A transverse wave, whose amplitude is 0.5 m wavelength is 1 m and frequency is 2Hz. is travelling along positive x direction. The equation of this wave will be-

Options

A. `y = 0.5 cos (2pix - 4pit)`
B. `y = 0.5 cos (2pix - 4pit)`
C. `y = 0.5 sin (pix - 2pit)`
D. `y = 0.5 cos (2pix - 2pit)`

Answer

To find the equation of the transverse wave traveling in the positive x-direction, we can follow these steps:

### Step 1: Write the general form of the wave equation
The equation of a transverse wave traveling in the positive x-direction can be expressed as:
\[ y = A \sin(kx - \omega t) \]
or
\[ y = A \cos(kx - \omega t) \]
where:
- \( A \) is the amplitude,
- \( k \) is the wave number,
- \( \omega \) is the angular frequency,
- \( x \) is the position,
- \( t \) is the time.

### Step 2: Identify the amplitude
From the problem, we know that the amplitude \( A \) is given as:
\[ A = 0.5 \, \text{m} \]

### Step 3: Calculate the wave number \( k \)
The wave number \( k \) is calculated using the formula:
\[ k = \frac{2\pi}{\lambda} \]
where \( \lambda \) is the wavelength. Given that the wavelength \( \lambda = 1 \, \text{m} \):
\[ k = \frac{2\pi}{1} = 2\pi \, \text{rad/m} \]

### Step 4: Calculate the angular frequency \( \omega \)
The angular frequency \( \omega \) is calculated using the formula:
\[ \omega = 2\pi f \]
where \( f \) is the frequency. Given that the frequency \( f = 2 \, \text{Hz} \):
\[ \omega = 2\pi \times 2 = 4\pi \, \text{rad/s} \]

### Step 5: Substitute the values into the wave equation
Now we can substitute the values of \( A \), \( k \), and \( \omega \) into the wave equation. We can choose either the sine or cosine form. Since the problem suggests using the cosine form, we write:
\[ y = 0.5 \cos(kx - \omega t) \]
Substituting the values of \( k \) and \( \omega \):
\[ y = 0.5 \cos(2\pi x - 4\pi t) \]

### Step 6: Final equation
Thus, the equation of the wave is:
\[ y = 0.5 \cos(2\pi x - 4\pi t) \]