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

2. **Time Period of SHM**:
- The time period \( T \) of the SHM is related to the angular frequency \( \omega \) by the formula:
\[ T = \frac{2\pi}{\omega} \]

3. **Time Period of Total Mechanical Energy**:
- The total mechanical energy \( E \) is constant and does not change with time. Since it does not oscillate, it can be considered to have a frequency of \( 0 \).

4. **Calculating the Time Period of Total Mechanical Energy**:
- If the frequency is \( 0 \), then the time period \( T' \) of the total mechanical energy can be calculated as:
\[ T' = \frac{1}{\text{frequency}} = \frac{1}{0} = \infty \]
Therefore, the time period of the total mechanical energy is infinite.

5. **Finding the Ratio \( \frac{T'}{T} \)**:
- Now, we need to find the ratio of the time period of total mechanical energy \( T' \) to the time period of SHM \( T \):
\[ \frac{T'}{T} = \frac{\infty}{T} \]
Since \( T \) is a finite value, the ratio becomes:
\[ \frac{T'}{T} = \infty \]

### Final Answer:
Thus, the ratio \( \frac{T'}{T} \) is \( \infty \).