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 find the phase difference between two particles that are 2 mm apart in the given wave equation \( y = 6 \sin(2\pi(2t - 0.1x)) \), we will follow these steps:

### Step 1: Identify the wave number \( k \)
The standard form of a wave equation is given by:
\[ y = a \sin(\omega t - kx) \]
From the given equation, we can identify:
\[ \omega = 2\pi \times 2 \quad \text{and} \quad k = 2\pi \times 0.1 \]
Thus, we can calculate \( k \):
\[ k = 2\pi \times 0.1 = 0.2\pi \, \text{mm}^{-1} \]

### Step 2: Determine the path difference \( \Delta x \)
The problem states that the separation between the two particles is \( \Delta x = 2 \, \text{mm} \).

### Step 3: Calculate the phase difference \( \Delta \phi \)
The phase difference \( \Delta \phi \) between two points separated by a distance \( \Delta x \) is given by:
\[ \Delta \phi = k \Delta x \]
Substituting the values we found:
\[ \Delta \phi = (0.2\pi) \times (2) = 0.4\pi \]

### Step 4: Convert the phase difference to degrees
To convert radians to degrees, we use the conversion factor \( \frac{180}{\pi} \):
\[ \Delta \phi = 0.4\pi \times \frac{180}{\pi} = 0.4 \times 180 = 72 \, \text{degrees} \]

### Final Answer
The phase difference between the two particles 2 mm apart at any instant is \( 72 \, \text{degrees} \).

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