Question

A 0.675 kg mass is attached to a

Answer 1

Answer:

T= 0.6 sec

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

$T= 2\pi \sqrt{\frac{m}{k} }$

Substituting the given data into the expression above we have

$T= 2*3.142\sqrt{\frac{0.675}{72.4} }\\T= 6.284*\sqrt{0.0093 }\\T= 0.6$

T= 0.6 sec