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

Options

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

Answer

To solve the problem, we need to find the phase difference between two simple harmonic motions (SHM) that have equal amplitudes and frequencies, given that the resultant amplitude is equal to the amplitude of the individual motions.

### Step-by-Step Solution:

1. **Define the SHM Equations**:
Let the two SHM equations be:
- \( x_1 = A \sin(\omega t) \)
- \( x_2 = A \sin(\omega t + \phi) \)
Here, \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase difference between the two motions.

2. **Resultant Amplitude**:
The resultant displacement \( x_r \) can be expressed as:
\[ x_r = x_1 + x_2 = A \sin(\omega t) + A \sin(\omega t + \phi) \]
Using the sine addition formula, we can rewrite \( x_2 \):
\[ x_2 = A \left( \sin(\omega t) \cos(\phi) + \cos(\omega t) \sin(\phi) \right) \]
Therefore,
\[ x_r = A \sin(\omega t) + A \left( \sin(\omega t) \cos(\phi) + \cos(\omega t) \sin(\phi) \right) \]

3. **Combine the Terms**:
Combine the terms involving \( \sin(\omega t) \):
\[ x_r = A \sin(\omega t) (1 + \cos(\phi)) + A \cos(\omega t) \sin(\phi) \]

4. **Resultant Amplitude Formula**:
The amplitude of the resultant motion can be found using the formula:
\[ A_r = \sqrt{(A(1 + \cos(\phi)))^2 + (A \sin(\phi))^2} \]
Simplifying this gives:
\[ A_r = A \sqrt{(1 + \cos(\phi))^2 + \sin^2(\phi)} \]

5. **Set Resultant Amplitude Equal to Individual Amplitude**:
According to the problem, the resultant amplitude is equal to the individual amplitude \( A \):
\[ A \sqrt{(1 + \cos(\phi))^2 + \sin^2(\phi)} = A \]
Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ \sqrt{(1 + \cos(\phi))^2 + \sin^2(\phi)} = 1 \]

6. **Square Both Sides**:
Squaring both sides gives:
\[ (1 + \cos(\phi))^2 + \sin^2(\phi) = 1 \]
Expanding the left side:
\[ 1 + 2\cos(\phi) + \cos^2(\phi) + \sin^2(\phi) = 1 \]
Using the identity \( \cos^2(\phi) + \sin^2(\phi) = 1 \):
\[ 1 + 2\cos(\phi) + 1 = 1 \]
Simplifying this gives:
\[ 2 + 2\cos(\phi) = 1 \]
Thus,
\[ 2\cos(\phi) = -1 \quad \Rightarrow \quad \cos(\phi) = -\frac{1}{2} \]

7. **Find Phase Difference**:
The angle \( \phi \) that satisfies \( \cos(\phi) = -\frac{1}{2} \) is:
\[ \phi = 120^\circ \quad \text{or} \quad \phi = 240^\circ \]
Since we are looking for the phase difference, we take \( \phi = 120^\circ \) (or \( \frac{2\pi}{3} \) radians).

### Final Answer:
The phase difference \( \phi \) between the two simple harmonic motions is \( 120^\circ \) or \( \frac{2\pi}{3} \) radians.

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