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 two simple harmonic motions (SHMs) acting in perpendicular directions with a phase difference of \(\frac{\pi}{2}\).

### Step-by-Step Solution:

1. **Define the SHM Equations**:
- Let the first SHM in the x-direction be represented as:
\[ x = A \sin(\omega t) \]
- The second SHM in the y-direction, with a phase difference of \(\frac{\pi}{2}\), can be represented as:
\[ y = A \sin\left(\omega t + \frac{\pi}{2}\right) \]

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

3. **Square both equations**:
- Now, we will square both equations:
- For the x-direction:
\[ x^2 = A^2 \sin^2(\omega t) \]
- For the y-direction:
\[ y^2 = A^2 \cos^2(\omega t) \]

4. **Use the Pythagorean Identity**:
- We know that \(\sin^2(\theta) + \cos^2(\theta) = 1\). Therefore, we can express \(y^2\) in terms of \(x^2\):
\[ y^2 = A^2 - A^2 \sin^2(\omega t) = A^2 - x^2 \]

5. **Combine the equations**:
- Rearranging gives us:
\[ x^2 + y^2 = A^2 \]
- This is the equation of a circle with radius \(A\).

6. **Conclusion**:
- The resultant motion of the two SHMs acting together is circular motion.

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

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