$A$ body of mass $100 \ g$ is tied to a spring of spring constant $8 \ N/m$,while the other end of the spring is fixed. If the body moves in a circular path on a smooth horizontal surface with a constant angular speed of $8 \ rad/s$,then the ratio of the extension in the spring to its natural length will be: (in $:1$)

When a mass $m$ is attached to a spring,it normally extends by $0.2\, m$. If the mass $m$ is given a slight additional extension and released,what will be its time period?

The total spring constant of the system as shown in the figure will be

$A$ spring having a spring constant $K$ is loaded with a mass $m$. The spring is cut into two equal parts and one of these is loaded again with the same mass. The new spring constant is

When a body of mass $8 \,kg$ is attached to a spring balance, the reading of the balance is $20 \,cm$. Instead of $8 \,kg$, if another body of mass $M$ is suspended from the spring balance and is made to oscillate vertically, the time period of oscillation is $\frac{\pi}{5} \,s$, then the value of $M$ is (Acceleration due to gravity $= 10 \,m/s^2$) (in $\,kg$)

$A$ mass $M$ is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes $S.H.M.$ of period $T$. If the mass is increased by $m$,the time period becomes $\frac{5T}{3}$. What is the ratio $\left(\frac{M}{m}\right)$?