Two linear SHM of equal amplitudes `A` and frequencies `omega` and `2omega` are impressed on a particle along `x` and `y - axes` respectively. If the initial phase difference between them is `pi//2`. Find the resultant path followed by the particle.

Explanation

To solve the problem of finding the resultant path followed by a particle under the influence of two simple harmonic motions (SHM) along the x and y axes, we can follow these steps:

### Step 1: Define the SHM equations
We have two SHMs:
1. Along the x-axis:
\[ x = A \sin(\omega t) \]
2. Along the y-axis:
\[ y = A \sin(2\omega t + \frac{\pi}{2}) \]

### Step 2: Simplify the y-axis equation
Using the trigonometric identity for sine, we can rewrite the y-axis equation. Since \(\sin(\theta + \frac{\pi}{2}) = \cos(\theta)\), we have:
\[ y = A \cos(2\omega t) \]

### Step 3: Express \(y\) in terms of \(x\)
From the x-axis equation, we can express \(\sin(\omega t)\) as:
\[ \sin(\omega t) = \frac{x}{A} \]
Now, we need to relate \(\cos(2\omega t)\) to \(x\). We know that:
\[ \cos(2\omega t) = 1 - 2\sin^2(\omega t) \]
Substituting \(\sin(\omega t) = \frac{x}{A}\) into this equation gives:
\[ \cos(2\omega t) = 1 - 2\left(\frac{x}{A}\right)^2 \]

### Step 4: Substitute back into the y equation
Now we can substitute this expression for \(\cos(2\omega t)\) back into the equation for \(y\):
\[ y = A \left(1 - 2\left(\frac{x}{A}\right)^2\right) \]
This simplifies to:
\[ y = A - \frac{2Ax^2}{A^2} \]
or:
\[ y = A - \frac{2}{A} x^2 \]

### Step 5: Rearranging the equation
To express the equation in a more standard form, we can rearrange it:
\[ y = A - \frac{2}{A} x^2 \]
This represents a downward-opening parabola in the xy-plane.

### Final Result
The resultant path followed by the particle is:
\[ y = A - \frac{2}{A} x^2 \]