To make the frequency of an oscillator double,we have to

$A$ spring stretches by $2 \text{ mm}$ when it is loaded with a mass of $200 \text{ g}$. From the equilibrium position, the mass is further pulled down by $2 \text{ mm}$ and released. The frequency associated with the system and the maximum energy in the spring are . . . . . . $\text{Hz}$ and . . . . . . $\text{J}$, respectively. (Take $g = 10 \text{ m/s}^2$)

$A$ particle at the end of a spring executes simple harmonic motion with a period $t_1$,while the corresponding period for another spring is $t_2$. If the period of oscillation with the two springs in series is $T$,then

An object of mass $2 \,kg$ is attached to a spring with spring constant $8 \,N/m$. If the object is executing simple harmonic motion, then the number of cycles it completes in $66 \,s$ is

$A$ particle is suspended from a vertical spring which is executing $S.H.M.$ of frequency $5 \ Hz$. The spring is unstretched at the highest point of oscillation. What is the maximum speed of the particle? (Take $g = 10 \ m/s^2$)

$A$ spring is stretched by $0.2 \ m$ when a mass of $0.5 \ kg$ is suspended to it. The time period of the spring when the $0.5 \ kg$ mass is replaced with a mass of $0.25 \ kg$ is (Acceleration due to gravity $= 10 \ m \ s^{-2}$) (in $s$)