If x, v and a denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period T, then, which of the following does not change with time?

Options

A. `a^(2) T^(2) + 4pi^(2) v^(2)`
B. `(aT)/(x)`
C. `aT + 2pi v`
D. `(aT)/(v)`

Answer

To determine which quantity does not change with time for a particle executing simple harmonic motion (SHM), we need to analyze the relationships between displacement (x), velocity (v), and acceleration (a) in SHM.

### Step-by-Step Solution:

1. **Understanding Simple Harmonic Motion (SHM)**:
- In SHM, the displacement \( x \) of a particle from its equilibrium position varies sinusoidally with time. The general equation for displacement can be written as:
\[ x(t) = A \cos(\omega t + \phi) \]
where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant.

2. **Velocity in SHM**:
- The velocity \( v \) is the time derivative of displacement:
\[ v(t) = \frac{dx}{dt} = -A \omega \sin(\omega t + \phi) \]
- The velocity changes with time as it depends on the sine function.

3. **Acceleration in SHM**:
- The acceleration \( a \) is the time derivative of velocity:
\[ a(t) = \frac{dv}{dt} = -A \omega^2 \cos(\omega t + \phi) \]
- Similar to velocity, acceleration also changes with time as it depends on the cosine function.

4. **Time Period (T)**:
- The time period \( T \) is the time taken for one complete cycle of motion and is given by:
\[ T = \frac{2\pi}{\omega} \]
- The time period is a constant for a given SHM and does not change with time.

5. **Conclusion**:
- Among the quantities \( x \), \( v \), and \( a \), the only quantity that remains constant over time is the time period \( T \). The displacement \( x \), velocity \( v \), and acceleration \( a \) all vary with time.

### Final Answer:
The quantity that does not change with time is the **time period \( T \)**.