To find the time period of oscillation for the spring described, we can apply principles from physics, particularly those related to simple harmonic motion (SHM). Given the mass of the spring and its spring constant, we can derive the time period using the formula for oscillating systems.

Understanding the System

The spring has a mass \( M = 3 \, \text{kg} \) and a spring constant \( k = 4\pi^2 \, \text{N/m} \). When the free end of the spring is pulled and released, it will oscillate around its equilibrium position. The motion of the spring can be modeled as a simple harmonic oscillator.

Key Formula for Time Period

The time period \( T \) of a mass-spring system is given by the formula:

• T = 2\pi \sqrt{\frac{M}{k}}

Substituting Values

Now, we can substitute the values of \( M \) and \( k \) into the formula:

• Mass \( M = 3 \, \text{kg} \)
• Spring constant \( k = 4\pi^2 \, \text{N/m} \)

Plugging these into the formula gives:

T = 2\pi \sqrt{\frac{3}{4\pi^2}}

Calculating the Time Period

We can simplify the expression step by step:

• First, calculate \( \frac{3}{4\pi^2} \):
• Then take the square root of that result.
• Finally, multiply by \( 2\pi \).

Let's compute it:

• Calculate \( 4\pi^2 \approx 39.478 \, \text{N/m} \)
• Then \( \frac{3}{39.478} \approx 0.076 \)
• Taking the square root gives \( \sqrt{0.076} \approx 0.275 \)
• Now, multiply by \( 2\pi \): \( 2\pi \times 0.275 \approx 1.73 \, \text{seconds} \)

Final Result

The time period of oscillation for the spring is approximately 1.73 seconds. This means that after every 1.73 seconds, the spring will complete one full cycle of oscillation.

In summary, by applying the formula for the time period of a mass-spring system and substituting the given values, we determined that the oscillation time period is about 1.73 seconds. This is a fundamental concept in mechanics that illustrates how mass and spring constant influence the behavior of oscillating systems.