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
[tex]T= 2\pi \sqrt{\frac{m}{k} }[/tex]
Substituting the given data into the expression above we have
[tex]T= 2*3.142\sqrt{\frac{0.675}{72.4} }\\T= 6.284*\sqrt{0.0093 }\\T= 0.6[/tex]
T= 0.6 sec
