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 find the phase difference between the oscillatory motion of two points separated by a distance of 0.5 m for the given wave equation \( y(x, t) = 2 \cos(2\pi(10t - 0.008x + 0.35)) \), we can follow these steps:

### Step 1: Identify the wave parameters
The wave equation is given in the form:
\[ y(x, t) = A \cos(\omega t - kx + \phi) \]
From the equation, we can identify:
- Amplitude \( A = 2 \) cm
- Angular frequency \( \omega = 2\pi \times 10 = 20\pi \) rad/s
- Wave number \( k = 0.008 \) rad/cm
- Phase constant \( \phi = 0.35 \)

### Step 2: Convert the distance to the same unit
The distance between the two points is given as 0.5 m. We need to convert this to centimeters since the wave number \( k \) is in rad/cm:
\[ 0.5 \text{ m} = 0.5 \times 100 = 50 \text{ cm} \]

### Step 3: Calculate the phase difference
The phase difference \( \Delta \phi \) between two points separated by a distance \( d \) can be calculated using the formula:
\[ \Delta \phi = k \cdot d \]
Substituting the values we have:
\[ \Delta \phi = 0.008 \, \text{rad/cm} \times 50 \, \text{cm} \]
\[ \Delta \phi = 0.4 \, \text{rad} \]

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

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