The composition of two simple harmonic motions of equal periods at right angle to each other and with a phase difference of `pi` result in the displacement of the particle along

Options

A. straight line
B. circle
C. ellipse
D. figure of eight

Explanation

To solve the problem of the composition of two simple harmonic motions (SHMs) at right angles to each other with a phase difference of π, we can follow these steps:

### Step-by-Step Solution

1. **Define the SHMs**:
Let the first SHM along the x-axis be represented as:
\[ x(t) = A_1 \sin(\omega t + \phi) \]
where \( A_1 \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase.

2. **Define the second SHM**:
The second SHM along the y-axis, which has a phase difference of π (180 degrees) with respect to the first, can be represented as:
\[ y(t) = A_2 \sin(\omega t + \phi + \pi) \]
Using the identity \( \sin(\theta + \pi) = -\sin(\theta) \), we can rewrite this as:
\[ y(t) = -A_2 \sin(\omega t + \phi) \]

3. **Express both SHMs**:
Now we have:
\[ x(t) = A_1 \sin(\omega t + \phi) \]
\[ y(t) = -A_2 \sin(\omega t + \phi) \]

4. **Relate x and y**:
To find the relationship between \( x \) and \( y \), we can express \( y \) in terms of \( x \):
\[ y = -\frac{A_2}{A_1} x \]

5. **Identify the trajectory**:
The equation \( y = -\frac{A_2}{A_1} x \) represents a straight line with a slope of \(-\frac{A_2}{A_1}\). This indicates that the path of the particle is a straight line.

6. **Conclusion**:
Therefore, the displacement of the particle results in a linear path. The trajectory of the particle is a straight line.

### Final Answer
The displacement of the particle along the trajectory is a straight line.