A particle executes `SHM` with a time period `T`. The time period with which its potential energy changes is

Options

A. `2T`
B. `T`
C. `T//2`
D. `3T//2`

Answer

To solve the problem, we need to understand the behavior of potential energy in Simple Harmonic Motion (SHM). Here's a step-by-step solution:

### Step 1: Understanding SHM
In SHM, a particle oscillates back and forth around an equilibrium position. The maximum displacement from the equilibrium position is called the amplitude (A). The time period (T) is the time taken to complete one full oscillation.

**Hint:** Recall that in SHM, the total mechanical energy is conserved and is the sum of kinetic and potential energy.

### Step 2: Potential Energy in SHM
The potential energy (PE) of a particle in SHM can be expressed as:
\[ PE = \frac{1}{2} k x^2 \]
where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position.

**Hint:** Remember that potential energy is maximum when the particle is at the maximum displacement (±A).

### Step 3: Analyzing the Motion
If we consider the particle starting from the extreme position (x = A), it will take time \( T/4 \) to reach the equilibrium position (x = 0) and another \( T/4 \) to reach the opposite extreme position (x = -A). Therefore, it takes \( T/2 \) to go from one extreme to the other.

**Hint:** Think about how the particle moves through its path and how potential energy changes as it moves.

### Step 4: Potential Energy Values
At the extreme position (x = A), the potential energy is:
\[ PE_{max} = \frac{1}{2} k A^2 \]
At the equilibrium position (x = 0), the potential energy is:
\[ PE_{eq} = 0 \]
When the particle moves from A to -A, it will pass through the equilibrium point where the potential energy is zero, and then it will reach -A where the potential energy will again be \( \frac{1}{2} k A^2 \).

**Hint:** Consider the symmetry of the motion and how potential energy values repeat.

### Step 5: Time for Potential Energy to Retain Value
The potential energy will be the same at two points during the oscillation: when the particle is at position A and when it is at position -A. The time taken to go from A to -A is \( T/2 \).

**Final Answer:** Therefore, the time period with which the potential energy changes is \( T/2 \).

### Summary
The potential energy retains its value after a time period of \( T/2 \) during the oscillation of the particle in SHM.

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