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. Here are the steps to derive the resultant motion:

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

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

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

### Step 3: Express the Resultant Motion
The resultant motion can be represented as a combination of the two motions:
\[ R^2 = x^2 + y^2 \]

Substituting the expressions for \( x(t) \) and \( y(t) \):
\[ R^2 = (A \sin(\omega t))^2 + (A \cos(\omega t))^2 \]

### Step 4: Simplify the Resultant Expression
Now, simplify the equation:
\[ R^2 = A^2 \sin^2(\omega t) + A^2 \cos^2(\omega t) \]
Using the Pythagorean identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \):
\[ R^2 = A^2 (\sin^2(\omega t) + \cos^2(\omega t)) = A^2 \]

### Step 5: Find the Resultant Amplitude
Taking the square root of both sides gives us the resultant amplitude:
\[ R = A \]

### Step 6: Determine the Nature of the Motion
Since the resultant motion is described by a constant amplitude \( R = A \) and the phase difference is \( \frac{\pi}{2} \), the motion will be circular. The particle will move in a circular path with a radius equal to the amplitude \( A \).

### Final Result
The resultant motion is circular with a radius \( A \).

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