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.

The diagrammatic representation of the given system is,
The force acting on the mass in terms of the spring constant is,

where x is the extended length of the spring,
Substituting the value of force,

where m is the mass, and g is the acceleration due to gravity,
Substituting the known values,

The period of oscillation in terms of the mass and spring constant is,
![T=2\pi\sqrt[]{(m)/(k)}](https://img.qammunity.org/2023/formulas/physics/high-school/59p5w267zpcbiffehn6l28fg1lc6criikc.png)
By substituting the known values except for mass, we get
![\begin{gathered} 0.67=\text{ 2}\pi*\sqrt[]{(m)/(50.63)} \\ \sqrt[]{(m)/(50.63)}=(0.67)/(2\pi) \\ \sqrt[]{(m)/(50.63)}=0.107 \\ (m)/(50.63)=0.01145 \\ m=0.58\text{ kg} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/d90jrqxvc3952e4mniictlciy0ttcexsnq.png)
Thus, the value of mass attached to the spring for 0.67 s period of the oscillation is 0.58 kg.