The equation of a wave is y =4 `sin {(pi)/(2) (2t +(x)/(8))}` ,where y,x are in cm and time is in second .The phase difference between two positions of the same particles which are occupied at time interval of 0.4 s is

Options

A. ` 0.2 pi`
B. ` 0.4 pi `
C. ` 0.6 pi `
D. ` 0.8 pi`

Answer

To find the phase difference between two positions of the same particle in the wave described by the equation \( y = 4 \sin \left( \frac{\pi}{2} \left( 2t + \frac{x}{8} \right) \right) \) at a time interval of 0.4 seconds, we can follow these steps:

### Step 1: Identify the wave equation
The given wave equation is:
\[ y = 4 \sin \left( \frac{\pi}{2} \left( 2t + \frac{x}{8} \right) \right) \]
Here, \( y \) is the displacement, \( x \) is the position, and \( t \) is the time.

### Step 2: Set the position of the particle
To simplify the calculation, we can choose a specific position for the particle. Let's set \( x = 0 \). This means we are observing the particle at the origin.

### Step 3: Calculate the phase at time \( t = 0 \)
At \( t = 0 \):
\[ \text{Phase at } t = 0, \phi_1 = \frac{\pi}{2} \left( 2(0) + \frac{0}{8} \right) = \frac{\pi}{2} \times 0 = 0 \]

### Step 4: Calculate the phase at time \( t = 0.4 \)
Now, we calculate the phase at \( t = 0.4 \) seconds:
\[ \text{Phase at } t = 0.4, \phi_2 = \frac{\pi}{2} \left( 2(0.4) + \frac{0}{8} \right) = \frac{\pi}{2} \times 0.8 = 0.4\pi \]

### Step 5: Find the phase difference
The phase difference \( \Delta \phi \) between the two positions is given by:
\[ \Delta \phi = \phi_2 - \phi_1 = 0.4\pi - 0 = 0.4\pi \]

### Conclusion
The phase difference between the two positions of the same particle at a time interval of 0.4 seconds is:
\[ \Delta \phi = 0.4\pi \]

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