A wave travelling in positive X-direction with `A=0.2m` has a velocity of `36 m//sec` if `lambda=60m`, then correct expression for the wave is

Options

A. `y=0.2sin[2pi(6t+(x)/(60))]`
B. `y=0.2sin[pi(6t+(x)/(60))]`
C. `y=0.2sin[2pi(6t-(x)/(60))]`
D. `y=0.2sin[pi(6t-(x)/(60))]`

Explanation

To find the correct expression for the wave traveling in the positive X-direction, we will follow these steps:

### Step 1: Identify the wave equation
The standard form of a wave traveling in the positive x-direction is given by:
\[ y = A \sin(\omega t - kx) \]
where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( k \) is the wave number.

### Step 2: Calculate the wave number \( k \)
The wave number \( k \) is calculated using the formula:
\[ k = \frac{2\pi}{\lambda} \]
Given \( \lambda = 60 \, m \):
\[ k = \frac{2\pi}{60} = \frac{\pi}{30} \, \text{m}^{-1} \]

### Step 3: Calculate the frequency \( f \)
The frequency \( f \) can be found using the relationship between velocity \( v \), wavelength \( \lambda \), and frequency:
\[ v = f \lambda \]
Rearranging gives:
\[ f = \frac{v}{\lambda} \]
Given \( v = 36 \, m/s \):
\[ f = \frac{36}{60} = 0.6 \, Hz \]

### Step 4: Calculate the angular frequency \( \omega \)
The angular frequency \( \omega \) is related to the frequency \( f \) by the formula:
\[ \omega = 2\pi f \]
Substituting the value of \( f \):
\[ \omega = 2\pi \times 0.6 = 1.2\pi \, \text{rad/s} \]

### Step 5: Substitute values into the wave equation
Now we can substitute the values of \( A \), \( \omega \), and \( k \) into the wave equation:
\[ y = 0.2 \sin(1.2\pi t - \frac{\pi}{30} x) \]

### Final Expression
Thus, the correct expression for the wave is:
\[ y = 0.2 \sin(1.2\pi t - \frac{\pi}{30} x) \]