Assertion : x-t equation of a particle in SHM is given as :`x=A "cos"omegat`
Reason : In the given equation the minimum potential energy is zero.

Options

A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
C. If Assertion is true but Reason is false.
D. If Assertion is false but Reason is true.

Explanation

To solve the given question, we need to analyze both the assertion and the reason provided.

### Step-by-Step Solution:

1. **Understanding the Assertion**:
- The assertion states that the displacement-time equation of a particle in Simple Harmonic Motion (SHM) is given by \( x = A \cos(\omega t) \).
- This equation is indeed a standard form of the equation for SHM, where:
- \( x \) is the displacement,
- \( A \) is the amplitude (maximum displacement),
- \( \omega \) is the angular frequency,
- \( t \) is the time.
- At \( t = 0 \), the displacement \( x \) equals \( A \), which means the particle starts from the maximum displacement (amplitude).

2. **Understanding the Reason**:
- The reason states that in the given equation, the minimum potential energy is zero.
- In SHM, potential energy (PE) is related to the displacement from the equilibrium position. The potential energy can be expressed as:
\[ PE = \frac{1}{2} k x^2 \]
where \( k \) is the spring constant.
- The potential energy is minimum when the displacement \( x = 0 \) (at the equilibrium position), and it is at this point that the potential energy is zero.
- However, the statement that "the minimum potential energy is zero" depends on the reference point chosen for potential energy. In many cases, we take the equilibrium position as the reference point where potential energy is zero. Thus, the statement can be misleading.

3. **Conclusion**:
- The assertion is true because \( x = A \cos(\omega t) \) is indeed the correct equation for SHM.
- The reason, however, is false because while the minimum potential energy can be zero, it is not universally true for all reference points.

### Final Answer:
- Assertion: True
- Reason: False