Question

A 0.675 kg mass is attached to a

spring of spring constant 72.4 N/m,
pulled, and released. What is the
period of the resulting oscillation?
(Unit = s)

Answer 1

0.607

Step-by-step explanation:

Trust me

Answer 2

T= 0.6 sec

Step-by-step explanation:

This problem bothers on the simple harmonic motion of a loaded spring

Given data

mass attached, m= 0-.675 kg

spring constant, k= 72.4 N/m

the period of oscillation can be solved for using the formula bellow

`T= 2\pi \sqrt{(m)/(k) }`

Substituting the given data into the expression above we have

`T= 2*3.142\sqrt{(0.675)/(72.4) }\\T= 6.284*√(0.0093 )\\T= 0.6`

T= 0.6 sec