Two linear simple harmonic motions of equal amplitude and frequency are impressed on a particle along x and y axis respectively. The initial phase difference between them is `pi/2`. The resultant path followed by the particle is

Options

A. a circle
B. a straight line
C. an ellipse
D. a parabola

Answer

To solve the problem, we need to analyze the two linear simple harmonic motions (SHM) acting on a particle along the x and y axes, given that they have equal amplitude and frequency and an initial phase difference of \( \frac{\pi}{2} \).

### Step-by-Step Solution:

1. **Define the SHM along x-axis:**
The motion along the x-axis can be described by the equation:
\[ x = A \sin(\omega t) \]
where \( A \) is the amplitude, and \( \omega \) is the angular frequency.

2. **Define the SHM along y-axis:**
The motion along the y-axis, with a phase difference of \( \frac{\pi}{2} \), can be described by:
\[ y = A \sin\left(\omega t + \frac{\pi}{2}\right) \]
Using the trigonometric identity \( \sin(\theta + \frac{\pi}{2}) = \cos(\theta) \), we can rewrite this as:
\[ y = A \cos(\omega t) \]

3. **Relate the two equations:**
We now have two equations:
\[ x = A \sin(\omega t) \]
\[ y = A \cos(\omega t) \]

4. **Express \( \sin(\omega t) \) and \( \cos(\omega t) \):**
From the first equation, we can express \( \sin(\omega t) \) as:
\[ \sin(\omega t) = \frac{x}{A} \]
And from the second equation, we can express \( \cos(\omega t) \) as:
\[ \cos(\omega t) = \frac{y}{A} \]

5. **Use the Pythagorean identity:**
We know from trigonometry that:
\[ \sin^2(\theta) + \cos^2(\theta) = 1 \]
Applying this to our expressions gives:
\[ \left(\frac{x}{A}\right)^2 + \left(\frac{y}{A}\right)^2 = 1 \]
Multiplying through by \( A^2 \) results in:
\[ x^2 + y^2 = A^2 \]

6. **Identify the resultant path:**
The equation \( x^2 + y^2 = A^2 \) represents a circle with radius \( A \). Therefore, the resultant path followed by the particle is a circle.

### Final Answer:
The resultant path followed by the particle is a **circle**.