Answer:
0.64 s
Explanation:
It's period of oscillation (T) can be determined by,
T = 2[tex]\pi[/tex][tex]\sqrt{\frac{l}{g} }[/tex]
Where l is the length (extension on the spring), and g the acceleration due to gravity.
But,
l = 10 cm = 0.1 m
g = 9.8 m/[tex]s^{2}[/tex]
Thus,
T = 2 x [tex]\frac{22}{7}[/tex] [tex]\sqrt{\frac{0.1}{9.8} }[/tex]
= 0.6350
T = 0.64 s
The period of oscillation would be 0.64 s.
