Two masses `m_(1) and m_(2)` are suspended together by a massless spring of constant k. When the masses are in equilibrium, `m_(1)` is removed without disturbing the system. Then the angular frequency of oscillation of `m_(2)` is

Options

A. `sqrt((k//m_(1)))`
B. `sqrt((k//m_(2)))`
C. `sqrt([k(m_(1) + m_(2))])`
D. `sqrt([k//(m_(1)m_(2))])`

Explanation

To solve the problem, we need to find the angular frequency of oscillation of mass \( m_2 \) after mass \( m_1 \) is removed from the system. Here are the steps to arrive at the solution:

### Step 1: Understand the System
Initially, both masses \( m_1 \) and \( m_2 \) are suspended together by a massless spring with spring constant \( k \). When both masses are present, they are in equilibrium under the influence of gravity.

### Step 2: Remove Mass \( m_1 \)
When mass \( m_1 \) is removed, only mass \( m_2 \) remains suspended from the spring. The spring will now oscillate with mass \( m_2 \).

### Step 3: Identify the Angular Frequency Formula
The angular frequency \( \omega \) of a mass-spring system is given by the formula:
\[ \omega = \sqrt{\frac{k_{\text{equivalent}}}{m_{\text{equivalent}}}} \]
where \( k_{\text{equivalent}} \) is the spring constant and \( m_{\text{equivalent}} \) is the mass attached to the spring.

### Step 4: Determine the Values
In this case:
- The spring constant \( k_{\text{equivalent}} \) remains \( k \) (since the spring is unchanged).
- The mass \( m_{\text{equivalent}} \) is now just \( m_2 \) (since \( m_1 \) has been removed).

### Step 5: Substitute the Values into the Formula
Substituting the values into the angular frequency formula, we get:
\[ \omega = \sqrt{\frac{k}{m_2}} \]

### Step 6: Conclusion
Thus, the angular frequency of oscillation of mass \( m_2 \) after mass \( m_1 \) is removed is:
\[ \omega = \sqrt{\frac{k}{m_2}} \]

### Final Answer
The angular frequency of oscillation of \( m_2 \) is \( \sqrt{\frac{k}{m_2}} \).

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