Two particles execute SHMs of the same amplitude and frequency along the same straight line. They cross one another when going in opposite direction. What is the phase difference between them when their displacements are half of their amplitudes ?

Options

A. `60^(@)`
B. `30^(@)`
C. `120^(@)`
D. `150^(@)`

Explanation

To solve the problem, we need to determine the phase difference between two particles executing Simple Harmonic Motion (SHM) when their displacements are half of their amplitudes. Here’s a step-by-step solution:

### Step 1: Write the SHM equation
The equation of motion for a particle in SHM can be expressed as:
\[ y = A \sin(\omega t + \phi) \]
where:
- \( y \) is the displacement,
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( t \) is time,
- \( \phi \) is the phase constant.

### Step 2: Set the displacement to half the amplitude
We are given that the displacement of both particles is half of their amplitude:
\[ y = \frac{A}{2} \]

### Step 3: Substitute the displacement into the SHM equation
Substituting \( y = \frac{A}{2} \) into the SHM equation gives:
\[ \frac{A}{2} = A \sin(\omega t + \phi) \]

### Step 4: Simplify the equation
Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ \frac{1}{2} = \sin(\omega t + \phi) \]

### Step 5: Find the angles corresponding to the sine value
The sine function equals \( \frac{1}{2} \) at two angles within one cycle:
1. \( \omega t + \phi = 30^\circ \) (or \( \frac{\pi}{6} \) radians)
2. \( \omega t + \phi = 150^\circ \) (or \( \frac{5\pi}{6} \) radians)

### Step 6: Assign phase values to the two particles
Let’s assume:
- For Particle 1, \( \omega t + \phi_1 = 30^\circ \)
- For Particle 2, \( \omega t + \phi_2 = 150^\circ \)

### Step 7: Calculate the phase difference
The phase difference \( \Delta \phi \) between the two particles is given by:
\[ \Delta \phi = \phi_2 - \phi_1 \]
Substituting the values we found:
\[ \Delta \phi = 150^\circ - 30^\circ = 120^\circ \]

### Conclusion
Thus, the phase difference between the two particles when their displacements are half of their amplitudes is:
\[ \Delta \phi = 120^\circ \]