When a 20 N can is hung from the bottom of a vertical spring, it causes the spring to stretch 20 cm. The spring is now placed horizontal on a frictionless table. One end of it is held fixed, and the other end is attached to a 5.0 N can. The can is then moved (stretching the spring) and released from rest. What is the period `("in" xx 10^(-2)s)` of the resulting oscillation?

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Determine the spring constant (k)
When a 20 N can is hung from the spring, it stretches the spring by 20 cm (0.2 m). We can use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement from its equilibrium position:

\[ F = kx \]

At equilibrium, the force exerted by the spring equals the weight of the can:

\[ k \cdot x = mg \]

Given:
- \( F = 20 \, \text{N} \)
- \( x = 0.2 \, \text{m} \)

We can rearrange the equation to find \( k \):

\[ k = \frac{F}{x} = \frac{20 \, \text{N}}{0.2 \, \text{m}} = 100 \, \text{N/m} \]

### Step 2: Find the mass (m) of the 5 N can
The weight of the can is given as 5 N. We can find the mass using the relation:

\[ mg = 5 \, \text{N} \]

Using \( g \approx 9.8 \, \text{m/s}^2 \):

\[ m = \frac{5 \, \text{N}}{9.8 \, \text{m/s}^2} \approx 0.51 \, \text{kg} \]

### Step 3: Calculate the period of oscillation (T)
The formula for the period of a mass-spring system is given by:

\[ T = 2\pi \sqrt{\frac{m}{k}} \]

Substituting the values of \( m \) and \( k \):

\[ T = 2\pi \sqrt{\frac{0.51 \, \text{kg}}{100 \, \text{N/m}}} \]

Calculating the term inside the square root:

\[ \frac{0.51}{100} = 0.0051 \]

Now, taking the square root:

\[ \sqrt{0.0051} \approx 0.0714 \]

Now, substituting back into the period formula:

\[ T = 2\pi \cdot 0.0714 \approx 0.448 \, \text{s} \]

### Step 4: Convert to the required format
The problem asks for the period in the format of \( xx \times 10^{-2} \, \text{s} \):

\[ T \approx 44.8 \times 10^{-2} \, \text{s} \]

### Final Answer
The period of the resulting oscillation is approximately:

\[ \boxed{44.8 \times 10^{-2} \, \text{s}} \]

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