frac{3}{2} \text{ Hz}\right) \cdot (16 \text{ cm}) = 24 \text{ cm/s} \]

### Final Answer
Thus, the wave velocity is:
\[ \boxed{24 \text{ cm/s}} \]