When a mass $m$ is attached to a spring,it extends by $0.2 \ m$. The mass $m$ is given a slight additional extension and released. Find the time period (take $g = \pi^2 \ m/s^2$).

$A$ silver atom in a solid oscillates in simple harmonic motion in a specific direction with a frequency of $10^{12} \ s^{-1}$. What is the force constant of the bonds connecting one atom to the others? (Molar mass of silver $= 108 \ g/mol$ and Avogadro number $= 6.02 \times 10^{23} \ mol^{-1}$)

As shown in the figure, a spring with spring constant $k = 150 \ N/m$ is kept in a stretched position by holding two masses of $1 \ kg$ and $200 \ g$ at a separation greater than the natural length of the spring. When the masses are released, assuming the horizontal surface to be frictionless, the angular frequency (in $SI$ units) of the system is:

$A$ mass $m$ is suspended from a spring of negligible mass. The spring is pulled a little and then released,so that the mass executes $S.H.M.$ with a time period $T$. If the mass is increased by $m_0$,the time period becomes $\frac{5T}{4}$. The ratio $\frac{m_0}{m}$ is

$A$ uniform circular disc of mass $12 \ kg$ is held by two identical springs. When the disc is slightly pressed down and released,it executes $S.H.M.$ of period $2 \ s$. The force constant of each spring is (nearly) (Take $\pi^2=10$) (in $Nm^{-1}$)

Two particles $A$ and $B$ of equal masses are suspended from two massless springs of spring constants $K_A$ and $K_B$ respectively. The maximum velocities of the particles during oscillations are equal. The ratio of the amplitude of '$B$' to that of '$A$' is