Two particle execute `SHM` of same amplitude of `20 cm` with same period along the same line about the same equilibrium position. The maximum distance between the two is `20 cm`. Their phase difference in radians is

Options

A. `(pi)/(3)`
B. `(pi)/(2)`
C. `(2pi)/(3)`
D. `(4pi)/(5)`

Answer

To solve the problem, we need to find the phase difference between two particles executing simple harmonic motion (SHM) given that their maximum distance apart is 20 cm.

### Step-by-Step Solution:

1. **Understand the Problem**:
We have two particles executing SHM with the same amplitude (A = 20 cm) and the same period. We need to find the phase difference (Φ) given that the maximum distance between the two particles is 20 cm.

2. **Write the Displacement Equations**:
Let the displacement of the first particle be:
\[ x_1 = A \sin(\omega t) = 20 \sin(\omega t) \]
Let the displacement of the second particle be:
\[ x_2 = A \sin(\omega t + \Phi) = 20 \sin(\omega t + \Phi) \]

3. **Calculate the Distance Between the Two Particles**:
The distance between the two particles is given by:
\[ d = |x_2 - x_1| = |20 \sin(\omega t + \Phi) - 20 \sin(\omega t)| \]
Simplifying, we have:
\[ d = 20 |\sin(\omega t + \Phi) - \sin(\omega t)| \]

4. **Use the Trigonometric Identity**:
We can use the identity for the difference of sines:
\[ \sin A - \sin B = 2 \cos\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right) \]
Applying this to our equation:
\[ d = 20 \cdot 2 \cos\left(\frac{(\omega t + \Phi) + \omega t}{2}\right) \sin\left(\frac{(\omega t + \Phi) - \omega t}{2}\right) \]
This simplifies to:
\[ d = 40 \cos\left(\omega t + \frac{\Phi}{2}\right) \sin\left(\frac{\Phi}{2}\right) \]

5. **Find Maximum Distance**:
The maximum value of \(d\) occurs when \(\cos\left(\omega t + \frac{\Phi}{2}\right) = 1\):
\[ d_{\text{max}} = 40 \sin\left(\frac{\Phi}{2}\right) \]
We know from the problem that the maximum distance is 20 cm:
\[ 20 = 40 \sin\left(\frac{\Phi}{2}\right) \]

6. **Solve for \(\sin\left(\frac{\Phi}{2}\right)\)**:
Dividing both sides by 40:
\[ \sin\left(\frac{\Phi}{2}\right) = \frac{1}{2} \]

7. **Find \(\frac{\Phi}{2}\)**:
The angle whose sine is \(\frac{1}{2}\) is \(\frac{\pi}{6}\):
\[ \frac{\Phi}{2} = \frac{\pi}{6} \]

8. **Calculate \(\Phi\)**:
Multiplying both sides by 2 gives:
\[ \Phi = \frac{\pi}{3} \]

### Final Answer:
The phase difference in radians is:
\[ \Phi = \frac{\pi}{3} \]