A mass of 0.2 kg is attached to the lower end of a massless spring of force constant `200(N)/(m)` the other 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 2cm before moving upwards.
C. The frequency of oscillation will be nearly 5 Hz.
D. If the system is taken to the moon, the frequency of oscillation will be the same as that on the earth.

Explanation

To solve the problem step by step, we will analyze the statements regarding the mass-spring system and verify their correctness.

### Step 1: Understanding the System
We have a mass \( m = 0.2 \, \text{kg} \) attached to a spring with a spring constant \( k = 200 \, \text{N/m} \). The mass is hanging from the spring, and we need to analyze its behavior under different conditions.

### Step 2: Finding the Equilibrium Position
At equilibrium, the weight of the mass is balanced by the restoring force of the spring. The weight \( W \) is given by:
\[ W = mg \]
where \( g \) (acceleration due to gravity) is approximately \( 10 \, \text{m/s}^2 \). Therefore,
\[ W = 0.2 \times 10 = 2 \, \text{N} \]

The restoring force of the spring is given by Hooke's Law:
\[ F = kx \]
At equilibrium, \( mg = kx \). Rearranging gives:
\[ x = \frac{mg}{k} \]
Substituting the values:
\[ x = \frac{2}{200} = 0.01 \, \text{m} = 1 \, \text{cm} \]
This means the spring stretches by 1 cm at equilibrium.

### Step 3: Analyzing the Second Statement
The second statement claims that if the mass is raised until the spring is unstretched and then released, it will go down by 2 cm before moving upwards.

Using the work-energy principle, we can analyze this:
1. When the mass is raised to the unstretched position, the potential energy due to gravity is converted into elastic potential energy of the spring when it is stretched.
2. The work done by gravity when the mass moves down by \( x \) is \( mgx \) and the work done against the spring is \( -\frac{1}{2}kx^2 \).

Setting the work done equal to the change in kinetic energy (which is zero at the starting and ending points):
\[ mgx - \frac{1}{2}kx^2 = 0 \]
This simplifies to:
\[ mgx = \frac{1}{2}kx^2 \]
Rearranging gives:
\[ x = \frac{2mg}{k} \]
Substituting the values:
\[ x = \frac{2 \times 0.2 \times 10}{200} = 0.02 \, \text{m} = 2 \, \text{cm} \]
Thus, the second statement is correct.

### Step 4: Finding the Frequency of Oscillation
The frequency of oscillation \( f \) for a mass-spring system is given by:
\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]
Calculating the time period \( T \):
\[ T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{0.2}{200}} = 2\pi \sqrt{0.001} = 2\pi \times 0.03162 \approx 0.198 \, \text{s} \]
Thus, the frequency is:
\[ f = \frac{1}{T} \approx \frac{1}{0.198} \approx 5.06 \, \text{Hz} \]
So, the third statement is also correct.

### Step 5: Analyzing the Fourth Statement
The fourth statement claims that if the system is taken to the moon, the frequency of oscillation will remain the same as on Earth.

The frequency of oscillation depends only on the mass and the spring constant, not on gravity. Therefore, the fourth statement is also correct.

### Conclusion
All four statements regarding the mass-spring system are correct.