If the mass attached to the spring mass system becomes 4 times then the time period of system become

Options

A. Four times
B. Double
C. Half
D. Remain same

Answer

To solve the problem of how the time period of a spring-mass system changes when the mass is increased to four times its original value, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Time Period**:
The time period \( T \) of a spring-mass system undergoing simple harmonic motion (SHM) is given by the formula:
\[ T = 2\pi \sqrt{\frac{M}{K}} \]
where \( M \) is the mass attached to the spring and \( K \) is the spring constant.

2. **Identify the Initial Conditions**:
Let the initial mass be \( M_1 \) and the initial time period be \( T_1 \). According to the formula:
\[ T_1 = 2\pi \sqrt{\frac{M_1}{K}} \]

3. **Change the Mass**:
If the mass is increased to four times its original value, the new mass \( M_2 \) can be expressed as:
\[ M_2 = 4M_1 \]

4. **Calculate the New Time Period**:
The new time period \( T_2 \) with the increased mass is:
\[ T_2 = 2\pi \sqrt{\frac{M_2}{K}} = 2\pi \sqrt{\frac{4M_1}{K}} \]

5. **Simplify the Expression**:
We can simplify the expression for \( T_2 \):
\[ T_2 = 2\pi \sqrt{\frac{4M_1}{K}} = 2\pi \cdot 2 \sqrt{\frac{M_1}{K}} = 2 \cdot T_1 \]

6. **Conclusion**:
Therefore, the new time period \( T_2 \) is double the original time period \( T_1 \):
\[ T_2 = 2T_1 \]

### Final Answer:
The time period of the system becomes **double** when the mass is increased to four times its original value.