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

Options

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Explanation

To determine the trajectory of the particle given the displacements \( x(t) = -a \sin(\omega t) \) and \( y(t) = a \sin(2\omega t) \), we can follow these steps:

### Step 1: Analyze the equations
The equations for the displacements are:
- \( x(t) = -a \sin(\omega t) \)
- \( y(t) = a \sin(2\omega t) \)

### Step 2: Find critical points
We will evaluate the displacements at specific time intervals to understand the trajectory.

**At \( t = 0 \):**
- \( x(0) = -a \sin(0) = 0 \)
- \( y(0) = a \sin(0) = 0 \)

So, at \( t = 0 \), the particle is at the origin (0,0).

### Step 3: Evaluate at \( t = \frac{\pi}{2\omega} \)
**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).

### Step 4: Evaluate at \( t = \frac{\pi}{4\omega} \)
**At \( t = \frac{\pi}{4\omega} \):**
- \( x\left(\frac{\pi}{4\omega}\right) = -a \sin\left(\frac{\pi}{4}\right) = -a \cdot \frac{\sqrt{2}}{2} = -\frac{a\sqrt{2}}{2} \)
- \( y\left(\frac{\pi}{4\omega}\right) = a \sin\left(\frac{\pi}{2}\right) = a \)

At this point, the particle is at \(-\frac{a\sqrt{2}}{2}, a\).

### Step 5: Evaluate at \( t = \frac{\pi}{2\omega} \)
**At \( t = \frac{\pi}{2\omega} \):**
- \( x\left(\frac{\pi}{2\omega}\right) = -a \sin(\pi) = 0 \)
- \( y\left(\frac{\pi}{2\omega}\right) = a \sin(2\pi) = 0 \)

The particle returns to the origin (0,0).

### Step 6: Evaluate at \( t = \frac{3\pi}{4\omega} \)
**At \( t = \frac{3\pi}{4\omega} \):**
- \( x\left(\frac{3\pi}{4\omega}\right) = -a \sin\left(\frac{3\pi}{4}\right) = -a \cdot \frac{\sqrt{2}}{2} = -\frac{a\sqrt{2}}{2} \)
- \( y\left(\frac{3\pi}{4\omega}\right) = a \sin\left(\frac{3\pi}{2}\right) = -a \)

At this point, the particle is at \(-\frac{a\sqrt{2}}{2}, -a\).

### Step 7: Evaluate at \( t = \frac{2\pi}{\omega} \)
**At \( t = \frac{2\pi}{\omega} \):**
- \( x\left(\frac{2\pi}{\omega}\right) = -a \sin(2\pi) = 0 \)
- \( y\left(\frac{2\pi}{\omega}\right) = a \sin(4\pi) = 0 \)

The particle returns to the origin (0,0).

### Conclusion
By plotting these points, we can see that the trajectory of the particle forms a closed loop. The trajectory will look like a sinusoidal wave pattern in the x-y plane.