Two SHM's with same amplitude and time period, when acting together in perpendicular directions with a phase difference of `(pi)/(2)` give rise to

Options

A. Straight motion
B. Elliptical motion
C. Circular motion
D. None of these

Answer

To solve the problem, we need to analyze the situation where two simple harmonic motions (SHMs) with the same amplitude and time period are acting together in perpendicular directions with a phase difference of \( \frac{\pi}{2} \).

### Step-by-Step Solution:

1. **Define the SHM Equations**:
Let the first SHM be represented by:
\[ x(t) = A \sin(\omega t) \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.

2. **Define the Second SHM**:
The second SHM, which is acting in a perpendicular direction with a phase difference of \( \frac{\pi}{2} \), can be represented as:
\[ y(t) = A \sin\left(\omega t + \frac{\pi}{2}\right) \]

3. **Simplify the Second SHM**:
Using the trigonometric identity \( \sin\left(\theta + \frac{\pi}{2}\right) = \cos(\theta) \), we can rewrite the second SHM as:
\[ y(t) = A \cos(\omega t) \]

4. **Relate \( x \) and \( y \)**:
We now have:
\[ x(t) = A \sin(\omega t) \]
\[ y(t) = A \cos(\omega t) \]
To find the relationship between \( x \) and \( y \), we can use the Pythagorean identity:
\[ \sin^2(\theta) + \cos^2(\theta) = 1 \]

5. **Express the Pythagorean Identity**:
Dividing both equations by \( A^2 \):
\[ \left(\frac{x}{A}\right)^2 + \left(\frac{y}{A}\right)^2 = \sin^2(\omega t) + \cos^2(\omega t) = 1 \]

6. **Final Equation**:
Rearranging gives us:
\[ x^2 + y^2 = A^2 \]
This equation represents a circle with a radius \( A \) centered at the origin.

7. **Conclusion**:
The two SHMs combine to produce circular motion in the \( xy \)-plane, where the radius of the circular path is equal to the amplitude \( A \).

### Final Answer:
The two SHMs give rise to circular motion.