A spring mass system preforms `S.H.M` if the mass is doubled keeping amplitude same, then the total energy of `S.H.M` will become :

Options

A. double
B. half
C. unchanged
D. 4 times

Explanation

To solve the problem, we need to analyze how the total energy of a spring-mass system performing Simple Harmonic Motion (SHM) changes when the mass is doubled while keeping the amplitude constant.

### Step-by-Step Solution:

1. **Understanding Total Energy in SHM**:
The total energy \( E \) of a spring-mass system performing SHM is given by the formula:
\[ E = \frac{1}{2} k A^2 \]
where:
- \( k \) is the spring constant,
- \( A \) is the amplitude of the motion.

2. **Effect of Doubling the Mass**:
When the mass \( m \) of the system is doubled (i.e., it becomes \( 2m \)), we need to analyze how this affects the total energy.

3. **Angular Frequency**:
The angular frequency \( \omega \) of the system is given by:
\[ \omega = \sqrt{\frac{k}{m}} \]
If the mass is doubled, the new angular frequency \( \omega' \) becomes:
\[ \omega' = \sqrt{\frac{k}{2m}} = \frac{\omega}{\sqrt{2}} \]

4. **Total Energy with New Mass**:
The total energy can also be expressed in terms of angular frequency:
\[ E = \frac{1}{2} m \omega^2 A^2 \]
If we substitute the new mass \( 2m \) and the new angular frequency \( \omega' \):
\[ E' = \frac{1}{2} (2m) \left(\frac{\omega}{\sqrt{2}}\right)^2 A^2 \]
Simplifying this gives:
\[ E' = \frac{1}{2} (2m) \left(\frac{\omega^2}{2}\right) A^2 = \frac{1}{2} m \omega^2 A^2 \]
This shows that the total energy remains:
\[ E' = E \]

5. **Conclusion**:
Since the amplitude \( A \) remains the same and the total energy does not depend on the mass when expressed in terms of amplitude and spring constant, the total energy of the SHM remains unchanged when the mass is doubled.

### Final Answer:
The total energy of the SHM will remain unchanged.