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 the relationship between the mass attached to a spring and the frequency of oscillation. The frequency of a mass-spring system is given by the formula:

\[ n = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]

where:
- \( n \) is the frequency of oscillation,
- \( k \) is the spring constant,
- \( m \) is the mass attached to the spring.

### Step-by-Step Solution:

1. **Identify the initial frequency**:
The initial frequency when mass \( m \) is suspended is given as \( n \).
\[ n = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]

2. **Determine the new mass**:
When the mass is increased to \( 4m \), we need to find the new frequency \( n_2 \).

3. **Write the formula for the new frequency**:
The frequency when the mass is \( 4m \) can be expressed as:
\[ n_2 = \frac{1}{2\pi} \sqrt{\frac{k}{4m}} \]

4. **Simplify the expression for \( n_2 \)**:
\[ n_2 = \frac{1}{2\pi} \sqrt{\frac{k}{4m}} = \frac{1}{2\pi} \cdot \frac{1}{2} \sqrt{\frac{k}{m}} = \frac{1}{2} \cdot \frac{1}{2\pi} \sqrt{\frac{k}{m}} = \frac{n}{2} \]

5. **Conclusion**:
Therefore, the frequency of the system when a mass \( 4m \) is suspended from the same spring is:
\[ n_2 = \frac{n}{2} \]

### Final Answer:
The frequency of the system when a mass \( 4m \) is suspended from the spring is \( \frac{n}{2} \).