Question

A mass of $0.2kg$ is attached to the lower end of a massless spring of force-constant $200N/m$,the upper end of which is fixed to a rigid support.Which of the following statements is /are true?

• A. In equilibrium, the spring will be stretched by $1cm$
• B. If the mass is raised till the spring is unstretched state and then released,it will go down by $2cm$ before moving upwards
• C. The frequency of oscillation will be nearly $5HZ$
• D. If the system is taken to the moon,the frequency of oscillation will be the same as on the earth

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A In equilibrium, the spring will be stretched by $1cm$
B If the mass is raised till the spring is unstretched state and then released,it will go down by $2cm$ before moving upwards
C If the system is taken to the moon,the frequency of oscillation will be the same as on the earth
D The frequency of oscillation will be nearly $5HZ$

The equilibrium stretching is given by equating weight with spring force.
$\therefore 0.2\times 10=200\times x$
$x=0.01m=1cm$
When taken back to unstretched state, the sum of spring force and weight together is converted into spring force.
$\therefore 200\times 0.01+0.2\times 10=200\times x$
$x=0.02m=2cm$
The frequency is given by
$f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}=\frac{1}{2\pi }\sqrt{\frac{200}{0.2}}\approx 5.05Hz$
Since frequency is independent of acceleration due to gravity, it will be same on moon. Hence, all options, A,B.C,D are correct.