When a wave travels in a medium displacement of a particle is given by ` y(x,t) =0.03sin pi (2t-0.01x)` where y and x are in metres and in seconds,The phase difference at a given instant of time between two paticles 25 m apart in the medium is

Options

A. ` pi //8`
B. `pi//4`
C. `pi//2`
D. `pi`

Answer

To find the phase difference between two particles that are 25 meters apart in a medium where the displacement of a particle is given by the wave equation \( y(x,t) = 0.03 \sin(\pi(2t - 0.01x)) \), we can follow these steps:

### Step 1: Identify the wave parameters
The given wave equation can be compared with the standard wave equation:
\[ y(x,t) = A \sin(\omega t - kx) \]
From the equation \( y(x,t) = 0.03 \sin(\pi(2t - 0.01x)) \), we can identify:
- Amplitude \( A = 0.03 \) m
- Angular frequency \( \omega = 2\pi \) (since \( \pi(2t) = 2\pi t \))
- Wave number \( k = 0.01\pi \)

### Step 2: Calculate the wavelength \( \lambda \)
The wave number \( k \) is related to the wavelength \( \lambda \) by the formula:
\[ k = \frac{2\pi}{\lambda} \]
Substituting the value of \( k \):
\[ 0.01\pi = \frac{2\pi}{\lambda} \]
Solving for \( \lambda \):
\[ \lambda = \frac{2\pi}{0.01\pi} = \frac{2}{0.01} = 200 \text{ m} \]

### Step 3: Use the phase difference formula
The phase difference \( \Delta \phi \) between two points separated by a distance \( \Delta x \) is given by:
\[ \Delta \phi = k \Delta x \]
We know \( \Delta x = 25 \) m and \( k = 0.01\pi \):
\[ \Delta \phi = (0.01\pi)(25) = 0.25\pi \]

### Step 4: Express the phase difference in terms of \( \pi \)
The phase difference can also be expressed as:
\[ \Delta \phi = 0.25\pi = \frac{\pi}{4} \]

### Final Answer
The phase difference at a given instant of time between two particles 25 m apart in the medium is:
\[ \Delta \phi = \frac{\pi}{4} \]

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