`x` and `y` displacements of a particle are given as `x(t)=-asinomegat` and `y(t)=asin2omegat`. Its trajectory will look like.

Options

A.
B.
C.
D.

Explanation

To solve the problem, we need to analyze the given equations for the displacements \( x(t) \) and \( y(t) \):

1. **Given Equations:**
\[ x(t) = -a \sin(\omega t) \]
\[ y(t) = a \sin(2\omega t) \]

2. **Finding Points on the Trajectory:**
- We will evaluate the trajectory at specific time intervals to understand how \( x \) and \( y \) change.

3. **At \( t = 0 \):**
\[ x(0) = -a \sin(0) = 0 \]
\[ y(0) = a \sin(0) = 0 \]
- So, the particle starts at the origin \( (0, 0) \).

4. **At \( t = \frac{\pi}{2\omega} \):**
\[ x\left(\frac{\pi}{2\omega}\right) = -a \sin\left(\omega \cdot \frac{\pi}{2\omega}\right) = -a \sin\left(\frac{\pi}{2}\right) = -a \]
\[ y\left(\frac{\pi}{2\omega}\right) = a \sin\left(2\omega \cdot \frac{\pi}{2\omega}\right) = a \sin(\pi) = 0 \]
- At this point, the particle is at \( (-a, 0) \).

5. **At \( t = \frac{\pi}{4\omega} \):**
\[ x\left(\frac{\pi}{4\omega}\right) = -a \sin\left(\frac{\pi}{4}\right) = -a \cdot \frac{\sqrt{2}}{2} = -\frac{a\sqrt{2}}{2} \]
\[ y\left(\frac{\pi}{4\omega}\right) = a \sin\left(\frac{\pi}{2}\right) = a \]
- The particle is at \( \left(-\frac{a\sqrt{2}}{2}, a\right) \).

6. **At \( t = \frac{\pi}{\omega} \):**
\[ x\left(\frac{\pi}{\omega}\right) = -a \sin(\pi) = 0 \]
\[ y\left(\frac{\pi}{\omega}\right) = a \sin(2\pi) = 0 \]
- The particle returns to the origin \( (0, 0) \).

7. **At \( t = \frac{3\pi}{4\omega} \):**
\[ x\left(\frac{3\pi}{4\omega}\right) = -a \sin\left(\frac{3\pi}{4}\right) = -a \cdot \frac{\sqrt{2}}{2} = -\frac{a\sqrt{2}}{2} \]
\[ y\left(\frac{3\pi}{4\omega}\right) = a \sin\left(\frac{3\pi}{2}\right) = -a \]
- The particle is at \( \left(-\frac{a\sqrt{2}}{2}, -a\right) \).

8. **At \( t = \frac{3\pi}{2\omega} \):**
\[ x\left(\frac{3\pi}{2\omega}\right) = -a \sin\left(\frac{3\pi}{2}\right) = a \]
\[ y\left(\frac{3\pi}{2\omega}\right) = a \sin(3\pi) = 0 \]
- The particle is at \( (a, 0) \).

9. **At \( t = 2\pi/\omega \):**
\[ x\left(2\pi/\omega\right) = -a \sin(2\pi) = 0 \]
\[ y\left(2\pi/\omega\right) = a \sin(4\pi) = 0 \]
- The particle returns to the origin \( (0, 0) \).

10. **Conclusion:**
- The trajectory of the particle will form a closed loop, oscillating between the points \( (-a, 0) \), \( (0, a) \), \( (a, 0) \), and \( (0, -a) \). The shape of the trajectory is a vertical ellipse.