Question

40) When a 0.62-kg mass is attached to a vertical spring, the spring stretches by 12 cm. How much mass must be attached to the spring to result in a 0.67-s period of oscillation?

Answer 1

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.

`F=mg`

The diagrammatic representation of the given system is,

The force acting on the mass in terms of the spring constant is,

`F=kx`

where x is the extended length of the spring,

Substituting the value of force,

`mg=kx`

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

Substituting the known values,

`\begin{gathered} 0.62*9.8=k*0.12 \\ 6.076=k*0.12 \\ k=(6.076)/(0.12) \\ k=50.63\text{ N/m} \end{gathered}`

The period of oscillation in terms of the mass and spring constant is,

`T=2\pi\sqrt[]{(m)/(k)}`

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}`

Thus, the value of mass attached to the spring for 0.67 s period of the oscillation is 0.58 kg.