A particle is executing SHM with time period T. If time period of its total mechanical energy isT' then `(T')/(T) ` is

Options

A. 2
B. `(1)/(2)`
C. Zero
D. Infinite

Answer

To solve the problem, we need to analyze the relationship between the time period of a particle executing Simple Harmonic Motion (SHM) and the time period of its total mechanical energy.

### Step-by-Step Solution:

1. **Understanding SHM**:
- A particle in SHM oscillates back and forth around an equilibrium position. The motion can be described by a sinusoidal function, and it has a specific time period \( T \).

2. **Total Mechanical Energy in SHM**:
- The total mechanical energy \( E \) of a particle in SHM is given by the formula:
\[ E = \frac{1}{2} m \omega^2 A^2 \]
where \( m \) is the mass of the particle, \( \omega \) is the angular frequency, and \( A \) is the amplitude of the motion.

3. **Constant Nature of Total Mechanical Energy**:
- In SHM, both \( m \), \( \omega \), and \( A \) are constants. Therefore, the total mechanical energy \( E \) remains constant throughout the motion.

4. **Time Period of Total Mechanical Energy**:
- Since the total mechanical energy \( E \) is constant, it does not oscillate or vary with time. Hence, we can say that the time period of the total mechanical energy \( T' \) is infinite.

5. **Finding the Ratio**:
- We are asked to find the ratio \( \frac{T'}{T} \). Since \( T' \) is infinite and \( T \) is a finite value (the time period of the SHM), we have:
\[ \frac{T'}{T} = \frac{\infty}{T} = \infty \]

### Conclusion:
The ratio \( \frac{T'}{T} \) is infinite.