As a body performs SHM its potential energy U. varies with time as indicated in

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Explanation

To solve the problem, we need to analyze how the potential energy \( U \) of a body performing Simple Harmonic Motion (SHM) varies with time.

### Step-by-Step Solution:

1. **Understanding Potential Energy in SHM**:
The potential energy \( U \) in SHM is given by the formula:
\[ U = \frac{1}{2} k x^2 \]
where \( k \) is the spring constant and \( x \) is the displacement from the mean position.

2. **Displacement in SHM**:
In SHM, the displacement \( x \) can be expressed as:
\[ x = A \sin(\omega t) \]
where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( t \) is time.

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

4. **Relating Spring Constant to Mass and Angular Frequency**:
We know that the spring constant \( k \) is related to the mass \( m \) and angular frequency \( \omega \) by:
\[ k = m \omega^2 \]
Substituting this into the potential energy equation gives:
\[ U = \frac{1}{2} m \omega^2 A^2 \sin^2(\omega t) \]

5. **Analyzing the Function**:
The function \( \sin^2(\omega t) \) varies between 0 and 1 as \( t \) changes. At \( t = 0 \), \( \sin(0) = 0 \), so \( U = 0 \). The potential energy reaches its maximum when \( \sin^2(\omega t) = 1 \).

6. **Graph of Potential Energy**:
The graph of \( U \) as a function of time \( t \) will be a sinusoidal wave squared, which is an even function. This means it will be symmetric about the vertical axis and will oscillate between 0 and its maximum value.

7. **Conclusion**:
Based on the analysis, we can conclude that the potential energy \( U \) varies with time in a sinusoidal manner, specifically as \( U \propto \sin^2(\omega t) \).