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:
- \( m \) is the mass of the particle,
- \( \omega \) is the angular frequency,
- \( A \) is the amplitude.

2. **Relating Angular Frequency to Time Period**:
The angular frequency \( \omega \) is related to the time period \( T \) by the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting this into the energy formula gives:
\[ E = \frac{1}{2} m \left(\frac{2\pi}{T}\right)^2 A^2 = \frac{2\pi^2 m A^2}{T^2} \]

3. **Doubling Amplitude and Time Period**:
If the amplitude \( A \) is doubled, then:
\[ A' = 2A \]
If the time period \( T \) is doubled, then:
\[ T' = 2T \]

4. **Calculating New Energy**:
Now, substituting the new values into the energy formula:
\[ E' = \frac{2\pi^2 m (A')^2}{(T')^2} = \frac{2\pi^2 m (2A)^2}{(2T)^2} \]
Simplifying this:
\[ E' = \frac{2\pi^2 m (4A^2)}{4T^2} = \frac{2\pi^2 m A^2}{T^2} \]
Notice that:
\[ E' = E \]

5. **Conclusion**:
Therefore, the new total energy \( E' \) remains the same as the original total energy \( E \):
\[ E' = E \]

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

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