Two simple harmonic motions with the same frequency act on a particle at right angles i.e., along X-axis and Y-axis. If the two amplitudes are equal and the phase difference is `pi//2`, the resultant motion will be

Options

A. a circle
B. an ellipse with the major axis along Y-axis
C. an ellipse with the major axis along X-axis
D. a straight line inclined at `45^(@)` to the X-axis

Answer

To solve the problem, we need to analyze the two simple harmonic motions acting on a particle along the X-axis and Y-axis. Let's break it down step by step:

### Step 1: Define the Simple Harmonic Motions
Let the first simple harmonic motion along the X-axis be represented as:
\[ x(t) = A \sin(\omega t) \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.

Let the second simple harmonic motion along the Y-axis be represented as:
\[ y(t) = A \sin\left(\omega t + \frac{\pi}{2}\right) \]

### Step 2: Simplify the Y-axis Motion
Using the trigonometric identity \( \sin\left(\theta + \frac{\pi}{2}\right) = \cos(\theta) \), we can rewrite the Y-axis motion:
\[ y(t) = A \cos(\omega t) \]

### Step 3: Write the Equations for Both Motions
Now we have:
- For the X-axis: \( x(t) = A \sin(\omega t) \)
- For the Y-axis: \( y(t) = A \cos(\omega t) \)

### Step 4: Square and Add the Two Equations
To find the resultant motion, we will square both equations and add them:
\[ x^2 + y^2 = (A \sin(\omega t))^2 + (A \cos(\omega t))^2 \]
\[ x^2 + y^2 = A^2 \sin^2(\omega t) + A^2 \cos^2(\omega t) \]

### Step 5: Use the Pythagorean Identity
We know from trigonometry that:
\[ \sin^2(\theta) + \cos^2(\theta) = 1 \]
Applying this identity:
\[ x^2 + y^2 = A^2 (\sin^2(\omega t) + \cos^2(\omega t)) = A^2 \cdot 1 \]
\[ x^2 + y^2 = A^2 \]

### Step 6: Interpret the Result
The equation \( x^2 + y^2 = A^2 \) represents a circle with radius \( A \) in the XY-plane. Therefore, the resultant motion of the particle is circular.

### Final Answer
The resultant motion will be a **circle**.

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