Answer:
The period is [tex]T = 0.700 \ s[/tex]
Explanation:
From the question we are told that
The mass is [tex]m = 0.350 \ kg[/tex]
The extension of the spring is [tex]x = 12.0 \ cm = 0.12 \ m[/tex]
The spring constant for this is mathematically represented as
[tex]k = \frac{F}{x}[/tex]
Where F is the force on the spring which is mathematically evaluated as
[tex]F = mg = 0.350 * 9.8[/tex]
[tex]F =3.43 \ N[/tex]
So
[tex]k = \frac{3.43 }{ 0.12}[/tex]
[tex]k = 28.583 \ N/m[/tex]
The period of oscillation is mathematically evaluated as
[tex]T = 2 \pi \sqrt{\frac{m}{k} }[/tex]
substituting values
[tex]T = 2 * 3.142* \sqrt{\frac{0.35 }{28.583} }[/tex]
[tex]T = 0.700 \ s[/tex]
