Two points are located at a distance of `10 m` and `15 m` from the source of oscillation. The period of oscillation is `0.05 `s and the velocity of the wave is `300m//s`. What is the phase difference between the oscillation of two points?

Options

A. `(pi)/(3)`
B. `(2pi)/(3)`
C. `pi`
D. `(pi)/(6)`

Explanation

To find the phase difference between the oscillations at two points located at distances of 10 m and 15 m from the source of oscillation, we can follow these steps:

### Step 1: Identify the distances from the source
Let:
- \( x_1 = 10 \, \text{m} \) (distance of the first point from the source)
- \( x_2 = 15 \, \text{m} \) (distance of the second point from the source)

### Step 2: Calculate the path difference
The path difference \( \Delta x \) between the two points is given by:
\[ \Delta x = x_2 - x_1 = 15 \, \text{m} - 10 \, \text{m} = 5 \, \text{m} \]

### Step 3: Determine the wave properties
Given:
- Period \( T = 0.05 \, \text{s} \)
- Wave velocity \( v = 300 \, \text{m/s} \)

### Step 4: Calculate the frequency
The frequency \( f \) can be calculated using the formula:
\[ f = \frac{1}{T} = \frac{1}{0.05 \, \text{s}} = 20 \, \text{Hz} \]

### Step 5: Calculate the wavelength
Using the relationship between wave velocity, frequency, and wavelength:
\[ v = f \lambda \implies \lambda = \frac{v}{f} = \frac{300 \, \text{m/s}}{20 \, \text{Hz}} = 15 \, \text{m} \]

### Step 6: Calculate the phase difference
The phase difference \( \Delta \phi \) can be calculated using the formula:
\[ \Delta \phi = \frac{2 \pi}{\lambda} \Delta x \]
Substituting the values:
\[ \Delta \phi = \frac{2 \pi}{15 \, \text{m}} \times 5 \, \text{m} \]
\[ \Delta \phi = \frac{2 \pi \times 5}{15} = \frac{10 \pi}{15} = \frac{2 \pi}{3} \]

### Final Answer
The phase difference between the oscillation of the two points is:
\[ \Delta \phi = \frac{2 \pi}{3} \, \text{radians} \]

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