A wave travelling along the x-axis is described by the equation y (x, t) = 0.005 sin (`alphax - betat`). If the wavelength and time period of the wave are 0.08 m and 2 s respectively, then `alpha, beta` in appropriate units are

Options

A. `alpha=25pi,beta=pi`
B. `alpha=0.08/pi,beta=2/pi`
C. `alpha=0.04/pi,beta=1/pi`
D. `alpha=12.5pi,beta=pi/2`

Explanation

To find the values of α (alpha) and β (beta) in the wave equation \( y(x, t) = 0.005 \sin(\alpha x - \beta t) \), we can follow these steps:

### Step 1: Identify the wave parameters
The wave equation is given as:
\[ y(x, t) = 0.005 \sin(\alpha x - \beta t) \]
We know that:
- Wavelength (\( \lambda \)) = 0.08 m
- Time period (\( T \)) = 2 s

### Step 2: Relate α to the wavelength
In wave mechanics, the wave number \( k \) is related to the wavelength by the equation:
\[ k = \frac{2\pi}{\lambda} \]
Since \( k \) corresponds to \( \alpha \) in our equation, we can write:
\[ \alpha = \frac{2\pi}{\lambda} \]
Substituting the value of \( \lambda \):
\[ \alpha = \frac{2\pi}{0.08} \]

### Step 3: Calculate α
Now, we calculate \( \alpha \):
\[ \alpha = \frac{2\pi}{0.08} = \frac{2\pi}{0.08} = 25\pi \, \text{m}^{-1} \]

### Step 4: Relate β to the time period
The angular frequency \( \omega \) is related to the time period by the equation:
\[ \omega = \frac{2\pi}{T} \]
Since \( \omega \) corresponds to \( \beta \) in our equation, we can write:
\[ \beta = \frac{2\pi}{T} \]
Substituting the value of \( T \):
\[ \beta = \frac{2\pi}{2} \]

### Step 5: Calculate β
Now, we calculate \( \beta \):
\[ \beta = \frac{2\pi}{2} = \pi \, \text{s}^{-1} \]

### Final Result
Thus, the values of \( \alpha \) and \( \beta \) are:
- \( \alpha = 25\pi \, \text{m}^{-1} \)
- \( \beta = \pi \, \text{s}^{-1} \)