The total energy of a vibrating particle in `SHM` is `E`. If its amplitude and time period are doubled, its total energy will be :-

Options

A. `16E`
B. `8E`
C. `4E`
D. `E`

Explanation

To solve the problem, we need to understand how the total energy of a particle in Simple Harmonic Motion (SHM) is related to its amplitude and time period.

### Step-by-Step Solution:

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

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

3. **Substituting Angular Frequency in Total Energy**:
Substituting the expression for \( \omega \) into the total energy formula gives:
\[ E = \frac{1}{2} m \left(\frac{2\pi}{T}\right)^2 A^2 \]
Simplifying this, we have:
\[ E = \frac{1}{2} m \cdot \frac{4\pi^2}{T^2} A^2 \]
\[ E = \frac{2\pi^2 m A^2}{T^2} \]

4. **Effect of Doubling Amplitude and Time Period**:
If the amplitude \( A \) is doubled (i.e., \( A' = 2A \)) and the time period \( T \) is doubled (i.e., \( T' = 2T \)), we need to find the new total energy \( E' \):
\[ E' = \frac{2\pi^2 m (2A)^2}{(2T)^2} \]
Simplifying this:
\[ E' = \frac{2\pi^2 m \cdot 4A^2}{4T^2} \]
\[ E' = \frac{2\pi^2 m A^2}{T^2} \]

5. **Comparing New Energy with Original Energy**:
Notice that:
\[ E' = E \]
This means that the total energy remains the same even after doubling the amplitude and the time period.

### Conclusion:
The total energy of the vibrating particle remains the same, so:
\[ E' = E \]

### Final Answer:
The total energy will be \( E \).