A particle is executing two different simple harmonic motions, mutually perpendicular, of different amplitudes and having phase difference of `pi//2`. The path of the particle will be

Options

A. circular
B. straight line
C. parabolic
D. elliptical

Explanation

To solve the problem of a particle executing two different simple harmonic motions (SHMs) that are mutually perpendicular, with different amplitudes and a phase difference of \( \frac{\pi}{2} \), we can follow these steps:

### Step 1: Understand the Motion
We have two SHMs:
1. One along the x-axis with amplitude \( A_x \).
2. Another along the y-axis with amplitude \( A_y \).

The phase difference between these two motions is \( \frac{\pi}{2} \).

### Step 2: Write the Equations of Motion
The equations for the two SHMs can be expressed as:
- For the x-component:
\[ x(t) = A_x \cos(\omega t) \]
- For the y-component:
\[ y(t) = A_y \sin(\omega t + \frac{\pi}{2}) = A_y \cos(\omega t) \]
(since \( \sin(\theta + \frac{\pi}{2}) = \cos(\theta) \))

### Step 3: Relate x and y
From the equations, we can express \( y \) in terms of \( x \):
- Substitute \( \cos(\omega t) \) from the x equation into the y equation:
\[ y = A_y \cos(\omega t) \]
\[ x = A_x \cos(\omega t) \]

### Step 4: Eliminate Time
To eliminate time and find the relationship between \( x \) and \( y \), we can use the identity:
\[ \left(\frac{x}{A_x}\right)^2 + \left(\frac{y}{A_y}\right)^2 = \cos^2(\omega t) + \cos^2(\omega t) = 1 \]
This leads us to the equation of an ellipse:
\[ \frac{x^2}{A_x^2} + \frac{y^2}{A_y^2} = 1 \]

### Step 5: Conclusion
The path of the particle executing these two SHMs will be an ellipse. The specific shape of the ellipse will depend on the relative amplitudes \( A_x \) and \( A_y \).

### Final Answer
The path of the particle will be an **ellipse**.

---