A particle is subjected to two simple harmonic motion in the same direction having equal amplitudes and equal frequency. If the resultant amplitude is equal to the amplitude of the individual motions. Find the phase difference between the individual motions.

Explanation

To solve the problem, we need to analyze the situation where a particle is subjected to two simple harmonic motions (SHMs) with equal amplitudes and frequencies. We are tasked with finding the phase difference between these two motions given that the resultant amplitude is equal to the amplitude of the individual motions.

### Step-by-Step Solution:

1. **Define the Parameters**:
Let the amplitude of each simple harmonic motion be \( A \). Therefore, we have:
- Amplitude of first motion, \( A_1 = A \)
- Amplitude of second motion, \( A_2 = A \)

2. **Resultant Amplitude Formula**:
The resultant amplitude \( A_R \) of two SHMs can be calculated using the formula:
\[ A_R = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos \phi} \]
where \( \phi \) is the phase difference between the two motions.

3. **Substituting the Values**:
Since \( A_1 = A \) and \( A_2 = A \), we can substitute these values into the formula:
\[ A_R = \sqrt{A^2 + A^2 + 2 A A \cos \phi} \]
This simplifies to:
\[ A_R = \sqrt{2A^2 + 2A^2 \cos \phi} \]
or,
\[ A_R = \sqrt{2A^2(1 + \cos \phi)} \]

4. **Condition Given in the Problem**:
We are given that the resultant amplitude \( A_R \) is equal to the amplitude of the individual motions \( A \):
\[ A_R = A \]

5. **Setting Up the Equation**:
Now we can set up the equation:
\[ A = \sqrt{2A^2(1 + \cos \phi)} \]

6. **Squaring Both Sides**:
Squaring both sides to eliminate the square root gives:
\[ A^2 = 2A^2(1 + \cos \phi) \]

7. **Dividing by \( A^2 \)**:
Assuming \( A \neq 0 \), we can divide both sides by \( A^2 \):
\[ 1 = 2(1 + \cos \phi) \]

8. **Solving for \( \cos \phi \)**:
Rearranging the equation gives:
\[ 1 = 2 + 2\cos \phi \]
\[ 2\cos \phi = 1 - 2 \]
\[ 2\cos \phi = -1 \]
\[ \cos \phi = -\frac{1}{2} \]

9. **Finding the Phase Difference**:
The value of \( \phi \) for which \( \cos \phi = -\frac{1}{2} \) corresponds to:
\[ \phi = 120^\circ \quad \text{or} \quad \phi = 240^\circ \]
However, since we are looking for the phase difference in the context of SHM, we take \( \phi = 120^\circ \).

### Final Answer:
The phase difference between the two individual motions is \( \phi = 120^\circ \).