Two longitudinal waves propagating in the X and Y directions superimpose. The wave equations are as below `Phi_(1)=A(omegat-kx)` and `Phi_(2)=Acos(omegat-ky)`. Trajectory of the motion of a particle lying on the line `y=x((2n+1)lamda)/(2)` will be

Options

A. straight line
B. circle
C. ellipse
D. none of these

Explanation

To solve the problem, we need to analyze the superposition of the two longitudinal waves given by their equations:

1. **Wave Equations**:
- \( \Phi_1 = A \cos(\omega t - kx) \)
- \( \Phi_2 = A \cos(\omega t - ky) \)

2. **Line of Interest**:
We are interested in a particle lying on the line defined by:
\[ y = x \frac{(2n + 1) \lambda}{2} \]
where \( n \) is an integer.

3. **Substituting the Line Equation into \( \Phi_2 \)**:
We substitute \( y \) into the wave equation \( \Phi_2 \):
\[ \Phi_2 = A \cos(\omega t - k y) = A \cos\left(\omega t - k \left(x \frac{(2n + 1) \lambda}{2}\right)\right) \]
Simplifying this gives:
\[ \Phi_2 = A \cos\left(\omega t - kx \frac{(2n + 1) \lambda}{2}\right) \]

4. **Using the Relation Between \( k \) and \( \lambda \)**:
Recall that \( k = \frac{2\pi}{\lambda} \). Therefore:
\[ k \frac{\lambda}{2} = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{2} = \pi \]
Substituting this back into \( \Phi_2 \):
\[ \Phi_2 = A \cos\left(\omega t - kx - (2n + 1)\pi\right) \]

5. **Using the Cosine Property**:
The cosine function has the property:
\[ \cos(\theta + \pi) = -\cos(\theta) \]
Thus:
\[ \Phi_2 = A \cos(\omega t - kx) \cdot (-1) = -A \cos(\omega t - kx) = -\Phi_1 \]

6. **Conclusion**:
This means that the two wave functions are in opposite phases along the line \( y = x \frac{(2n + 1) \lambda}{2} \). The trajectory of the motion of a particle along this line will be a straight line, as the waves are perfectly out of phase.

Therefore, the trajectory of the motion of a particle lying on the specified line is a **straight line**.