Question

A mass of $0.2\text{kg}$ is attached to the lower end of a massless spring of force constant $200{\text{Nm}}^{-1}$, the upper end of which is fixed to a rigid support. Which of the following statements is true?

• A. In equilibrium, the spring will be stretched by $1\text{cm}$
• B. If the body is raised till the spring attains its natural length and then released, it will go down by $2\text{cm}$ before moving upwards
• C. The frequency of oscillation will be nearly $50\text{Hz}$
• D. All of the above

Answer 1

The correct option is D All of the above

Given,
Mass of body $\left(m\right)=0.2\text{kg}$
force constant $\left(K\right)=200\text{N/m}$
At equilibrium,
$mg=Kx$
$\Rightarrow x=\frac{mg}{K}=\frac{0.2\times 10}{200}=0.01\text{m}=1\text{cm}$
Frequency of oscillation $f=\frac{1}{2\pi }\sqrt{\frac{K}{m}}$
$=\frac{1}{2\pi }\sqrt{\frac{200}{0.2}}$
$=50\text{Hz}$
At equilibrium, the spring is $1\text{cm}$ below the unstretched position.
If it is raised to that position and released,
let $d$ be displacement of mass before moving up
$mgd=\frac{1}{2}K{d}^2$
$d=\frac{2mg}{K}=2\text{cm}$

Thus,spring will go $2\text{cm}$ below the equilibrium
position.

So, option d is the correct answer.