A plane wave propagates along positive x-direction in a homogeneous medium of density `p=200 kg//m^(3)`. Due to propagation of the wave medium particle oscillate. Space density of their oscillation energy is `E=0.16pi^(2) J//m^(3)` and maximum shear strain produced in the mendium is `phi_(0)=8pixx10^(-5)`. if at an instant, phase difference between two particles located at points `(1m,1m,1m)` and `(2m, 2m, 2m,)` is `Deltatheta=144^(@)`, assuming at `t=0` phase of particle at `x=0` to be zero,
Equation of wave is

Options

A. `y pi 10^(-4)sinpipi2000t-0.8xpi`
B. `y pi 10^(-4)sinpipi400t-0.8xpi`
C. `y pi 10^(-4)sinpipi100t-8xpi`
D. `y pi 10^(-4)sinpipi100t-2xpi`

Explanation

To find the equation of the wave propagating in a homogeneous medium, we will follow these steps:

### Step 1: Identify Given Data
- Density of the medium, \( p = 200 \, \text{kg/m}^3 \)
- Space density of oscillation energy, \( E = 0.16 \pi^2 \, \text{J/m}^3 \)
- Maximum shear strain, \( \phi_0 = 8 \pi \times 10^{-5} \)
- Phase difference between two particles, \( \Delta \theta = 144^\circ \)

### Step 2: Convert Phase Difference to Radians
Convert \( \Delta \theta \) from degrees to radians:
\[ \Delta \theta = 144^\circ \times \frac{\pi}{180} = \frac{144 \pi}{180} = \frac{4\pi}{5} \, \text{radians} \]

### Step 3: Calculate Wave Speed
The energy density \( E \) is related to the amplitude \( A \), frequency \( f \), and density \( p \) by the formula:
\[ E = \frac{1}{2} p A^2 \omega^2 \]
where \( \omega = 2 \pi f \).

Rearranging gives:
\[ A^2 = \frac{2E}{p \omega^2} \]

### Step 4: Relate Shear Strain to Amplitude
The maximum shear strain \( \phi_0 \) is given by:
\[ \phi_0 = A \omega \]
From this, we can express \( A \):
\[ A = \frac{\phi_0}{\omega} \]

### Step 5: Calculate Frequency and Angular Frequency
Using the wave speed \( v \):
\[ v = f \lambda \]
We need to find the wavelength \( \lambda \) using the phase difference:
\[ \Delta \theta = \frac{2\pi \Delta x}{\lambda} \]
Given \( \Delta x = 2 - 1 = 1 \, \text{m} \):
\[ \lambda = \frac{2\pi \Delta x}{\Delta \theta} = \frac{2\pi \times 1}{\frac{4\pi}{5}} = \frac{5}{2} = 2.5 \, \text{m} \]

Now, using the wave speed:
\[ v = \frac{\lambda}{T} = f \lambda \]
We can find \( f \):
\[ f = \frac{v}{\lambda} = \frac{500}{2.5} = 200 \, \text{Hz} \]
And the angular frequency:
\[ \omega = 2\pi f = 2\pi \times 200 = 400\pi \, \text{rad/s} \]

### Step 6: Substitute Values into the Wave Equation
The general form of the wave equation is:
\[ y(x, t) = A \sin(\omega t - kx) \]
Where \( k = \frac{2\pi}{\lambda} = \frac{2\pi}{2.5} = \frac{4\pi}{5} \).

### Step 7: Calculate Amplitude
Using the relationship between \( A \) and \( \phi_0 \):
\[ A = \frac{\phi_0}{\omega} = \frac{8\pi \times 10^{-5}}{400\pi} = 2 \times 10^{-7} \, \text{m} \]

### Step 8: Final Wave Equation
Substituting \( A \), \( \omega \), and \( k \) into the wave equation:
\[ y(x, t) = 2 \times 10^{-7} \sin(400\pi t - \frac{4\pi}{5} x) \]

### Final Answer
The equation of the wave is:
\[ y(x, t) = 2 \times 10^{-7} \sin(400\pi t - \frac{4\pi}{5} x) \]