A sinusoidal wave travels along a taut string of linear mass density `0.1g//cm`. The particles oscillate along `y-` direction and wave moves in the positive `x-` direction . The amplitude and frequency of oscillation are `2mm` and `50Hz` respectively. The minimum distance between two particles oscillating in the same phase is` 4m`.
If at `x=2m` and `t=2s`, the particle is at `y=1 m m ` and its velocity is in positive `y- ` direction, then the equation of this travelling wave is `: ( y` is in `m m, t` is in seconds and `x` is in metres `)`

Options

A. `y=2 sin ((pi x)/(2)-100 pi t + 30^(@))`
B. `y=2 sin ((pi x)/(2) - 100 pi t + 120 ^(@))`
C. `y=2 sin ((pi x)/(2)-100 pi t + 150^(@))`
D. None of these

Explanation

To find the equation of the traveling wave, we will follow these steps:

### Step 1: Identify the given parameters
- **Amplitude (A)**: Given as \(2 \, \text{mm} = 2 \times 10^{-3} \, \text{m}\)
- **Frequency (f)**: Given as \(50 \, \text{Hz}\)
- **Linear mass density**: Given as \(0.1 \, \text{g/cm} = 0.1 \times 10^{-3} \, \text{kg/m}\)
- **Minimum distance between two particles oscillating in the same phase**: Given as \(4 \, \text{m}\), which represents the wavelength (\(\lambda\)).

### Step 2: Calculate the wave number (k)
The wave number \(k\) is given by the formula:
\[ k = \frac{2\pi}{\lambda} \]
Substituting \(\lambda = 4 \, \text{m}\):
\[ k = \frac{2\pi}{4} = \frac{\pi}{2} \, \text{rad/m} \]

### Step 3: Calculate the angular frequency (\(\omega\))
The angular frequency \(\omega\) is given by:
\[ \omega = 2\pi f \]
Substituting \(f = 50 \, \text{Hz}\):
\[ \omega = 2\pi \times 50 = 100\pi \, \text{rad/s} \]

### Step 4: Use the wave equation
The general equation of a traveling wave moving in the positive x-direction is given by:
\[ y(x, t) = A \sin(kx - \omega t + \phi) \]
Where \(\phi\) is the phase constant that we need to determine.

### Step 5: Determine the phase constant (\(\phi\))
At \(x = 2 \, \text{m}\) and \(t = 2 \, \text{s}\), the particle's displacement \(y\) is given as \(1 \, \text{mm} = 1 \times 10^{-3} \, \text{m}\) and its velocity is in the positive y-direction.

The velocity \(v\) of the wave can be expressed as:
\[ v = \frac{dy}{dt} = A \omega \cos(kx - \omega t + \phi) \]
Since the velocity is positive, \(\cos(kx - \omega t + \phi)\) must be positive.

Substituting the values:
\[ 1 \times 10^{-3} = 2 \times 10^{-3} \sin\left(\frac{\pi}{2} \cdot 2 - 100\pi \cdot 2 + \phi\right) \]
This simplifies to:
\[ \frac{1}{2} = \sin\left(\pi - 200\pi + \phi\right) \]
\[ \frac{1}{2} = \sin(-199\pi + \phi) \]
Since \(\sin(-\theta) = -\sin(\theta)\):
\[ -\frac{1}{2} = \sin(199\pi - \phi) \]
The sine function equals \(-\frac{1}{2}\) at angles of \(-\frac{\pi}{6} + 2n\pi\) or \(\frac{7\pi}{6} + 2n\pi\).

For simplicity, we can take:
\[ 199\pi - \phi = -\frac{\pi}{6} \]
Solving for \(\phi\):
\[ \phi = 199\pi + \frac{\pi}{6} \]

### Step 6: Write the final equation
Now substituting \(A\), \(k\), \(\omega\), and \(\phi\) into the wave equation:
\[ y(x, t) = 2 \times 10^{-3} \sin\left(\frac{\pi}{2} x - 100\pi t + \left(199\pi + \frac{\pi}{6}\right)\right) \]

### Final Answer
Thus, the equation of the traveling wave is:
\[ y(x, t) = 2 \sin\left(\frac{\pi}{2} x - 100\pi t + \frac{199\pi}{6}\right) \, \text{mm} \]