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: Understanding the Problem
We have two particles executing SHM with the same frequency. They meet at the position \( x = +\frac{A}{2} \). Since they are moving in opposite directions, one particle must be moving towards \( +\frac{A}{2} \) from the left, while the other is moving towards it from the right.

### Step 2: Identifying the Positions of the Particles
Let’s denote the two particles as Particle 1 and Particle 2:
- Particle 1 is moving towards \( +\frac{A}{2} \) from the origin (0).
- Particle 2 is moving towards \( +\frac{A}{2} \) from the maximum amplitude \( A \).

### Step 3: Position of Each Particle
The position of a particle in SHM can be described by the equation:
\[ x(t) = A \sin(\omega t + \phi) \]
Where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( \phi \) is the phase constant.

For Particle 1, which starts from the origin:
\[ x_1(t) = A \sin(\omega t) \]
For Particle 2, which starts from the maximum amplitude \( A \):
\[ x_2(t) = A \sin(\omega t + \pi) = -A \sin(\omega t) \]

### Step 4: Meeting Condition
At the meeting point \( x = +\frac{A}{2} \):
\[ A \sin(\omega t) = \frac{A}{2} \]
This simplifies to:
\[ \sin(\omega t) = \frac{1}{2} \]
The angle \( \omega t \) corresponding to \( \sin(\omega t) = \frac{1}{2} \) is:
\[ \omega t = \frac{\pi}{6} \quad \text{or} \quad \omega t = \frac{5\pi}{6} \]

### Step 5: Finding the Phase of Each Particle
For Particle 1:
- If \( \omega t = \frac{\pi}{6} \), then the phase \( \phi_1 = 0 + \frac{\pi}{6} = \frac{\pi}{6} \).

For Particle 2:
- If \( \omega t = \frac{5\pi}{6} \), then the phase \( \phi_2 = \pi + \frac{5\pi}{6} = \frac{11\pi}{6} \).

### Step 6: Calculating the Phase Difference
The phase difference \( \Delta \phi \) between the two particles is given by:
\[ \Delta \phi = \phi_2 - \phi_1 \]
Substituting the values:
\[ \Delta \phi = \frac{11\pi}{6} - \frac{\pi}{6} = \frac{10\pi}{6} = \frac{5\pi}{3} \]

### Step 7: Converting to Degrees
To convert \( \frac{5\pi}{3} \) radians to degrees:
\[ \Delta \phi = \frac{5\pi}{3} \times \frac{180}{\pi} = 300^\circ \]

### Conclusion
Thus, the phase difference between the two particles is \( 300^\circ \) or equivalently \( \frac{5\pi}{3} \) radians.

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