A mass of 0.2 hg is attached to the lower end of a massles spring of force constant `200(N)/(m)` the opper end of which is fixed to a rigid support. Study the following statements.

Options

A. In equilibrium , the spring will be stretched by 1 cm
B. If the mass is raised till the spring becomes unstretched and then released , it will go down by 2 cm before moving upwards
C. The frequency of oscillation will be nearly 5 Hz
D. All of the above

Explanation

To solve the problem step by step, we need to analyze the given information about the spring-mass system and evaluate the statements provided.

### Step 1: Understand the system
We have a mass of 0.2 kg attached to a spring with a spring constant \( k = 200 \, \text{N/m} \). The system is in equilibrium when the gravitational force acting on the mass is balanced by the spring force.

### Step 2: Calculate the gravitational force
The gravitational force \( F_g \) acting on the mass can be calculated using the formula:
\[ F_g = m \cdot g \]
Where:
- \( m = 0.2 \, \text{kg} \)
- \( g = 10 \, \text{m/s}^2 \) (approximate value of acceleration due to gravity)

Calculating:
\[ F_g = 0.2 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 2 \, \text{N} \]

### Step 3: Set up the equilibrium condition
At equilibrium, the spring force \( F_s \) is equal to the gravitational force:
\[ F_s = k \cdot x \]
Where \( x \) is the stretch in the spring. Setting \( F_s = F_g \):
\[ k \cdot x = m \cdot g \]

### Step 4: Solve for the stretch \( x \)
Rearranging the equation to find \( x \):
\[ x = \frac{m \cdot g}{k} \]
Substituting the known values:
\[ x = \frac{2 \, \text{N}}{200 \, \text{N/m}} = 0.01 \, \text{m} = 1 \, \text{cm} \]

### Conclusion for Statement 1
The first statement is correct: In equilibrium, the spring will be stretched by 1 cm.

### Step 5: Analyze Statement 2
If the mass is raised until the spring is unstretched and then released, it will fall due to gravity. The maximum stretch will be the same as the equilibrium stretch, which is 1 cm downward.

When released, it will fall 1 cm to reach the equilibrium position and then continue to move an additional 1 cm downward (to the maximum displacement). Thus, it will move down by a total of 2 cm before moving upwards.

### Conclusion for Statement 2
The second statement is also correct: The mass will go down by 2 cm before moving upwards.

### Step 6: Analyze Statement 3
The frequency of oscillation \( f \) for a mass-spring system is given by:
\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]
Substituting the values:
\[ f = \frac{1}{2\pi} \sqrt{\frac{200 \, \text{N/m}}{0.2 \, \text{kg}}} \]
Calculating:
\[ f = \frac{1}{2\pi} \sqrt{1000} = \frac{10}{2\pi} \approx \frac{10}{6.28} \approx 1.59 \, \text{Hz} \]
This is approximately 5 Hz.

### Conclusion for Statement 3
The third statement is incorrect: The frequency of oscillations is not nearly 5 Hz.

### Final Conclusion
- Statement 1: Correct
- Statement 2: Correct
- Statement 3: Incorrect