A particle is executing simple harmonic motion with an amplitude A and time period T. The displacement of the particles after 2T period from its initial position is

Options

A. A
B. 4A
C. 8A
D. zero

Explanation

To solve the problem of finding the displacement of a particle executing simple harmonic motion (SHM) after a time period of \(2T\), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Time Period (T)**:
- The time period \(T\) is the time taken for the particle to complete one full cycle of motion. In simple harmonic motion, this means moving from the initial position to the maximum displacement (amplitude), back to the initial position, to the maximum displacement in the opposite direction, and then back to the initial position.

2. **Determine the Displacement After \(2T\)**:
- After one time period \(T\), the particle returns to its initial position. Therefore, after two time periods \(2T\), the particle will also return to its initial position.
- Hence, the displacement after \(2T\) is \(0\).

3. **Using the Displacement Formula**:
- The displacement \(X\) in simple harmonic motion can be expressed as:
\[ X = A \sin(\omega t) \]
- Here, \(A\) is the amplitude, \(\omega\) is the angular frequency, and \(t\) is the time.

4. **Calculate Angular Frequency (\(\omega\))**:
- The angular frequency \(\omega\) is given by:
\[ \omega = \frac{2\pi}{T} \]

5. **Substituting \(t = 2T\) into the Displacement Formula**:
- Substitute \(t = 2T\) into the displacement formula:
\[ X = A \sin\left(\omega \cdot 2T\right) \]
- This becomes:
\[ X = A \sin\left(2 \cdot \frac{2\pi}{T} \cdot T\right) = A \sin(4\pi) \]

6. **Evaluate \(\sin(4\pi)\)**:
- The sine function has a periodicity of \(2\pi\), so:
\[ \sin(4\pi) = 0 \]

7. **Conclusion**:
- Therefore, the displacement \(X\) after \(2T\) is:
\[ X = A \cdot 0 = 0 \]

### Final Answer:
The displacement of the particle after \(2T\) from its initial position is \(0\).

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