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

Options

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

Answer

To solve the problem of the composition of two simple harmonic motions (SHMs) of equal periods 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's denote the first SHM along the x-axis as:
\[ x = A \sin(\omega t) \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.

2. **Define the second SHM**:
The second SHM is along the y-axis and has a phase difference of π. Therefore, we can express it as:
\[ y = B \sin(\omega t + \pi) \]
Since \( \sin(\theta + \pi) = -\sin(\theta) \), we can rewrite this as:
\[ y = -B \sin(\omega t) \]

3. **Relate the two equations**:
From the first equation, we can express \( \sin(\omega t) \) in terms of \( x \):
\[ \sin(\omega t) = \frac{x}{A} \]
Substituting this into the equation for \( y \):
\[ y = -B \left(\frac{x}{A}\right) \]
This simplifies to:
\[ y = -\frac{B}{A} x \]

4. **Identify the type of relationship**:
The equation \( y = -\frac{B}{A} x \) is a linear equation in the form \( y = mx \), where \( m = -\frac{B}{A} \). This indicates that the relationship between \( x \) and \( y \) is linear.

5. **Conclusion**:
Since the relationship is linear, the displacement of the particle will be along a straight line.

### Final Answer:
The displacement of the particle along the composition of the two SHMs results in a straight line.