Three mass 0.1 kg ,0.3 kg and 0.4 kg are suspended at end of a spring. When is 0.4 kg mass is removed , the system oscillates with a period 2 s . When the 0.3 kg mass is also removed , the system will oscillates with a period

Options

A. 1 s
B. 2 s
C. 3 s
D. 4 s

Explanation

To solve this problem, we need to analyze the oscillation of the spring-mass system and how the removal of masses affects the period of oscillation. Let's break it down step by step.

### Step 1: Understand the formula for the period of oscillation
The period \( T \) of a mass-spring system is given by the formula:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
where \( m \) is the mass attached to the spring and \( k \) is the spring constant.

### Step 2: Analyze the first scenario
When the 0.4 kg mass is removed, the remaining mass is:
\[ m_1 = 0.1 \, \text{kg} + 0.3 \, \text{kg} = 0.4 \, \text{kg} \]
Given that the period \( T_1 \) is 2 seconds, we can write:
\[ T_1 = 2\pi \sqrt{\frac{0.4}{k}} = 2 \, \text{s} \]

### Step 3: Solve for the spring constant \( k \)
We can rearrange the equation to find \( k \):
\[ 2 = 2\pi \sqrt{\frac{0.4}{k}} \]
Dividing both sides by \( 2\pi \):
\[ \frac{1}{\pi} = \sqrt{\frac{0.4}{k}} \]
Squaring both sides:
\[ \frac{1}{\pi^2} = \frac{0.4}{k} \]
Thus,
\[ k = 0.4\pi^2 \]

### Step 4: Analyze the second scenario
Now, when the 0.3 kg mass is also removed, the remaining mass is:
\[ m_2 = 0.1 \, \text{kg} \]
We need to find the new period \( T_2 \):
\[ T_2 = 2\pi \sqrt{\frac{m_2}{k}} = 2\pi \sqrt{\frac{0.1}{0.4\pi^2}} \]

### Step 5: Simplify the expression for \( T_2 \)
Substituting \( k \) into the equation:
\[ T_2 = 2\pi \sqrt{\frac{0.1}{0.4\pi^2}} = 2\pi \sqrt{\frac{0.1}{0.4}} \cdot \frac{1}{\pi} \]
This simplifies to:
\[ T_2 = 2\sqrt{\frac{0.1}{0.4}} = 2\sqrt{\frac{1}{4}} = 2 \cdot \frac{1}{2} = 1 \, \text{s} \]

### Final Answer
Thus, when the 0.3 kg mass is also removed, the system will oscillate with a period of:
\[ \boxed{1 \, \text{s}} \]