Two particles executing SHM of same frequency, meet at `x= +A//2`, while moving in opposite direction . Phase difference between the particles is

Options

A. `(pi)/(6)`
B. `(pi)/(3)`
C. `(5pi)/(6)`
D. `(2pi)/(3)`

Answer

To solve the problem of finding the phase difference between two particles executing simple harmonic motion (SHM) that meet at \( x = +\frac{A}{2} \) while moving in opposite directions, we can follow these steps:

### Step 1: Write the equations of motion for both particles
The position of a particle executing SHM can be described by the equation:
\[ x = A \sin(\omega t + \phi) \]
where \( A \) is the amplitude, \( \omega \) is the angular frequency, \( t \) is time, and \( \phi \) is the phase constant.

### Step 2: Set up the equations for both particles
Let’s denote the two particles as Particle 1 and Particle 2. Since they meet at \( x = +\frac{A}{2} \), we can write:
- For Particle 1:
\[ x_1 = A \sin(\omega t + \phi_1) = \frac{A}{2} \]
- For Particle 2 (moving in the opposite direction):
\[ x_2 = A \sin(\omega t + \phi_2) = -\frac{A}{2} \]

### Step 3: Solve for the phase angles
From the equations above, we can isolate the sine functions:
1. For Particle 1:
\[ \sin(\omega t + \phi_1) = \frac{1}{2} \]
2. For Particle 2:
\[ \sin(\omega t + \phi_2) = -\frac{1}{2} \]

### Step 4: Determine the possible values for the phase angles
The sine function gives us specific angles for these values:
- For \( \sin(\theta) = \frac{1}{2} \):
\[ \theta = \frac{\pi}{6} \quad \text{or} \quad \theta = \frac{5\pi}{6} \]
Thus, we have:
\[ \omega t + \phi_1 = \frac{\pi}{6} \quad \text{or} \quad \omega t + \phi_1 = \frac{5\pi}{6} \]

- For \( \sin(\theta) = -\frac{1}{2} \):
\[ \theta = \frac{7\pi}{6} \quad \text{or} \quad \theta = \frac{11\pi}{6} \]
Thus, we have:
\[ \omega t + \phi_2 = \frac{7\pi}{6} \quad \text{or} \quad \omega t + \phi_2 = \frac{11\pi}{6} \]

### Step 5: Calculate the phase difference
Now, we can find the phase difference \( \Delta \phi = \phi_2 - \phi_1 \). We will consider the cases:
1. If \( \phi_1 = \frac{\pi}{6} \) and \( \phi_2 = \frac{7\pi}{6} \):
\[ \Delta \phi = \frac{7\pi}{6} - \frac{\pi}{6} = \frac{6\pi}{6} = \pi \]

2. If \( \phi_1 = \frac{5\pi}{6} \) and \( \phi_2 = \frac{7\pi}{6} \):
\[ \Delta \phi = \frac{7\pi}{6} - \frac{5\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \]

3. If \( \phi_1 = \frac{\pi}{6} \) and \( \phi_2 = \frac{11\pi}{6} \):
\[ \Delta \phi = \frac{11\pi}{6} - \frac{\pi}{6} = \frac{10\pi}{6} = \frac{5\pi}{3} \]

4. If \( \phi_1 = \frac{5\pi}{6} \) and \( \phi_2 = \frac{11\pi}{6} \):
\[ \Delta \phi = \frac{11\pi}{6} - \frac{5\pi}{6} = \frac{6\pi}{6} = \pi \]

### Step 6: Determine the valid phase difference
The valid phase differences that we calculated are \( \pi \), \( \frac{\pi}{3} \), and \( \frac{5\pi}{3} \). However, since the particles are moving in opposite directions, we consider the effective phase difference, which is:
\[ \Delta \phi = \frac{2\pi}{3} \]

### Conclusion
Thus, the phase difference between the two particles is:
\[ \Delta \phi = \frac{2\pi}{3} \]