A particle is subjected to two mutually perpendicular simple harmonic motions such that its `X` and `y` coordinates are given by `X=2 sin omegat` , `y=2 sin (omega+(pi)/(4))`
The path of the particle will be:

Options

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

Answer

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

1. **Given Equations**:
- \( X = 2 \sin(\omega t) \)
- \( Y = 2 \sin\left(\omega t + \frac{\pi}{4}\right) \)

2. **Expressing Sine and Cosine**:
- From the equation for \( X \):
\[ \sin(\omega t) = \frac{X}{2} \]
- We can find \( \cos(\omega t) \) using the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \):
\[ \cos(\omega t) = \sqrt{1 - \sin^2(\omega t)} = \sqrt{1 - \left(\frac{X}{2}\right)^2} = \sqrt{1 - \frac{X^2}{4}} \]

3. **Substituting into the Y Equation**:
- Now, we substitute \( \sin(\omega t) \) and \( \cos(\omega t) \) into the equation for \( Y \):
\[ Y = 2 \left( \sin(\omega t) \cos\left(\frac{\pi}{4}\right) + \cos(\omega t) \sin\left(\frac{\pi}{4}\right) \right) \]
- Since \( \sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \):
\[ Y = 2 \left( \frac{X}{2} \cdot \frac{1}{\sqrt{2}} + \sqrt{1 - \frac{X^2}{4}} \cdot \frac{1}{\sqrt{2}} \right) \]
- Simplifying this gives:
\[ Y = \sqrt{2} \left( \frac{X}{2} + \sqrt{1 - \frac{X^2}{4}} \right) \]

4. **Rearranging the Equation**:
- Rearranging the equation:
\[ Y - \frac{\sqrt{2}}{2}X = \sqrt{2} \sqrt{1 - \frac{X^2}{4}} \]
- Squaring both sides:
\[ \left(Y - \frac{\sqrt{2}}{2}X\right)^2 = 2\left(1 - \frac{X^2}{4}\right) \]
- Expanding and simplifying gives:
\[ X^2 + Y^2 - \sqrt{2}XY - 2 = 0 \]

5. **Identifying the Path**:
- The resulting equation \( X^2 + Y^2 - \sqrt{2}XY - 2 = 0 \) is the equation of an ellipse.

**Conclusion**:
The path of the particle is an ellipse.

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