College

3. A ball of mass 2.0 kg is attached to a vertical spring. The spring constant is 196 N/m. What is the period if it were to be in SHM?

A. 0.063 s
B. 0.635 s
C. 6.3 s
D. 63 s

Answer :

The period of the ball in simple harmonic motion is 0.634 s.

What is period?

Period can be defined as the time ( in seconds) taken for a body to complete one oscillation.

To calculate the period of the ball in SHM, we use the formula below.

Formula:

  • T = 2π√(m/k)............... Equation 1

Where:

  • T = Period of the ball
  • m = mass of the ball
  • k = spring constant
  • π = pie

From the question,

Given:

  • m = 2 kg
  • k = 196 N/m
  • π = 3.14

Substitute these values into equation 1

  • T = 2(3.14)√(2/196)
  • T = 6.28√(0.010204)
  • T = 0.634 s

Hence, The period of the ball in simple harmonic motion is 0.634 s.

Learn more about period here: https://brainly.com/question/21924087

Final answer:

The period of a ball attached to a vertical spring in simple harmonic motion can be calculated using the formula T = 2π√(m/k), where T is the period, m is the mass of the object, and k is the spring constant. Plugging in the values for mass and spring constant, the period is approximately 0.633 s.

Explanation:

The period of a simple harmonic motion (SHM) of an object attached to a vertical spring can be calculated using the formula:

T = 2π√(m/k)

where T is the period, m is the mass of the object, and k is the spring constant.

For the given problem, the mass of the ball is 2.0 kg and the spring constant is 196 N/m. Plugging these values into the formula:

T = 2π√(2.0/196) ≈ 0.633 s

Therefore, the period of the ball in SHM is approximately 0.633 s.