A mass m is vertically suspended from a spring of negligible mass, the system oscillates with a frequency n. what will be the frequency of the system, if a mass `4m` is suspended from the same spring?

Options

A. `(n)/(4)`
B. `4n`
C. `(n)/(2)`
D. `2n`

Answer

To solve the problem, we need to understand how the frequency of oscillation of a mass-spring system changes when the mass is altered.

### Step-by-Step Solution:

1. **Understanding the Frequency Formula**:
The frequency \( n \) of a mass-spring system is given by the formula:
\[ n = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]
where \( k \) is the spring constant and \( m \) is the mass attached to the spring.

2. **Case 1 - Original Mass**:
For the original mass \( m \), the frequency is:
\[ n = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \tag{1} \]

3. **Case 2 - New Mass**:
Now, if we suspend a mass \( 4m \) from the same spring, the new frequency \( n' \) can be expressed as:
\[ n' = \frac{1}{2\pi} \sqrt{\frac{k}{4m}} \tag{2} \]

4. **Simplifying the New Frequency**:
We can simplify the expression for \( n' \):
\[ n' = \frac{1}{2\pi} \sqrt{\frac{k}{4m}} = \frac{1}{2\pi} \cdot \frac{1}{2} \sqrt{\frac{k}{m}} = \frac{1}{2} \cdot \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]
Thus, we can relate \( n' \) to \( n \):
\[ n' = \frac{1}{2} n \tag{3} \]

5. **Final Result**:
Therefore, the frequency of the system when a mass \( 4m \) is suspended from the same spring is:
\[ n' = \frac{n}{2} \]

### Conclusion:
The frequency of the system with mass \( 4m \) is half of the original frequency \( n \).