In S.H.M., potential energy (U) V/s, time (t) . Graph is

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Explanation

To determine the potential energy versus time graph for a particle undergoing simple harmonic motion (S.H.M.), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Equation of S.H.M.**:
The displacement \( x \) of a particle in S.H.M. can be described by the equation:
\[ x(t) = A \sin(\omega t + \phi) \]
where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( \phi \) is the phase constant.

2. **Potential Energy in S.H.M.**:
The potential energy \( U \) of a particle in S.H.M. is given by:
\[ U = \frac{1}{2} k x^2 \]
where \( k \) is the spring constant.

3. **Substituting the Displacement**:
Substitute the expression for \( x(t) \) into the potential energy formula:
\[ U(t) = \frac{1}{2} k (A \sin(\omega t + \phi))^2 \]
This simplifies to:
\[ U(t) = \frac{1}{2} k A^2 \sin^2(\omega t + \phi) \]

4. **Analyzing the Graph**:
The term \( \frac{1}{2} k A^2 \) is a constant. The potential energy \( U(t) \) is proportional to \( \sin^2(\omega t + \phi) \). The graph of \( \sin^2(x) \) oscillates between 0 and 1, meaning \( U(t) \) will oscillate between 0 and \( \frac{1}{2} k A^2 \).

5. **Graph Characteristics**:
- The graph will always be above the time axis (since \( \sin^2 \) is always non-negative).
- If \( \phi = 0 \), the graph starts at \( U = 0 \) when \( t = 0 \).
- If \( \phi \) is not zero, the graph will be shifted horizontally.

6. **Final Graph**:
The resulting graph of potential energy versus time will be a sinusoidal wave that oscillates between 0 and \( \frac{1}{2} k A^2 \), resembling a sine squared function.