The equation of a simple harmonic wave is given by `y = 6 sin 2pi ( 2t – 0.1x)` ,where x and y are in mm and t is in second. The phase difference between two particles 2 mm apart at any instant is

Options

A. `18^@`
B. `36^@`
C. `54^@`
D. `72^@`

Answer

To solve the problem, we need to find the phase difference between two particles that are 2 mm apart in a simple harmonic wave described by the equation \( y = 6 \sin(2\pi(2t - 0.1x)) \).

### Step-by-Step Solution:

1. **Identify the Wave Equation**:
The given wave equation is:
\[ y = 6 \sin(2\pi(2t - 0.1x)) \]
Here, we can identify the angular frequency \( \omega \) and the wave number \( k \).

2. **Extract Parameters**:
- From the equation, we can rewrite it as:
\[ y = 6 \sin(\omega t - kx) \]
where \( \omega = 4\pi \) (since \( 2\pi \times 2 = 4\pi \)) and \( k = 0.2\pi \) (since \( 2\pi \times 0.1 = 0.2\pi \)).

3. **Calculate Wavelength (\( \lambda \))**:
The relationship between \( k \) and \( \lambda \) is given by:
\[ k = \frac{2\pi}{\lambda} \]
Rearranging gives:
\[ \lambda = \frac{2\pi}{k} \]
Substituting \( k = 0.2\pi \):
\[ \lambda = \frac{2\pi}{0.2\pi} = 10 \text{ mm} \]

4. **Determine Path Difference**:
The path difference (\( \Delta x \)) between the two particles is given as 2 mm.

5. **Calculate Phase Difference (\( \Delta \phi \))**:
The phase difference is calculated using the formula:
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \]
Substituting \( \lambda = 10 \text{ mm} \) and \( \Delta x = 2 \text{ mm} \):
\[ \Delta \phi = \frac{2\pi}{10} \times 2 = \frac{4\pi}{10} = \frac{2\pi}{5} \text{ radians} \]

6. **Convert Phase Difference to Degrees**:
To convert radians to degrees, we use the conversion factor \( \frac{180}{\pi} \):
\[ \Delta \phi = \frac{2\pi}{5} \times \frac{180}{\pi} = \frac{2 \times 180}{5} = 72 \text{ degrees} \]

### Final Answer:
The phase difference between the two particles 2 mm apart is \( 72 \) degrees.