The length of a weigtless spring increaes by 2cm when a weight of 1.0kg in the period of oscillation of the spring and its kinetic energy of oscillation. Take `g=10ms^(-2)`.

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the Spring Constant (k)
Given that the spring stretches by 2 cm (0.02 m) when a weight of 1 kg is applied, we can use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement from its equilibrium position.

The force exerted by the weight is:
\[ F = m \cdot g \]
Where:
- \( m = 1 \, \text{kg} \)
- \( g = 10 \, \text{m/s}^2 \)

Calculating the force:
\[ F = 1 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 10 \, \text{N} \]

According to Hooke's Law:
\[ F = k \cdot x \]
Where:
- \( x = 0.02 \, \text{m} \) (the extension of the spring)

Rearranging for \( k \):
\[ k = \frac{F}{x} = \frac{10 \, \text{N}}{0.02 \, \text{m}} = 500 \, \text{N/m} \]

### Step 2: Calculate the Time Period of Oscillation (T)
The formula for the time period of a mass-spring system is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]

Substituting the values:
- \( m = 1 \, \text{kg} \)
- \( k = 500 \, \text{N/m} \)

Calculating the time period:
\[ T = 2\pi \sqrt{\frac{1 \, \text{kg}}{500 \, \text{N/m}}} \]
\[ T = 2\pi \sqrt{0.002} \]
\[ T = 2\pi \cdot 0.04472 \approx 0.28 \, \text{s} \]

### Step 3: Calculate the Kinetic Energy (KE)
The kinetic energy of the oscillating mass can be calculated using the formula:
\[ KE = \frac{1}{2} k x^2 \]

Substituting the values:
- \( k = 500 \, \text{N/m} \)
- \( x = 0.02 \, \text{m} \)

Calculating the kinetic energy:
\[ KE = \frac{1}{2} \cdot 500 \, \text{N/m} \cdot (0.02 \, \text{m})^2 \]
\[ KE = \frac{1}{2} \cdot 500 \cdot 0.0004 \]
\[ KE = 250 \cdot 0.0004 = 0.1 \, \text{J} \]

### Final Results
- Time Period (T) = 0.28 seconds
- Kinetic Energy (KE) = 0.1 Joules