A particle simultaneously participates in two mutually perpendicular oscillations `x = sin pi t` & `y = 2cos 2 pit`. Write the equation of trajectory of the particle.

Explanation

To find the equation of the trajectory of the particle participating in two mutually perpendicular oscillations given by \( x = \sin(\pi t) \) and \( y = 2\cos(2\pi t) \), we need to eliminate the time variable \( t \) from these equations.

### Step-by-Step Solution:

1. **Identify the equations**:
We have two equations:
\[ x = \sin(\pi t) \quad \text{(1)} \]
\[ y = 2\cos(2\pi t) \quad \text{(2)} \]

2. **Express \( \cos(2\pi t) \) in terms of \( \sin(\pi t) \)**:
We know the trigonometric identity:
\[ \cos(2\theta) = 1 - 2\sin^2(\theta) \]
By substituting \( \theta = \pi t \), we can rewrite \( \cos(2\pi t) \):
\[ \cos(2\pi t) = 1 - 2\sin^2(\pi t) \]

3. **Substitute \( \sin(\pi t) \) from equation (1)**:
From equation (1), we have:
\[ \sin(\pi t) = x \]
Therefore, we can substitute this into the expression for \( \cos(2\pi t) \):
\[ \cos(2\pi t) = 1 - 2x^2 \]

4. **Substitute into equation (2)**:
Now, substitute \( \cos(2\pi t) \) into equation (2):
\[ y = 2\cos(2\pi t) = 2(1 - 2x^2) \]
Simplifying this gives:
\[ y = 2 - 4x^2 \]

5. **Rearranging the equation**:
Rearranging the equation to a standard form:
\[ 4x^2 + y - 2 = 0 \]
or, equivalently,
\[ 4x^2 + y = 2 \]

6. **Final form of the equation**:
To express it in a more recognizable form, we can write:
\[ 2x^2 + \frac{y}{2} = 1 \]

Thus, the equation of the trajectory of the particle is:
\[ 2x^2 + \frac{y}{2} = 1 \]