$A$ body of mass $4.9 \, kg$ hangs from a spring and oscillates with a period $0.5 \, s$. On the removal of the body, the spring is shortened by (take $g=10 \, m/s^2, \pi^2=10$)

Two massless springs with spring constants $2k$ and $9k$ carry $50\,g$ and $100\,g$ masses at their free ends. These two masses oscillate vertically such that their maximum velocities are equal. Then,the ratio of their respective amplitudes 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:

$A$ 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 $\left(\frac{M}{m}\right)$?

$A$ 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 $= 6.02 \times 10^{23} \ mol^{-1}$)

The motion of a mass on a spring,with spring constant $K$ is as shown in the figure. The equation of motion is given by $x(t) = A \sin \omega t + B \cos \omega t$ with $\omega = \sqrt{\frac{K}{m}}$. Suppose that at time $t = 0$,the position of the mass is $x(0)$ and velocity is $v(0)$,then its displacement can also be represented as $x(t) = C \cos (\omega t - \phi)$,where $C$ and $\phi$ are: