The equation of a wave is `y = 4 sin {pi/2(2t + x/8)}` where y , x are in cm and time in seconds. The phase difference between two position of the same particle which are occupied at time interval of 0.4 s is `x pi xx10^(-1)` , what is the value of x ?

Answer

To solve the problem, we will follow these steps:

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

### Step 2: Determine the phase of the wave
The phase of the wave can be expressed as:
\[ \phi = \frac{\pi}{2} (2t + \frac{x}{8}) \]

### Step 3: Calculate the phase at two different times
1. **At time \( t = 0 \)**:
\[ \phi_1 = \frac{\pi}{2} \left( 2(0) + \frac{x}{8} \right) = \frac{\pi}{2} \cdot \frac{x}{8} = \frac{\pi x}{16} \]

2. **At time \( t = 0.4 \, \text{s} \)**:
\[ \phi_2 = \frac{\pi}{2} \left( 2(0.4) + \frac{x}{8} \right) = \frac{\pi}{2} \left( 0.8 + \frac{x}{8} \right) = \frac{\pi}{2} \cdot 0.8 + \frac{\pi x}{16} = 0.4\pi + \frac{\pi x}{16} \]

### Step 4: Calculate the phase difference
The phase difference \( \Delta \phi \) between the two positions of the same particle is given by:
\[ \Delta \phi = \phi_2 - \phi_1 = \left( 0.4\pi + \frac{\pi x}{16} \right) - \frac{\pi x}{16} \]
This simplifies to:
\[ \Delta \phi = 0.4\pi \]

### Step 5: Set the phase difference equal to the given expression
According to the problem, the phase difference is also given as:
\[ \Delta \phi = x \pi \times 10^{-1} \]
Setting the two expressions for phase difference equal gives:
\[ 0.4\pi = x \pi \times 10^{-1} \]

### Step 6: Solve for \( x \)
Dividing both sides by \( \pi \):
\[ 0.4 = x \times 10^{-1} \]
Multiplying both sides by \( 10 \):
\[ x = 4 \]

### Final Answer
The value of \( x \) is:
\[ \boxed{4} \]