A particle is subjected to two mutually perendicular simple harmonic motion such that its x and y coordinates are given by :
`x = 2 sin omega " "& " "y = 2 sin (omega t + (pi)/(2))`
The path of the particle will be

Options

A. an ellipse
B. a straight line
C. a parabola
D. a circle

Answer

To find the path of the particle subjected to two mutually perpendicular simple harmonic motions, we start with the given equations for the x and y coordinates:

1. **Given Equations:**
- \( x = 2 \sin(\omega t) \)
- \( y = 2 \sin\left(\omega t + \frac{\pi}{2}\right) \)

2. **Simplifying the y-coordinate:**
The term \( \sin\left(\omega t + \frac{\pi}{2}\right) \) can be simplified using the sine addition formula:
\[ \sin\left(a + b\right) = \sin a \cos b + \cos a \sin b \]
Here, \( a = \omega t \) and \( b = \frac{\pi}{2} \):
\[ y = 2 \left(\sin(\omega t) \cdot 0 + \cos(\omega t) \cdot 1\right) = 2 \cos(\omega t) \]

3. **Now we have:**
- \( x = 2 \sin(\omega t) \)
- \( y = 2 \cos(\omega t) \)

4. **Squaring both equations:**
\[ x^2 = (2 \sin(\omega t))^2 = 4 \sin^2(\omega t) \]
\[ y^2 = (2 \cos(\omega t))^2 = 4 \cos^2(\omega t) \]

5. **Adding the squared equations:**
\[ x^2 + y^2 = 4 \sin^2(\omega t) + 4 \cos^2(\omega t) \]
Factor out the 4:
\[ x^2 + y^2 = 4(\sin^2(\omega t) + \cos^2(\omega t)) \]

6. **Using the Pythagorean identity:**
We know that \( \sin^2(\theta) + \cos^2(\theta) = 1 \):
\[ x^2 + y^2 = 4 \cdot 1 = 4 \]

7. **Final equation:**
The equation \( x^2 + y^2 = 4 \) represents a circle centered at the origin with a radius of 2.

**Conclusion:**
The path of the particle is a circle with a radius of 2.

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