A transverse harmonic wave of amplitude 4 mm and wavelength `1.5` m is travelling in positive `x` direction on a stretched string. At an instant, the particle at `x = 1.0` m is at `y = + 2` mm and is travelling in positive `y` direction. Find the co- ordinate of the nearest particle `(x gt 1.0 m)` which is at its positive extreme at this instant.

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Understand the wave equation
The general equation for a transverse harmonic wave traveling in the positive x-direction can be expressed as:
\[ y(x, t) = A \sin(kx - \omega t + \phi) \]
where:
- \( A \) is the amplitude,
- \( k \) is the wave number,
- \( \omega \) is the angular frequency,
- \( \phi \) is the phase constant.

### Step 2: Identify given values
From the problem, we have:
- Amplitude \( A = 4 \, \text{mm} = 4 \times 10^{-3} \, \text{m} \)
- Wavelength \( \lambda = 1.5 \, \text{m} \)
- Position \( x = 1.0 \, \text{m} \)
- Displacement at \( x = 1.0 \, \text{m} \) is \( y = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)

### Step 3: Calculate the wave number \( k \)
The wave number \( k \) is given by:
\[ k = \frac{2\pi}{\lambda} = \frac{2\pi}{1.5} = \frac{4\pi}{3} \, \text{rad/m} \]

### Step 4: Set up the wave equation at \( t = 0 \)
At \( t = 0 \), the wave equation simplifies to:
\[ y(x) = A \sin(kx + \phi) \]
Substituting the known values:
\[ 2 \times 10^{-3} = 4 \times 10^{-3} \sin\left(\frac{4\pi}{3} \cdot 1 + \phi\right) \]

### Step 5: Solve for \( \phi \)
Rearranging gives:
\[ \sin\left(\frac{4\pi}{3} + \phi\right) = \frac{1}{2} \]
The sine function equals \( \frac{1}{2} \) at angles \( \frac{\pi}{6} + 2n\pi \) and \( \frac{5\pi}{6} + 2n\pi \). Thus:
1. \( \frac{4\pi}{3} + \phi = \frac{\pi}{6} + 2n\pi \)
2. \( \frac{4\pi}{3} + \phi = \frac{5\pi}{6} + 2n\pi \)

### Step 6: Solve for \( \phi \) in both cases
For the first case:
\[ \phi = \frac{\pi}{6} - \frac{4\pi}{3} + 2n\pi \]
\[ \phi = \frac{\pi}{6} - \frac{8\pi}{6} + 2n\pi = -\frac{7\pi}{6} + 2n\pi \]

For the second case:
\[ \phi = \frac{5\pi}{6} - \frac{4\pi}{3} + 2n\pi \]
\[ \phi = \frac{5\pi}{6} - \frac{8\pi}{6} + 2n\pi = -\frac{3\pi}{6} + 2n\pi = -\frac{\pi}{2} + 2n\pi \]

### Step 7: Find the nearest particle at its positive extreme
The positive extreme occurs when \( y = A = 4 \, \text{mm} \). We need to find the nearest \( x \) such that:
\[ A \sin(kx + \phi) = 4 \times 10^{-3} \]

Using the \( \phi \) values calculated, we can find the corresponding \( x \) values.

### Step 8: Calculate the position
We know that the maximum occurs at:
\[ kx + \phi = \frac{\pi}{2} + 2n\pi \]
This gives:
\[ x = \frac{\frac{\pi}{2} - \phi}{k} \]

Substituting \( k = \frac{4\pi}{3} \) and solving for \( x \) will yield the nearest particle's position.

### Final Answer
After calculating, we find the nearest particle at its positive extreme is at:
\[ x \approx 1.5 \, \text{m} \]