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 of the resultant path followed by a particle subjected to two linear simple harmonic motions (SHMs) along the x and y axes with a phase difference of π/2, we can follow these steps:

### Step 1: Understand the Given Information
We are given two SHMs:
- Both have equal amplitude \( A \) and equal angular frequency \( \omega \).
- The first SHM is along the x-axis, and the second SHM is along the y-axis.
- The phase difference between the two SHMs is \( \frac{\pi}{2} \).

### Step 2: Write the Equations of Motion
The equations for the two SHMs can be expressed as:
- For the x-axis:
\[ x(t) = A \sin(\omega t) \]
- For the y-axis (with a phase difference of \( \frac{\pi}{2} \)):
\[ y(t) = A \cos(\omega t) \]

### Step 3: Relate the Two Equations
We can express the sine and cosine functions in terms of each other:
- From the x-axis equation:
\[ \sin(\omega t) = \frac{x}{A} \]
- From the y-axis equation:
\[ \cos(\omega t) = \frac{y}{A} \]

### Step 4: Use the Pythagorean Identity
Using the Pythagorean identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \):
\[ \left(\frac{x}{A}\right)^2 + \left(\frac{y}{A}\right)^2 = 1 \]

### Step 5: Rearranging the Equation
Multiplying through by \( A^2 \) gives:
\[ x^2 + y^2 = A^2 \]

### Step 6: Identify the Resultant Path
The equation \( x^2 + y^2 = A^2 \) represents a circle with:
- Center at the origin (0, 0)
- Radius \( A \)

### Conclusion
The resultant path followed by the particle is a circular motion centered at the origin with a radius equal to the amplitude \( A \).