A particle exectuing simple harmonic motion of time period T has x V and as its displacement , velocity and ecceleration , rexpectively . Which of the following quantities does not change with time ?

Options

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

Explanation

To solve the problem, we need to analyze the quantities involved in simple harmonic motion (SHM) and determine which of them remains constant over time.

### Step-by-Step Solution:

1. **Understanding the Variables**:
- In SHM, we have the following variables:
- \( x \): Displacement
- \( v \): Velocity
- \( a \): Acceleration
- \( T \): Time period of the motion

2. **Acceleration in SHM**:
- The acceleration \( a \) of a particle in SHM can be expressed as:
\[ a = -\omega^2 x \]
- Here, \( \omega \) (angular frequency) is related to the time period \( T \) by:
\[ \omega = \frac{2\pi}{T} \]

3. **Substituting for Angular Frequency**:
- Substituting \( \omega \) into the acceleration equation gives:
\[ a = -\left(\frac{2\pi}{T}\right)^2 x \]
- This shows that acceleration is dependent on displacement \( x \), which changes over time.

4. **Analyzing the Ratio \( \frac{aT}{x} \)**:
- We need to analyze the expression \( \frac{aT}{x} \):
\[ \frac{aT}{x} = \frac{-\left(\frac{2\pi}{T}\right)^2 x \cdot T}{x} \]
- Simplifying this expression:
\[ \frac{aT}{x} = -\frac{(2\pi)^2}{T} \]
- Here, \( -\frac{(2\pi)^2}{T} \) is a constant because \( T \) is a constant time period.

5. **Conclusion**:
- Since \( \frac{aT}{x} \) does not change with time, the answer to the question is that this quantity remains constant.

### Final Answer:
The quantity that does not change with time is \( \frac{aT}{x} \).

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