A sinusoidal wave is given by y=A sin (kx-`omegat`). The ratio of its maximum particle velocity to wave velocity is

Options

A. 1
B. `omega`
C. `omegaA`
D. `kA`

Answer

To find the ratio of the maximum particle velocity to the wave velocity for the given sinusoidal wave \( y = A \sin(kx - \omega t) \), we can follow these steps:

### Step 1: Identify the wave parameters
The wave is represented by the equation:
\[ y = A \sin(kx - \omega t) \]
where:
- \( A \) is the amplitude of the wave,
- \( k \) is the wave number,
- \( \omega \) is the angular frequency.

### Step 2: Calculate the maximum particle velocity
The maximum particle velocity (\( v_{\text{max}} \)) of a wave can be calculated using the formula:
\[ v_{\text{max}} = \omega A \]
This is derived from the fact that the particle velocity is the derivative of the displacement with respect to time.

### Step 3: Calculate the wave velocity
The wave velocity (\( v \)) is given by the formula:
\[ v = \frac{\omega}{k} \]
This relationship comes from the definition of wave speed in terms of angular frequency and wave number.

### Step 4: Find the ratio of maximum particle velocity to wave velocity
Now, we can find the ratio of the maximum particle velocity to the wave velocity:
\[ \text{Ratio} = \frac{v_{\text{max}}}{v} = \frac{\omega A}{\frac{\omega}{k}} \]

### Step 5: Simplify the ratio
When we simplify the ratio, we get:
\[ \text{Ratio} = \frac{\omega A \cdot k}{\omega} = kA \]

### Conclusion
Thus, the ratio of the maximum particle velocity to the wave velocity is:
\[ kA \]

### Final Answer
The ratio of maximum particle velocity to wave velocity is \( kA \).

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