Given data:
* The value of the initial mass attached is 0.62 kg.
* The spring is stretched to 12 cm = -0.12 m
* The period of oscillation required is 0.67 s.
Solution:
The force acting on the mass hanging vertically is equal to the weight of the body.
[tex]F=mg[/tex]The diagrammatic representation of the given system is,
The force acting on the mass in terms of the spring constant is,
[tex]F=kx[/tex]where x is the extended length of the spring,
Substituting the value of force,
[tex]mg=kx[/tex]where m is the mass, and g is the acceleration due to gravity,
Substituting the known values,
[tex]\begin{gathered} 0.62\times9.8=k\times0.12 \\ 6.076=k\times0.12 \\ k=\frac{6.076}{0.12} \\ k=50.63\text{ N/m} \end{gathered}[/tex]The period of oscillation in terms of the mass and spring constant is,
[tex]T=2\pi\sqrt[]{\frac{m}{k}}[/tex]By substituting the known values except for mass, we get
[tex]\begin{gathered} 0.67=\text{ 2}\pi\times\sqrt[]{\frac{m}{50.63}} \\ \sqrt[]{\frac{m}{50.63}}=\frac{0.67}{2\pi} \\ \sqrt[]{\frac{m}{50.63}}=0.107 \\ \frac{m}{50.63}=0.01145 \\ m=0.58\text{ kg} \end{gathered}[/tex]Thus, the value of mass attached to the spring for 0.67 s period of the oscillation is 0.58 kg.
