A transverse wave is propagating along `+x` direction. At `t=2` sec the particle at `x=4`m is at `y=2` mm. With the passage of time its `y` coordinate increases and reaches to a maximum of 4 mm. The wave equation is (using `omega` and `k` with their usual meanings)

Options

A. `y=4sin(omega(t+2)+k(x-2)+(pi)/(6))`
B. `y=4sin(omega(t+2)+k(x)+(pi)/(6))`
C. `y=4sin(omega(t-2)-k(x-4)+(5pi)/(6))`
D. `y=4sin(omega(t-2)-k(x-4)+(5pi)/(6))`

Answer

To solve the problem step by step, we will derive the wave equation based on the given information.

### Step 1: Write the standard wave equation
The standard equation for a transverse wave propagating in the positive x-direction is given by:
\[ y = A \sin(\omega t - kx + \phi) \]
where:
- \(y\) is the displacement,
- \(A\) is the amplitude,
- \(\omega\) is the angular frequency,
- \(k\) is the wave number,
- \(t\) is time,
- \(x\) is the position along the wave,
- \(\phi\) is the phase constant.

### Step 2: Identify the given values
From the problem statement:
- At \(t = 2\) seconds and \(x = 4\) m, the particle's displacement \(y = 2\) mm = \(2 \times 10^{-3}\) m.
- The maximum displacement (amplitude) \(A = 4\) mm = \(4 \times 10^{-3}\) m.

### Step 3: Substitute known values into the wave equation
Substituting the known values into the wave equation:
\[ 2 \times 10^{-3} = 4 \times 10^{-3} \sin(2\omega - 4k + \phi) \]

### Step 4: Simplify the equation
Dividing both sides by \(4 \times 10^{-3}\):
\[ \frac{2 \times 10^{-3}}{4 \times 10^{-3}} = \sin(2\omega - 4k + \phi) \]
This simplifies to:
\[ \frac{1}{2} = \sin(2\omega - 4k + \phi) \]

### Step 5: Determine the angle corresponding to sine
The sine function equals \(\frac{1}{2}\) at:
\[ 2\omega - 4k + \phi = \frac{\pi}{6} \quad \text{(1st quadrant)} \]
or
\[ 2\omega - 4k + \phi = \frac{5\pi}{6} \quad \text{(2nd quadrant)} \]
Since the problem states that the displacement is increasing, we will use the first case:
\[ 2\omega - 4k + \phi = \frac{\pi}{6} \]

### Step 6: Solve for \(\phi\)
Rearranging gives:
\[ \phi = \frac{\pi}{6} - 2\omega + 4k \]

### Step 7: Substitute \(\phi\) back into the wave equation
Now substituting \(\phi\) back into the standard wave equation:
\[ y = 4 \times 10^{-3} \sin(\omega t - kx + \left(\frac{\pi}{6} - 2\omega + 4k\right)) \]

### Step 8: Simplify the wave equation
This can be simplified to:
\[ y = 4 \times 10^{-3} \sin\left(\omega t - kx + \frac{\pi}{6} - 2\omega + 4k\right) \]
Rearranging gives:
\[ y = 4 \times 10^{-3} \sin\left(\omega t - 2\omega - kx + 4k + \frac{\pi}{6}\right) \]

### Final Wave Equation
Thus, the final wave equation is:
\[ y = 4 \times 10^{-3} \sin\left(\omega t - kx + \frac{\pi}{6} - 2\omega + 4k\right) \]