In order that the resultant path on superimposing two mutually perpendicular SHM be a circle, the conditions are that

Options

A. the amplitudes on both SHM should be equal and they should have a phase difference of `(pi)/(2)`
B. the amplitude should be in the ratio 1:2 and the phase difference should be zero
C. the amplitude should be in the ratio 1:2 and the phase difference should be `(pi)/(2)`
D. the amplitudes should be equal and the phase difference should be zero

Explanation

To determine the conditions under which the resultant path of two mutually perpendicular simple harmonic motions (SHM) is a circle, we can follow these steps:

### Step-by-Step Solution:

1. **Define the SHM Equations**:
Let's consider two simple harmonic motions along the x and y axes:
- For the x-axis: \( x = A \sin(\omega t) \)
- For the y-axis: \( y = A \cos(\omega t) \)

2. **Square and Add the Equations**:
To find the resultant motion, we can square both equations and add them:
\[ x^2 + y^2 = (A \sin(\omega t))^2 + (A \cos(\omega t))^2 \]

3. **Use the Pythagorean Identity**:
Applying the Pythagorean identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \):
\[ x^2 + y^2 = A^2 (\sin^2(\omega t) + \cos^2(\omega t)) = A^2 \]

4. **Resultant Motion**:
The equation \( x^2 + y^2 = A^2 \) represents a circle with radius \( A \). This shows that the resultant path is indeed a circle.

5. **Conditions for Circular Motion**:
- **Same Amplitude**: The amplitudes of both SHMs must be equal, i.e., \( A_x = A_y = A \).
- **Phase Difference**: The phase difference between the two SHMs must be \( 90^\circ \) or \( \frac{\pi}{2} \) radians. This means if one SHM is represented by \( \sin(\omega t) \), the other must be represented by \( \cos(\omega t) \).

### Conclusion:
For the resultant path of two mutually perpendicular SHMs to be a circle, the conditions are:
- The amplitudes of both SHMs must be equal.
- The phase difference between the two SHMs must be \( 90^\circ \) (or \( \frac{\pi}{2} \) radians).