A object of mass m is suspended from a spring and it executes SHM with frequency n. If the mass is increased 4 times , the new frequency will be

Options

A. 2 n
B. n/2
C. n
D. n/4

Explanation

To solve the problem, we need to understand how the frequency of a mass-spring system changes when the mass is altered. The frequency of simple harmonic motion (SHM) for a mass-spring system is given by the formula:

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

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

### Step-by-Step Solution:

1. **Identify the initial frequency**:
The initial frequency of the system is given as \( n \). Therefore, we can write:
\[ n = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]

2. **Determine the new mass**:
The problem states that the mass is increased 4 times. If the original mass is \( m \), the new mass \( m_2 \) will be:
\[ m_2 = 4m \]

3. **Write the formula for the new frequency**:
The new frequency \( f_2 \) with the new mass \( m_2 \) can be expressed as:
\[ f_2 = \frac{1}{2\pi} \sqrt{\frac{k}{m_2}} = \frac{1}{2\pi} \sqrt{\frac{k}{4m}} \]

4. **Simplify the expression for the new frequency**:
We can simplify the expression for \( f_2 \):
\[ f_2 = \frac{1}{2\pi} \sqrt{\frac{k}{4m}} = \frac{1}{2\pi} \cdot \frac{1}{2} \sqrt{\frac{k}{m}} = \frac{1}{2} \cdot \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]
Therefore, we can relate \( f_2 \) to the original frequency \( n \):
\[ f_2 = \frac{1}{2} n \]

5. **Conclusion**:
The new frequency when the mass is increased 4 times is:
\[ f_2 = \frac{n}{2} \]

### Final Answer:
The new frequency will be \( \frac{n}{2} \).

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