The equation of a transverse wave is given by ` psi =10 ^(-2) sin pi [30t -x sqrt3 -y] `
where x,y and `psi` are in metre and t in second If phase difference between two points ` A( 2sqrt3m ,2m ) and B ( 3sqrt 3m ,3m ) be npi ` .Find the value of n.

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to find the phase difference between two points A and B given the wave equation. The wave equation is:

\[ \psi = 10^{-2} \sin \left( \pi \left( 30t - x\sqrt{3} - y \right) \right) \]

### Step 1: Identify the phase of the wave
The phase of the wave can be expressed as:

\[ \phi = \pi \left( 30t - x\sqrt{3} - y \right) \]

### Step 2: Calculate the phase at point A
Point A has coordinates \( A(2\sqrt{3}, 2) \). We substitute these values into the phase equation:

\[ \phi_A = \pi \left( 30t - (2\sqrt{3})\sqrt{3} - 2 \right) \]

Calculating \( (2\sqrt{3})\sqrt{3} \):

\[ (2\sqrt{3})\sqrt{3} = 2 \cdot 3 = 6 \]

Now substituting this back into the phase equation:

\[ \phi_A = \pi \left( 30t - 6 - 2 \right) = \pi \left( 30t - 8 \right) \]

### Step 3: Calculate the phase at point B
Point B has coordinates \( B(3\sqrt{3}, 3) \). We substitute these values into the phase equation:

\[ \phi_B = \pi \left( 30t - (3\sqrt{3})\sqrt{3} - 3 \right) \]

Calculating \( (3\sqrt{3})\sqrt{3} \):

\[ (3\sqrt{3})\sqrt{3} = 3 \cdot 3 = 9 \]

Now substituting this back into the phase equation:

\[ \phi_B = \pi \left( 30t - 9 - 3 \right) = \pi \left( 30t - 12 \right) \]

### Step 4: Find the phase difference
The phase difference \( \Delta \phi \) between points A and B is given by:

\[ \Delta \phi = \phi_B - \phi_A \]

Substituting the expressions we found:

\[ \Delta \phi = \pi \left( 30t - 12 \right) - \pi \left( 30t - 8 \right) \]

This simplifies to:

\[ \Delta \phi = \pi \left( -12 + 8 \right) = \pi \left( -4 \right) = -4\pi \]

### Step 5: Relate phase difference to \( n\pi \)
According to the problem, the phase difference is given as \( n\pi \). Therefore, we have:

\[ -4\pi = n\pi \]

Dividing both sides by \( \pi \):

\[ n = -4 \]

### Final Answer
The value of \( n \) is:

\[ \boxed{-4} \]