$A$ body of mass $4 \ kg$ attached to a spring of force constant $64 \ N \ m^{-1}$ executes simple harmonic motion on a frictionless horizontal surface. The time period of oscillation is

$A$ spring-mass system (mass $m$,spring constant $k$,and natural length $\ell_{0}$) rests in equilibrium on a horizontal disc. The free end of the spring is fixed at the center of the disc. If the disc,together with the spring-mass system,rotates about its axis with an angular velocity $\omega$ (where $k >> m \omega^{2}$),the relative change in the length of the spring is best given by which option?

$A$ 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 $g = 10 \text{ m/s}^2$)

$A$ mass $m$ is suspended from a spring of length $l$ and force constant $K$. The frequency of vibration of the mass is $f_1$. The spring is cut into two equal parts and the same mass is suspended from one of the parts. The new frequency of vibration of the mass is $f_2$. Which of the following relations between the frequencies is correct?

$A$ particle of mass $m$ is attached to four springs with spring constants $k, k, 2k$ and $2k$ as shown in the figure. Four springs are attached to the four corners of a square and a particle is placed at the center. If the particle is pushed slightly towards any side of the square and released, the period of oscillation will be

$A$ mass $m$ is suspended from two coupled springs connected in series. The force constants for the springs are $K_1$ and $K_2$. The time period of the suspended mass will be: