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 step by step, we will use the relationship between the frequency of simple harmonic motion (SHM) and the mass attached to the spring.

### Step 1: Understand the formula for frequency in SHM
The frequency \( n \) of an object executing simple harmonic motion (SHM) when attached to a spring is given by the formula:
\[ n = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]
where:
- \( n \) is the frequency,
- \( k \) is the spring constant,
- \( m \) is the mass attached to the spring.

### Step 2: Identify the initial conditions
In the problem, we are given that the initial mass is \( m \) and the initial frequency is \( n \). Therefore, we can write:
\[ n = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]

### Step 3: Determine the new mass
According to the problem, the mass is increased to 4 times its original value. Thus, the new mass \( m_2 \) is:
\[ m_2 = 4m \]

### Step 4: Write the formula for the new frequency
We need to find the new frequency \( n_2 \) when the mass is \( m_2 \). Using the same frequency formula, we have:
\[ n_2 = \frac{1}{2\pi} \sqrt{\frac{k}{m_2}} = \frac{1}{2\pi} \sqrt{\frac{k}{4m}} \]

### Step 5: Simplify the expression for the new frequency
Now, we can simplify the expression for \( n_2 \):
\[ n_2 = \frac{1}{2\pi} \sqrt{\frac{k}{4m}} = \frac{1}{2\pi} \cdot \frac{1}{2} \sqrt{\frac{k}{m}} = \frac{1}{2} \left(\frac{1}{2\pi} \sqrt{\frac{k}{m}}\right) \]
This shows that:
\[ n_2 = \frac{n}{2} \]

### Step 6: Conclusion
Thus, the new frequency \( n_2 \) when the mass is increased to four times its original value is:
\[ n_2 = \frac{n}{2} \]

### Final Answer
The new frequency will be half of the original frequency \( n \).

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