For the travelling harmonic wave
y(x, t) = 2 cos2`pi`(10t - 0.008x + 0.35) where X and Y are in cm and t is in s. The phase difference between oscillatory motion of two points separated by distance of 0.5 m is

Options

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

Explanation

To solve the problem, we need to find the phase difference between two points separated by a distance of 0.5 m for the given wave equation:

**Given wave equation:**
\[ y(x, t) = 2 \cos(2\pi(10t - 0.008x) + 0.35) \]

### Step 1: Identify the wave parameters
The general form of a traveling wave is:
\[ y(x, t) = A \cos(2\pi ft - kx + \phi) \]
where:
- \( A \) is the amplitude,
- \( f \) is the frequency,
- \( k \) is the wave number,
- \( \phi \) is the phase constant.

From the given equation, we can identify:
- \( A = 2 \) cm,
- \( f = 10 \) Hz,
- \( k = 0.008 \) cm\(^{-1}\).

### Step 2: Calculate the wavelength
The wave number \( k \) is related to the wavelength \( \lambda \) by the equation:
\[ k = \frac{2\pi}{\lambda} \]

Rearranging gives:
\[ \lambda = \frac{2\pi}{k} \]

Substituting \( k = 0.008 \) cm\(^{-1}\):
\[ \lambda = \frac{2\pi}{0.008} \]

Calculating \( \lambda \):
\[ \lambda \approx \frac{6.2832}{0.008} \approx 785.4 \text{ cm} \]

### Step 3: Calculate the phase difference
The phase difference \( \phi \) between two points separated by a distance \( x \) is given by:
\[ \phi = k \cdot x \]

Here, \( x = 0.5 \) m, which needs to be converted to centimeters:
\[ x = 0.5 \text{ m} = 50 \text{ cm} \]

Now substituting the values:
\[ \phi = k \cdot x = 0.008 \cdot 50 \]

Calculating \( \phi \):
\[ \phi = 0.008 \cdot 50 = 0.4 \text{ radians} \]

### Step 4: Conclusion
The phase difference between the oscillatory motion of two points separated by a distance of 0.5 m is:
\[ \phi = 0.4 \text{ radians} \]